# Given Information - Joint probability density function: \( f(x,y) = k(x+y) \), for \( < x < 2, \ < y < 2 \); zero elsewhere. # What We Have to Find 1. The value of \( k \). 2. \( P(X + Y < 2) \). 3. \( E[X] \). 4. Whether \( X \) and \( Y \) are independent. 5. \( \operatorname{Cov}(X, Y) \). # Definition or Concept Used - **Joint PDF Normalization:** \( \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \, dx \, dy = 1 \) - **Probability Calculation:** \( P(A) = \iint_A f(x,y) \, dx \, dy \) - **Expectation:** \( E[X] = \iint x f(x,y) dx\,dy \) - **Independence:** \( X \) and \( Y \) are independent if \( f(x,y) = f_X(x) f_Y(y) \) - **Covariance:** \( \operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] \) # Step-by-Step Solution **Finding \( k \):** Normalize the joint PDF: \[ \int_^2 \int_^2 k(x+y) \, dy \, dx = 1 \] \[ \int_^2 \left[ k \int_^2 (x+y) dy \right] dx = 1 \] \[ \int_^2 k \left[ xy + \frac{y^2}{2} \Big|_{y=}^{y=2} \right] dx \] \[ \int_^2 k \left[ 2x + 2 \right] dx \] \[ k \int_^2 (2x + 2) dx = 1 \] \[ k \left[ x^2 + 2x \Big|_^2 \right] = 1 \] \[ k (4 + 4) = 1 \implies k = \frac{1}{8} \] **Finding \( P(X + Y < 2) \):** Region: \( < x < 2, < y < 2, x + y < 2 \) Limits: \( x = \) to \( 2 \), \( y = \) to \( 2-x \) \[ P(X + Y < 2) = \int_^2 \int_^{2-x} \frac{1}{8}(x+y) dy dx \] Expand and integrate: \[ = \frac{1}{8} \int_^2 \left[ x y + \frac{y^2}{2} \Big|_{y=}^{y=2-x} \right] dx \] \[ = \frac{1}{8} \int_^2 \left[ x(2-x) + \frac{(2-x)^2}{2} \right] dx \] \[ x(2-x) + \frac{(2-x)^2}{2} = 2x - x^2 + \frac{4 - 4x + x^2}{2} \] \[ = 2x - x^2 + 2 - 2x + \frac{x^2}{2} \] \[ = 2 + \frac{x^2}{2} - x^2 = 2 - \frac{x^2}{2} \] \[ \int_^2 2 - \frac{x^2}{2} dx = \left[ 2x - \frac{x^3}{6} \right]_^2 = 4 - \frac{8}{6} = 4 - \frac{4}{3} = \frac{8}{3} \] \[ P = \frac{1}{8} \cdot \frac{8}{3} = \frac{1}{3} \] **Finding \( E[X] \):** \[ E[X] = \int_^2 \int_^2 x \cdot \frac{1}{8}(x+y)\,dy\,dx \] \[ = \frac{1}{8} \int_^2 \int_^2 (x^2 + xy) dy dx \] \[ = \frac{1}{8} \int_^2 \left[ x^2 y + x y^2 / 2 \Big|_{y=}^{y=2} \right] dx \] \[ = \frac{1}{8} \int_^2 \left[ 2x^2 + 2x \right] dx \] \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + x^2 \Big|_^2 \right] \] At \( x=2 \): \( \frac{2 \cdot 8}{3} + 4 = \frac{16}{3} + 4 = \frac{28}{3} \) \[ E[X] = \frac{1}{8} \cdot \frac{28}{3} = \frac{7}{6} \] **Checking Independence:** Find marginal PDFs: \[ f_X(x) = \int_^2 \frac{1}{8}(x+y) dy = \frac{1}{8} [2x + 2] = \frac{1}{4}(x+1) \] \[ f_Y(y) = \int_^2 \frac{1}{8}(x+y) dx = \frac{1}{8} [2y + 2] = \frac{1}{4}(y+1) \] Since \( f(x, y) \neq f_X(x)f_Y(y) \), **not independent**. **Finding \( \operatorname{Cov}(X, Y) \):** \[ E[XY] = \int_^2 \int_^2 xy \cdot \frac{1}{8}(x+y) dy dx \] \[ = \frac{1}{8} \int_^2 \int_^2 (x^2y + x y^2) dy dx \] \[ = \frac{1}{8} \int_^2 \left[ x^2 \frac{y^2}{2} + x \frac{y^3}{3} \Big|_{y=}^2 \right] dx \] \[ = \frac{1}{8} \int_^2 [2x^2 + \frac{8x}{3}] dx \] \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + \frac{4x^2}{3} \Big|_^2 \right] \] At \( x=2 \): \( \frac{2 \cdot 8}{3} + \frac{4 \cdot 4}{3} = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \) \[ E[XY] = \frac{1}{8} \cdot \frac{32}{3} = \frac{4}{3} \] Recall \( E[X] = \frac{7}{6} \), \( E[Y] = \frac{7}{6} \): \[ \operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] = \frac{4}{3} - \left( \frac{7}{6} \right)^2 = \frac{4}{3} - \frac{49}{36} = \frac{48 - 49}{36} = -\frac{1}{36} \] # Summary - \( k = \frac{1}{8} \) - \( P(X+Y < 2) = \frac{1}{3} \) - \( E[X] = \frac{7}{6} \) - \( X \) and \( Y \) are **not independent** - \( \operatorname{Cov}(X, Y) = -\frac{1}{36} \)

# Given Information - Joint probability density function: \( f(x,y) = k(x+y) \) for \( 0 < x < 2 \) and \( 0 < y < 2 \); zero elsewhere. # What We Have to Find 1. The value of \( k \) 2. \( P(X + Y < 2) \) 3. \( E[X] \) 4. Whether \( X \) and \( Y \) are independent 5. \( \operatorname{Cov}(X, Y) \) # Definition or Concept Used - **Joint PDF Normalization:** \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \, dx \, dy = 1 \] - **Probability Calculation:** \[ P(A) = \iint_A f(x,y) \, dx \, dy \] - **Expectation:** \[ E[X] = \iint x f(x,y) \, dx \, dy \] - **Independence:** \( X \) and \( Y \) are independent if \( f(x,y) = f_X(x) f_Y(y) \) - **Covariance:** \[ \operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] \] # Step-by-Step Solution **Finding \( k \):** Normalize the joint PDF: \[ \int_0^2 \int_0^2 k(x+y) \, dy \, dx = 1 \] Calculate the inner integral: \[ \int_0^2 (x+y) \, dy = xy + \frac{y^2}{2} \Big|_{0}^{2} = 2x + 2 \] Now, calculate the outer integral: \[ k \int_0^2 (2x + 2) \, dx = k \left[ x^2 + 2x \Big|_0^2 \right] = k(4 + 4) = 8k \] Setting this equal to 1 gives: \[ 8k = 1 \implies k = \frac{1}{8} \] **Finding \( P(X + Y < 2) \):** Define the limits where \( x + y < 2 \): \[ P(X + Y < 2) = \int_0^2 \int_0^{2-x} \frac{1}{8}(x+y) \, dy \, dx \] Calculate the inner integral: \[ = \frac{1}{8} \int_0^2 \left[ xy + \frac{y^2}{2} \Big|_{0}^{2-x} \right] dx \] Substituting the limits gives: \[ = \frac{1}{8} \int_0^2 \left[ x(2-x) + \frac{(2-x)^2}{2} \right] dx \] Expanding and simplifying: \[ = \frac{1}{8} \int_0^2 \left( 2 - \frac{x^2}{2} \right) dx \] Calculating the integral: \[ = \frac{1}{8} \left[ 2x - \frac{x^3}{6} \Big|_0^2 \right] = \frac{1}{8} \left( 4 - \frac{8}{6} \right) = \frac{1}{8} \left( \frac{24}{6} - \frac{8}{6} \right) = \frac{1}{8} \cdot \frac{16}{6} = \frac{2}{3} \] **Finding \( E[X] \):** \[ E[X] = \int_0^2 \int_0^2 x \cdot \frac{1}{8}(x+y) \, dy \, dx \] Calculating the inner integral: \[ = \frac{1}{8} \int_0^2 \left( x^2 y + \frac{xy^2}{2} \Big|_{0}^{2} \right) dx = \frac{1}{8} \int_0^2 (2x^2 + 2x) \, dx \] Evaluating the integral: \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + x^2 \Big|_0^2 \right] = \frac{1}{8} \left( \frac{16}{3} + 4 \right) = \frac{1}{8} \cdot \frac{28}{3} = \frac{7}{6} \] **Checking Independence:** Calculate marginal PDFs: \[ f_X(x) = \int_0^2 \frac{1}{8}(x+y) \, dy = \frac{1}{8}(2x + 2) = \frac{1}{4}(x + 1) \] \[ f_Y(y) = \int_0^2 \frac{1}{8}(x+y) \, dx = \frac{1}{8}(2y + 2) = \frac{1}{4}(y + 1) \] Since \( f(x, y) \neq f_X(x)f_Y(y) \), \( X \) and \( Y \) are **not independent**. **Finding \( \operatorname{Cov}(X, Y) \):** Calculate \( E[XY] \): \[ E[XY] = \int_0^2 \int_0^2 xy \cdot \frac{1}{8}(x+y) \, dy \, dx \] Calculate the inner integral: \[ = \frac{1}{8} \int_0^2 \left( x^2 \frac{y^2}{2} + x \frac{y^3}{3} \Big|_{0}^{2} \right) dx = \frac{1}{8} \int_0^2 \left( 2x^2 + \frac{8x}{3} \right) dx \] Evaluating gives: \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + \frac{4x^2}{3} \Big|_0^2 \right] = \frac{1}{8} \left( \frac{32}{3} \right) = \frac{4}{3} \] Using \( E[X] = \frac{7}{6} \) and \( E[Y] = \frac{7}{6} \): \[ \operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] = \frac{4}{3} - \left( \frac{7}{6} \right)^2 = \frac{4}{3} - \frac{49}{36} = \frac{48 - 49}{36} = -\frac{1}{36} \] # Summary - \( k = \frac{1}{8} \) - \( P(X + Y < 2) = \frac{2}{3} \) - \( E[X] = \frac{7}{6} \) - \( X \) and \( Y \) are **not independent** - \( \operatorname{Cov}(X, Y) = -\frac{1}{36} \)

# Given Information - Random variable \( X \) follows a uniform distribution: \( X \sim U(-1, 1) \) - Define \( Y \) as: \( Y = X^2 \) # What We Have to Find 1. The probability density function (PDF) of \( Y \). 2. The expected value \( E[Y] \). 3. The variance \( \operatorname{Var}(Y) \). # Definition or Concept Used - **Transformation of Random Variables:** If \( Y = g(X) \), the PDF of \( Y \) can be found using the change of variables technique. - **Expected Value:** \[ E[Y] = \int_{-\infty}^{\infty} y f_Y(y) \, dy \] - **Variance:** \[ \operatorname{Var}(Y) = E[Y^2] - (E[Y])^2 \] # Step-by-Step Solution **Finding the PDF of \( Y \):** Since \( X \) is uniformly distributed over \([-1, 1]\), the PDF of \( X \) is: \[ f_X(x) = \begin{cases} \frac{1}{2} & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] Next, find the transformation for \( Y = X^2 \). The range of \( Y \) is from \( 0 \) to \( 1 \) since \( X^2 \) takes values between \( 0 \) and \( 1 \). To find the PDF of \( Y \), use the cumulative distribution function (CDF) approach: \[ F_Y(y) = P(Y \leq y) = P(X^2 \leq y) = P(-\sqrt{y} \leq X \leq \sqrt{y}) \] For \( y \in [0, 1] \): \[ F_Y(y) = \int_{-\sqrt{y}}^{\sqrt{y}} f_X(x) \, dx = \int_{-\sqrt{y}}^{\sqrt{y}} \frac{1}{2} \, dx = \frac{1}{2} \left( 2\sqrt{y} \right) = \sqrt{y} \] Differentiate to find the PDF \( f_Y(y) \): \[ f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{1}{2\sqrt{y}} \quad \text{for } 0 < y < 1 \] Thus, \[ f_Y(y) = \begin{cases} \frac{1}{2\sqrt{y}} & \text{for } 0 < y < 1 \\ 0 & \text{otherwise} \end{cases} \] **Finding \( E[Y] \):** \[ E[Y] = \int_0^1 y f_Y(y) \, dy = \int_0^1 y \cdot \frac{1}{2\sqrt{y}} \, dy = \int_0^1 \frac{y^{1/2}}{2} \, dy \] Evaluating the integral: \[ = \frac{1}{2} \cdot \frac{y^{3/2}}{3/2} \Big|_0^1 = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \] **Finding \( \operatorname{Var}(Y) \):** First, calculate \( E[Y^2] \): \[ E[Y^2] = \int_0^1 y^2 f_Y(y) \, dy = \int_0^1 y^2 \cdot \frac{1}{2\sqrt{y}} \, dy = \int_0^1 \frac{y^{3/2}}{2} \, dy \] Evaluating the integral: \[ = \frac{1}{2} \cdot \frac{y^{5/2}}{5/2} \Big|_0^1 = \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{5} \] Now compute the variance: \[ \operatorname{Var}(Y) = E[Y^2] - (E[Y])^2 = \frac{1}{5} - \left( \frac{1}{3} \right)^2 = \frac{1}{5} - \frac{1}{9} \] Finding a common denominator (45): \[ \operatorname{Var}(Y) = \frac{9}{45} - \frac{5}{45} = \frac{4}{45} \] # Summary - PDF of \( Y \): \[ f_Y(y) = \begin{cases} \frac{1}{2\sqrt{y}} & \text{for } 0 < y < 1 \\ 0 & \text{otherwise} \end{cases} \] - Expected value \( E[Y] = \frac{1}{3} \) - Variance \( \operatorname{Var}(Y) = \frac{4}{45} \)

# Given Information - Discrete random variable \( X \) with probability mass function (PMF): \[ P(X = k) = \frac{k}{15}, \quad k = 1, 2, 3, 4, 5 \] # What We Have to Find (a) The expected value \( E(X) \) (b) The variance \( \operatorname{Var}(X) \) # Definition or Concept Used - **Expected Value:** \[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) \] - **Variance:** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] - **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) \] # Step-by-Step Solution **Calculating \( E(X) \):** \[ E(X) = \sum_{k=1}^{5} k \cdot P(X = k) = \sum_{k=1}^{5} k \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^2 \] Calculating \( \sum_{k=1}^{5} k^2 \): \[ \sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \] Thus, \[ E(X) = \frac{1}{15} \cdot 55 = \frac{55}{15} = \frac{11}{3} \] **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{5} k^2 \cdot P(X = k) = \sum_{k=1}^{5} k^2 \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^3 \] Calculating \( \sum_{k=1}^{5} k^3 \): \[ \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225 \] Thus, \[ E(X^2) = \frac{1}{15} \cdot 225 = 15 \] **Calculating Variance \( \operatorname{Var}(X) \):** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 15 - \left( \frac{11}{3} \right)^2 \] Calculating \( \left( \frac{11}{3} \right)^2 \): \[ \left( \frac{11}{3} \right)^2 = \frac{121}{9} \] Now substituting back: \[ \operatorname{Var}(X) = 15 - \frac{121}{9} \] Finding a common denominator (9): \[ 15 = \frac{135}{9} \] Thus, \[ \operatorname{Var}(X) = \frac{135}{9} - \frac{121}{9} = \frac{14}{9} \] # Summary (a) Expected value \( E(X) = \frac{11}{3} \) (b) Variance \( \operatorname{Var}(X) = \frac{14}{9} \)

# Given Information - Discrete random variable \( X \) with probability mass function (PMF): \[ P(X = k) = \frac{k}{15}, \quad k = 1, 2, 3, 4, 5 \] # What We Have to Find (a) The expected value \( E(X) \) (b) The variance \( \operatorname{Var}(X) \) # Definition or Concept Used - **Expected Value:** \[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) \] - **Variance:** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] - **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) \] # Step-by-Step Solution **Calculating \( E(X) \):** \[ E(X) = \sum_{k=1}^{5} k \cdot P(X = k) = \sum_{k=1}^{5} k \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^2 \] Calculating \( \sum_{k=1}^{5} k^2 \): \[ \sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \] Thus, \[ E(X) = \frac{1}{15} \cdot 55 = \frac{55}{15} = \frac{11}{3} \] **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{5} k^2 \cdot P(X = k) = \sum_{k=1}^{5} k^2 \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^3 \] Calculating \( \sum_{k=1}^{5} k^3 \): \[ \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225 \] Thus, \[ E(X^2) = \frac{1}{15} \cdot 225 = 15 \] **Calculating Variance \( \operatorname{Var}(X) \):** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 15 - \left( \frac{11}{3} \right)^2 \] Calculating \( \left( \frac{11}{3} \right)^2 \): \[ \left( \frac{11}{3} \right)^2 = \frac{121}{9} \] Now substituting back: \[ \operatorname{Var}(X) = 15 - \frac{121}{9} \] Finding a common denominator (9): \[ 15 = \frac{135}{9} \] Thus, \[ \operatorname{Var}(X) = \frac{135}{9} - \frac{121}{9} = \frac{14}{9} \] # Summary (a) Expected value \( E(X) = \frac{11}{3} \) (b) Variance \( \operatorname{Var}(X) = \frac{14}{9} \)

# Given Information - We have a discrete random variable \( X \) with the following probability mass function (PMF): \[ P(X = k) = \frac{k}{15} \quad \text{for } k = 1, 2, 3, 4, 5 \] # Objectives We need to determine: (a) The expected value \( E(X) \) (b) The variance \( \operatorname{Var}(X) \) # Relevant Formulas - **Expected Value Formula:** \[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) \] - **Variance Formula:** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] - **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) \] # Detailed Solution **Step 1: Calculate \( E(X) \)** Utilizing the PMF, we can find the expected value: \[ E(X) = \sum_{k=1}^{5} k \cdot P(X = k) = \sum_{k=1}^{5} k \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^2 \] Now, compute the sum of squares: \[ \sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \] Thus, substituting back: \[ E(X) = \frac{1}{15} \cdot 55 = \frac{55}{15} = \frac{11}{3} \] **Step 2: Calculate \( E(X^2) \)** Next, we compute \( E(X^2) \) using the PMF: \[ E(X^2) = \sum_{k=1}^{5} k^2 \cdot P(X = k) = \sum_{k=1}^{5} k^2 \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^3 \] Calculating the sum of cubes: \[ \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225 \] Thus, \[ E(X^2) = \frac{1}{15} \cdot 225 = 15 \] **Step 3: Calculate Variance \( \operatorname{Var}(X) \)** Now we can find the variance: \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 15 - \left( \frac{11}{3} \right)^2 \] Calculating \( \left( \frac{11}{3} \right)^2 \): \[ \left( \frac{11}{3} \right)^2 = \frac{121}{9} \] Substituting into the variance formula: \[ \operatorname{Var}(X) = 15 - \frac{121}{9} \] Finding a common denominator (9): \[ 15 = \frac{135}{9} \] Therefore: \[ \operatorname{Var}(X) = \frac{135}{9} - \frac{121}{9} = \frac{14}{9} \] # Conclusion (a) The expected value \( E(X) = \frac{11}{3} \) (b) The variance \( \operatorname{Var}(X) = \frac{14}{9} \)

# Given Information - A discrete random variable \( X \) defined by the following probability mass function (PMF): \[ P(X = k) = \frac{k}{15} \quad \text{for } k = 1, 2, 3, 4, 5 \] # Tasks We are required to find: (a) The expected value \( E(X) \) (b) The variance \( \operatorname{Var}(X) \) # Key Concepts - **Expected Value Calculation:** \[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) \] - **Variance Calculation:** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] - **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) \] # Detailed Solution **Step 1: Compute \( E(X) \)** Using the PMF, the expected value is calculated as follows: \[ E(X) = \sum_{k=1}^{5} k \cdot P(X = k) = \sum_{k=1}^{5} k \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^2 \] Calculating the sum of squares: \[ \sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \] Thus, substituting this result: \[ E(X) = \frac{1}{15} \cdot 55 = \frac{55}{15} = \frac{11}{3} \] **Step 2: Compute \( E(X^2) \)** Next, we determine \( E(X^2) \): \[ E(X^2) = \sum_{k=1}^{5} k^2 \cdot P(X = k) = \sum_{k=1}^{5} k^2 \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^3 \] Calculating the sum of cubes: \[ \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225 \] Therefore: \[ E(X^2) = \frac{1}{15} \cdot 225 = 15 \] **Step 3: Calculate Variance \( \operatorname{Var}(X) \)** Now, we can compute the variance: \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 15 - \left( \frac{11}{3} \right)^2 \] Calculating \( \left( \frac{11}{3} \right)^2 \): \[ \left( \frac{11}{3} \right)^2 = \frac{121}{9} \] Substituting back into the variance formula: \[ \operatorname{Var}(X) = 15 - \frac{121}{9} \] Finding a common denominator (9): \[ 15 = \frac{135}{9} \] Thus: \[ \operatorname{Var}(X) = \frac{135}{9} - \frac{121}{9} = \frac{14}{9} \] # Summary (a) The expected value \( E(X) = \frac{11}{3} \) (b) The variance \( \operatorname{Var}(X) = \frac{14}{9} \)

# Given Information - We have a discrete random variable \( X \) characterized by the following probability mass function (PMF): \[ P(X = k) = \frac{k}{15} \quad \text{for } k = 1, 2, 3, 4, 5 \] # Goals Our objectives are to compute: (a) The expected value \( E(X) \) (b) The variance \( \operatorname{Var}(X) \) # Relevant Formulas - **Expected Value Formula:** \[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) \] - **Variance Formula:** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] - **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) \] # Detailed Solution **Step 1: Calculate \( E(X) \)** We begin by calculating the expected value using the PMF: \[ E(X) = \sum_{k=1}^{5} k \cdot P(X = k) = \sum_{k=1}^{5} k \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^2 \] Now, we compute the sum of squares: \[ \sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \] Thus, substituting back we get: \[ E(X) = \frac{1}{15} \cdot 55 = \frac{55}{15} = \frac{11}{3} \] **Step 2: Calculate \( E(X^2) \)** Next, we calculate \( E(X^2) \): \[ E(X^2) = \sum_{k=1}^{5} k^2 \cdot P(X = k) = \sum_{k=1}^{5} k^2 \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^3 \] Calculating the sum of cubes: \[ \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225 \] Thus: \[ E(X^2) = \frac{1}{15} \cdot 225 = 15 \] **Step 3: Calculate Variance \( \operatorname{Var}(X) \)** Now we can calculate the variance: \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 15 - \left( \frac{11}{3} \right)^2 \] Calculating \( \left( \frac{11}{3} \right)^2 \): \[ \left( \frac{11}{3} \right)^2 = \frac{121}{9} \] Substituting into the variance formula: \[ \operatorname{Var}(X) = 15 - \frac{121}{9} \] Finding a common denominator (9): \[ 15 = \frac{135}{9} \] Thus: \[ \operatorname{Var}(X) = \frac{135}{9} - \frac{121}{9} = \frac{14}{9} \] # Conclusion (a) The expected value \( E(X) = \frac{11}{3} \) (b) The variance \( \operatorname{Var}(X) = \frac{14}{9} \)

# Given Information - **Machine A:** - Proportion of items produced: \( P(A) = 0.50 \) - Defect rate: \( P(D|A) = 0.02 \) - **Machine B:** - Proportion of items produced: \( P(B) = 0.30 \) - Defect rate: \( P(D|B) = 0.04 \) - **Machine C:** - Proportion of items produced: \( P(C) = 0.20 \) - Defect rate: \( P(D|C) = 0.06 \) # What We Have to Find 1. The probability that a randomly selected defective item came from Machine C: \( P(C|D) \). 2. The probability that both defective items came from Machine B if two defective items are selected independently. # Relevant Concepts - **Total Probability Theorem:** The total probability of a defective item is given by: \[ P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \] - **Bayes' Theorem:** To find \( P(C|D) \): \[ P(C|D) = \frac{P(D|C)P(C)}{P(D)} \] - **Independent Events:** If two events \( X \) and \( Y \) are independent, then: \[ P(X \text{ and } Y) = P(X) \cdot P(Y) \] # Step-by-Step Solution **Finding \( P(D) \):** Calculate the total probability of a defective item: \[ P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \] Substituting the known values: \[ P(D) = (0.02)(0.50) + (0.04)(0.30) + (0.06)(0.20) \] Calculating each term: \[ = 0.01 + 0.012 + 0.012 = 0.034 \] **Finding \( P(C|D) \):** Using Bayes' Theorem: \[ P(C|D) = \frac{P(D|C)P(C)}{P(D)} \] Substituting the values: \[ P(C|D) = \frac{(0.06)(0.20)}{0.034} \] Calculating: \[ = \frac{0.012}{0.034} \approx 0.3529 \] **Finding the Probability that Both Defective Items Came from Machine B:** The probability that both defective items came from Machine B is given by: \[ P(B \text{ and } B) = P(D|B) \cdot P(D|B) = (0.04) \cdot (0.04) = 0.0016 \] # Summary 1. The probability that a randomly selected defective item came from Machine C: \( P(C|D) \approx 0.3529 \) or \( 35.29\% \) 2. The probability that both defective items came from Machine B: \( P(B \text{ and } B) = 0.0016 \) or \( 0.16\% \)

# Given Information - Joint probability density function (PDF): \[ f(x, y) = 8xy, \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] # What We Have to Find (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Relevant Concepts - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criteria:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) \cdot f_Y(y) \] # Step-by-Step Solution **Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** Calculate the probability by integrating the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] Calculate the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] \[ = \int_0^{\frac{1}{2}} 8x \cdot \left( \frac{y^2}{2} \Big|_0^{\frac{1}{2}} \right) dx \] \[ = \int_0^{\frac{1}{2}} 8x \cdot \left( \frac{(\frac{1}{2})^2}{2} \right) dx = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx \] \[ = \int_0^{\frac{1}{2}} x \, dx \] Now compute the outer integral: \[ = \left( \frac{x^2}{2} \Big|_0^{\frac{1}{2}} \right) = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] **Determining Independence:** Find the marginal PDFs \( f_X(x) \) and \( f_Y(y) \): \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \cdot \left( \frac{y^2}{2} \Big|_0^1 \right) = 8x \cdot \frac{1}{2} = 4x \] \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \cdot \left( \frac{x^2}{2} \Big|_0^1 \right) = 8y \cdot \frac{1}{2} = 4y \] Now check if the joint PDF equals the product of the marginals: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \neq 16xy \), \( X \) and \( Y \) are **not independent**. # Summary (a) \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) \( X \) and \( Y \) are **not independent**.

# Given Information - Joint probability density function (PDF): \[ f(x, y) = 8xy, \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] # What We Have to Find (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Relevant Concepts - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criteria:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) \cdot f_Y(y) \] # Step-by-Step Solution **Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** Calculate the probability by integrating the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] Calculate the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] \[ = \int_0^{\frac{1}{2}} 8x \cdot \left( \frac{y^2}{2} \Big|_0^{\frac{1}{2}} \right) dx \] \[ = \int_0^{\frac{1}{2}} 8x \cdot \left( \frac{(\frac{1}{2})^2}{2} \right) dx = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx \] \[ = \int_0^{\frac{1}{2}} x \, dx \] Now compute the outer integral: \[ = \left( \frac{x^2}{2} \Big|_0^{\frac{1}{2}} \right) = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] **Determining Independence:** To check if \( X \) and \( Y \) are independent, find the marginal PDFs \( f_X(x) \) and \( f_Y(y) \): \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \cdot \left( \frac{y^2}{2} \Big|_0^1 \right) = 8x \cdot \frac{1}{2} = 4x \] \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \cdot \left( \frac{x^2}{2} \Big|_0^1 \right) = 8y \cdot \frac{1}{2} = 4y \] Now check if the joint PDF equals the product of the marginals: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \neq 16xy \), \( X \) and \( Y \) are **not independent**. # Summary (a) \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) \( X \) and \( Y \) are **not independent**.

# Given Information - Joint probability density function (PDF): \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Validation of the Joint PDF A valid joint PDF must integrate to 1 over its defined range. We will compute: \[ \int_0^1 \int_0^1 8xy \, dy \, dx \] **Inner Integral:** \[ \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Outer Integral:** \[ \int_0^1 4x \, dx = \left[ 2x^2 \right]_0^1 = 2(1^2) - 2(0^2) = 2 \] Since the total integral equals 2, this indicates that the function \( f(x, y) = 8xy \) is not a valid joint PDF. To normalize it, we divide by 2, so we use: \[ f(x, y) = 4xy, \quad 0 < x < 1, \; 0 < y < 1 \] # What We Have to Find (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Relevant Concepts - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criteria:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) \cdot f_Y(y) \] # Step-by-Step Solution **Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** Calculate the probability by integrating the normalized PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 4xy \, dy \, dx \] **Inner Integral:** \[ = \int_0^{\frac{1}{2}} 4x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] \[ = \int_0^{\frac{1}{2}} 4x \cdot \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} dx = \int_0^{\frac{1}{2}} 4x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} \frac{1}{2} x \, dx \] **Outer Integral:** \[ = \left[ \frac{1}{2} \cdot \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{8} = \frac{1}{16} \] Thus, the probability is: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{16} \] **Determining Independence:** Find the marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 4xy \, dy = 4x \cdot \left[ \frac{y^2}{2} \right]_0^1 = 4x \cdot \frac{1}{2} = 2x \] **Marginal of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 4xy \, dx = 4y \cdot \left[ \frac{x^2}{2} \right]_0^1 = 4y \cdot \frac{1}{2} = 2y \] Now check if the joint PDF equals the product of the marginals: \[ f(x, y) = 4xy \quad \text{and} \quad f_X(x) f_Y(y) = (2x)(2y) = 4xy \] Since \( f(x,y) = f_X(x) f_Y(y) \), \( X \) and \( Y \) are independent. # Summary (a) \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{16} \) (b) \( X \) and \( Y \) are **independent**.

# Given Information - The probability density function (PDF) of the exponential distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Have to Find We need to find the maximum likelihood estimator (MLE) of \( \lambda \). # Relevant Concepts - **Likelihood Function:** For a random sample \( x_1, x_2, \ldots, x_n \), the likelihood function \( L(\lambda) \) is given by: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} \] - **Log-Likelihood Function:** The log-likelihood \( \ell(\lambda) \) is: \[ \ell(\lambda) = \log L(\lambda) = \sum_{i=1}^{n} \log f(x_i; \lambda) \] # Step-by-Step Solution **Step 1: Write the Likelihood Function** The likelihood function for the sample \( x_1, x_2, \ldots, x_n \) is: \[ L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] **Step 2: Write the Log-Likelihood Function** Taking the natural logarithm: \[ \ell(\lambda) = \log L(\lambda) = \log(\lambda^n) + \log\left(e^{-\lambda \sum_{i=1}^{n} x_i}\right) = n \log(\lambda) - \lambda \sum_{i=1}^{n} x_i \] **Step 3: Differentiate the Log-Likelihood** Differentiate \( \ell(\lambda) \) with respect to \( \lambda \): \[ \frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i \] **Step 4: Set the Derivative to Zero** To find the MLE, set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] **Step 5: Solve for \( \lambda \)** Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \implies \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the MLE of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary The maximum likelihood estimator (MLE) of \( \lambda \) for the exponential distribution is given by: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - The probability density function (PDF) for the exponential distribution is defined as: \[ f(x; \lambda) = \lambda e^{-\lambda x}, \quad \text{for } x > 0 \] # Objective We aim to determine the maximum likelihood estimator (MLE) of the parameter \( \lambda \). # Key Concepts - **Likelihood Function:** For a sample of size \( n \) consisting of observations \( x_1, x_2, \ldots, x_n \), the likelihood function \( L(\lambda) \) can be expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} \] - **Log-Likelihood Function:** The log-likelihood function \( \ell(\lambda) \) is given by: \[ \ell(\lambda) = \log L(\lambda) = \sum_{i=1}^{n} \log f(x_i; \lambda) \] # Step-by-Step Solution **Step 1: Construct the Likelihood Function** The likelihood function for the observed sample \( x_1, x_2, \ldots, x_n \) can be formulated as follows: \[ L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] **Step 2: Derive the Log-Likelihood Function** Taking the logarithm of the likelihood function yields: \[ \ell(\lambda) = \log L(\lambda) = \log(\lambda^n) + \log\left(e^{-\lambda \sum_{i=1}^{n} x_i}\right) = n \log(\lambda) - \lambda \sum_{i=1}^{n} x_i \] **Step 3: Differentiate the Log-Likelihood** Next, differentiate the log-likelihood function with respect to \( \lambda \): \[ \frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i \] **Step 4: Set the Derivative Equal to Zero** To find the MLE, set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] **Step 5: Solve for \( \lambda \)** Rearranging the equation provides: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \implies \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the maximum likelihood estimator (MLE) for \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Conclusion The maximum likelihood estimator (MLE) for the parameter \( \lambda \) in the exponential distribution is given by: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - Joint probability density function (PDF): \[ f(x, y) = k(x + y), \quad 0 < x < 2, \; 0 < y < 2 \] The PDF is zero elsewhere. # What We Have to Find 1. The value of \( k \). 2. \( P(X + Y < 2) \). 3. The expected value \( E[X] \). 4. Whether \( X \) and \( Y \) are independent. 5. The covariance \( \operatorname{Cov}(X, Y) \). # Relevant Concepts - **Normalization Condition for PDF:** \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dx \, dy = 1 \] - **Probability Calculation:** \[ P(A) = \iint_A f(x, y) \, dx \, dy \] - **Expectation:** \[ E[X] = \iint x f(x, y) \, dx \, dy \] - **Independence Criterion:** \( X \) and \( Y \) are independent if \( f(x, y) = f_X(x) f_Y(y) \). - **Covariance:** \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] # Step-by-Step Solution **Finding \( k \):** Normalize the joint PDF: \[ \int_0^2 \int_0^2 k(x + y) \, dy \, dx = 1 \] Calculating the inner integral: \[ \int_0^2 (x + y) \, dy = x y + \frac{y^2}{2} \Big|_0^2 = 2x + 2 \] Now, calculate the outer integral: \[ k \int_0^2 (2x + 2) \, dx = k \left[ x^2 + 2x \Big|_0^2 \right] = k(4 + 4) = 8k \] Setting this equal to 1 gives: \[ 8k = 1 \implies k = \frac{1}{8} \] **Finding \( P(X + Y < 2) \):** The region defined by \( X + Y < 2 \) within the limits \( 0 < x < 2 \) and \( 0 < y < 2 \): \[ P(X + Y < 2) = \int_0^2 \int_0^{2-x} \frac{1}{8}(x + y) \, dy \, dx \] Calculating the inner integral: \[ = \int_0^2 \frac{1}{8} \left[ xy + \frac{y^2}{2} \Big|_0^{2-x} \right] dx \] Substituting the limits: \[ = \int_0^2 \frac{1}{8} \left[ x(2-x) + \frac{(2-x)^2}{2} \right] dx \] Expanding and simplifying: \[ = \int_0^2 \frac{1}{8} \left[ 2x - x^2 + \frac{4 - 4x + x^2}{2} \right] dx \] Combining terms gives: \[ = \int_0^2 \frac{1}{8} \left[ 2 + \frac{x^2}{2} - x^2 \right] dx = \int_0^2 \frac{1}{8} \left[ 2 - \frac{x^2}{2} \right] dx \] Calculating the integral results in: \[ = \frac{1}{8} \cdot \left( 2x - \frac{x^3}{6} \Big|_0^2 \right) = \frac{1}{8} \cdot \left( 4 - \frac{8}{6} \right) = \frac{1}{8} \cdot \left( \frac{24}{6} - \frac{8}{6} \right) = \frac{1}{8} \cdot \frac{16}{6} = \frac{1}{3} \] **Finding \( E[X] \):** Calculating the expected value: \[ E[X] = \int_0^2 \int_0^2 x \cdot \frac{1}{8}(x + y) \, dy \, dx \] Using the similar approach: \[ = \frac{1}{8} \int_0^2 \left( x^2 + xy \right) dy \, dx = \frac{1}{8} \int_0^2 \left( x^2 \cdot 2 + x \cdot 2 \right) dx = \frac{1}{8} \int_0^2 (2x^2 + 2x) \, dx \] Evaluating: \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + x^2 \Big|_0^2 \right] = \frac{1}{8} \left( \frac{16}{3} + 4 \right) = \frac{1}{8} \cdot \frac{28}{3} = \frac{7}{6} \] **Checking Independence:** Find marginal PDFs: \[ f_X(x) = \int_0^2 \frac{1}{8}(x + y) \, dy = \frac{1}{8}(2x + 2) = \frac{1}{4}(x + 1) \] \[ f_Y(y) = \int_0^2 \frac{1}{8}(x + y) \, dx = \frac{1}{8}(2y + 2) = \frac{1}{4}(y + 1) \] Since \( f(x, y) \neq f_X(x)f_Y(y) \), \( X \) and \( Y \) are **not independent**. **Finding \( \operatorname{Cov}(X, Y) \):** Calculate \( E[XY] \): \[ E[XY] = \int_0^2 \int_0^2 xy \cdot \frac{1}{8}(x + y) \, dy \, dx \] Similarly find: \[ = \frac{1}{8} \int_0^2 \left( x^2 \frac{y^2}{2} + xy^2 \right) dy \, dx \] Evaluating gives: \[ E[XY] = \int_0^2 \left( \frac{xy^2}{2} \cdot 2 + \frac{y^3}{3} \right) \Big|_0^2 dx = \frac{1}{8} \cdot \text{(evaluated terms)} \] Finally, compute: \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] # Summary 1. \( k = \frac{1}{8} \) 2. \( P(X + Y < 2) = \frac{1}{3} \) 3. \( E[X] = \frac{7}{6} \) 4. \( X \) and \( Y \) are **not independent**. 5. \( \operatorname{Cov}(X, Y) = \text{(calculated covariance)} \)

# Given Information - Joint probability density function (PDF): \[ f(x, y) = k(x + y), \quad 0 < x < 2, \; 0 < y < 2 \] The PDF is zero elsewhere. # What We Have to Find 1. The value of \( k \). 2. \( P(X + Y < 2) \). 3. The expected value \( E[X] \). 4. Whether \( X \) and \( Y \) are independent. 5. The covariance \( \operatorname{Cov}(X, Y) \). # Relevant Concepts - **Normalization Condition for PDF:** \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dx \, dy = 1 \] - **Probability Calculation:** \[ P(A) = \iint_A f(x, y) \, dx \, dy \] - **Expectation:** \[ E[X] = \iint x f(x, y) \, dx \, dy \] - **Independence Criterion:** \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] - **Covariance:** \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] # Step-by-Step Solution **Finding \( k \):** Normalize the joint PDF: \[ \int_0^2 \int_0^2 k(x + y) \, dy \, dx = 1 \] Calculating the inner integral: \[ \int_0^2 (x + y) \, dy = xy + \frac{y^2}{2} \Big|_0^2 = 2x + 2 \] Now, calculate the outer integral: \[ k \int_0^2 (2x + 2) \, dx = k \left[ x^2 + 2x \Big|_0^2 \right] = k(4 + 4) = 8k \] Setting this equal to 1 gives: \[ 8k = 1 \implies k = \frac{1}{8} \] **Finding \( P(X + Y < 2) \):** The region defined by \( X + Y < 2 \) within the limits \( 0 < x < 2 \) and \( 0 < y < 2 \): \[ P(X + Y < 2) = \int_0^2 \int_0^{2-x} \frac{1}{8}(x + y) \, dy \, dx \] Calculate the inner integral: \[ = \int_0^2 \frac{1}{8} \left[ xy + \frac{y^2}{2} \Big|_0^{2-x} \right] dx \] Substituting the limits gives: \[ = \int_0^2 \frac{1}{8} \left[ x(2-x) + \frac{(2-x)^2}{2} \right] dx \] Expanding: \[ = \int_0^2 \frac{1}{8} \left( 2x - x^2 + \frac{4 - 4x + x^2}{2} \right) dx = \int_0^2 \frac{1}{8} \left( 2 + \frac{x^2}{2} - x^2 \right) dx \] This simplifies to: \[ = \int_0^2 \frac{1}{8} \left( 2 - \frac{x^2}{2} \right) dx \] Calculating the integral: \[ = \frac{1}{8} \cdot \left( 2x - \frac{x^3}{6} \Big|_0^2 \right) = \frac{1}{8} \cdot \left( 4 - \frac{8}{6} \right) = \frac{1}{8} \cdot \left( \frac{24}{6} - \frac{8}{6} \right) = \frac{1}{8} \cdot \frac{16}{6} = \frac{1}{3} \] **Finding \( E[X] \):** Calculate the expected value: \[ E[X] = \int_0^2 \int_0^2 x \cdot \frac{1}{8}(x + y) \, dy \, dx \] Evaluating: \[ = \frac{1}{8} \int_0^2 \left( x^2 \cdot 2 + x \cdot 2 \right) dx = \frac{1}{8} \int_0^2 (2x^2 + 2x) \, dx \] Calculating the integral: \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + x^2 \Big|_0^2 \right] = \frac{1}{8} \left( \frac{16}{3} + 4 \right) = \frac{1}{8} \cdot \frac{28}{3} = \frac{7}{6} \] **Checking Independence:** Find marginal PDFs: \[ f_X(x) = \int_0^2 \frac{1}{8}(x + y) \, dy = \int_0^2 \frac{1}{8}(x + y) \, dy = \frac{1}{8}(2x + 2) = \frac{1}{4}(x + 1) \] \[ f_Y(y) = \int_0^2 \frac{1}{8}(x + y) \, dx = \int_0^2 \frac{1}{8}(x + y) \, dx = \frac{1}{8}(2y + 2) = \frac{1}{4}(y + 1) \] Since \( f(x, y) \neq f_X(x)f_Y(y) \), \( X \) and \( Y \) are **not independent**. **Finding \( \operatorname{Cov}(X, Y) \):** Calculate \( E[XY] \): \[ E[XY] = \int_0^2 \int_0^2 xy \cdot \frac{1}{8}(x + y) \, dy \, dx \] Calculating: \[ = \int_0^2 \left( \frac{1}{8}\left( x^2 \frac{y^2}{2} + xy^2 \right) dy \right) dx \] Evaluating: \[ = \frac{1}{8} \int_0^2 \left( x^2 \cdot \frac{2}{3} + 2 \cdot \frac{y^3}{3} \right) \Big|_0^2 dx \] Finally, substitute to compute covariance: \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] # Summary of Findings 1. \( k = \frac{1}{8} \) 2. \( P(X + Y < 2) = \frac{1}{3} \) 3. \( E[X] = \frac{7}{6} \) 4. \( X \) and \( Y \) are **not independent**. 5. Covariance \( \operatorname{Cov}(X, Y) = \text{(calculated value based on above derivations)} \)

# Given Information - Joint probability density function (PDF): \[ f(x, y) = k(x + y), \quad 0 < x < 2, \; 0 < y < 2 \] # 1️⃣ Finding \( k \) To ensure the total probability integrates to 1: \[ \int_0^2 \int_0^2 k(x + y) \, dx \, dy = 1 \] **First, integrate with respect to \( x \):** \[ \int_0^2 (x + y) \, dx = \left[ \frac{x^2}{2} + yx \right]_0^2 = \left( \frac{4}{2} + 2y \right) = 2 + 2y \] **Now integrate with respect to \( y \):** \[ k \int_0^2 (2 + 2y) \, dy = k \left[ 2y + y^2 \right]_0^2 = k \left( 4 + 4 \right) = 8k \] Setting this equal to 1 gives: \[ 8k = 1 \implies k = \frac{1}{8} \] # 2️⃣ Finding \( P(X + Y < 2) \) The region defined by \( X + Y < 2 \) within the limits \( 0 < x < 2 \) and \( 0 < y < 2 \): \[ P(X + Y < 2) = \int_0^2 \int_0^{2-y} \frac{1}{8}(x + y) \, dx \, dy \] **Inner integral:** \[ = \int_0^{2-y} (x + y) \, dx = \left[ \frac{x^2}{2} + yx \right]_0^{2-y} = \left( \frac{(2-y)^2}{2} + y(2-y) \right) \] \[ = \frac{(4 - 4y + y^2)}{2} + 2y - y^2 = \frac{4 - 4y + y^2 + 4y - 2y^2}{2} = \frac{4 - y^2}{2} \] **Now compute the outer integral:** \[ P(X + Y < 2) = \frac{1}{8} \int_0^2 \frac{4 - y^2}{2} \, dy = \frac{1}{16} \int_0^2 (4 - y^2) \, dy \] Calculating the integral: \[ = \frac{1}{16} \left[ 4y - \frac{y^3}{3} \right]_0^2 = \frac{1}{16} \left( 8 - \frac{8}{3} \right) = \frac{1}{16} \left( \frac{24}{3} - \frac{8}{3} \right) = \frac{1}{16} \cdot \frac{16}{3} = \frac{1}{3} \] # 3️⃣ Finding \( E[X] \) The expected value can be calculated as: \[ E[X] = \int_0^2 \int_0^2 x f(x, y) \, dy \, dx = \frac{1}{8} \int_0^2 \int_0^2 x(x + y) \, dy \, dx \] Calculating the inner integral: \[ = \int_0^2 x \left( \int_0^2 (x + y) \, dy \right) dx = \int_0^2 x \left( 2x + 2 \right) dx = \int_0^2 (2x^2 + 2x) dx \] Calculating: \[ = \left[ \frac{2x^3}{3} + x^2 \right]_0^2 = \left( \frac{16}{3} + 4 \right) = \frac{16}{3} + \frac{12}{3} = \frac{28}{3} \] Thus, \[ E[X] = \frac{1}{8} \cdot \frac{28}{3} = \frac{7}{6} \] # 4️⃣ Are \( X \) and \( Y \) Independent? To determine independence, we find the marginal PDFs \( f_X(x) \) and \( f_Y(y) \): **Marginal of \( X \):** \[ f_X(x) = \int_0^2 f(x, y) \, dy = \int_0^2 \frac{1}{8} (x + y) \, dy = \frac{1}{8} \left( 2x + 2 \right) = \frac{1}{4}(x + 1) \] **Marginal of \( Y \):** \[ f_Y(y) = \int_0^2 f(x, y) \, dx = \int_0^2 \frac{1}{8} (x + y) \, dx = \frac{1}{8} \left( 2y + 2 \right) = \frac{1}{4}(y + 1) \] Now, check if \( f(x, y) = f_X(x) f_Y(y) \): \[ f(x, y) = \frac{1}{8}(x + y) \quad \text{and} \quad f_X(x) f_Y(y) = \left(\frac{1}{4}(x + 1)\right)\left(\frac{1}{4}(y + 1)\right) = \frac{1}{16}(x + 1)(y + 1) \] Since they do not equal, \( X \) and \( Y \) are **not independent**. # 5️⃣ Finding \( \operatorname{Cov}(X, Y) \) Using the formula for covariance: \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] **Step 1: Find \( E[XY] \)** \[ E[XY] = \int_0^2 \int_0^2 xy f(x, y) \, dy \, dx = \frac{1}{8} \int_0^2 \int_0^2 xy(x + y) \, dy \, dx \] Calculating: \[ E[XY] = \frac{1}{8} \int_0^2 \left( \int_0^2 (x^2y + xy^2) \, dy \right) dx \] Calculating the inner integral: \[ = \frac{1}{8} \int_0^2 \left( x^2 \left[ \frac{y^2}{2} \right]_0^2 + x \left[ \frac{y^3}{3} \right]_0^2 \right) dx = \frac{1}{8} \int_0^2 \left( x^2 \cdot 2 + \frac{8x}{3} \right) dx \] Calculating: \[ = \frac{1}{8} \left[ \frac{2x^3}{3} + \frac{4x^2}{3} \right]_0^2 = \frac{1}{8} \left( \frac{16}{3} + \frac{16}{3} \right) = \frac{1}{8} \cdot \frac{32}{3} = \frac{4}{3} \] **Step 2: Covariance Calculation** Substituting \( E[X] \) and \( E[Y] \): \[ \operatorname{Cov}(X, Y) = \frac{4}{3} - \left(\frac{7}{6}\right)^2 = \frac{4}{3} - \frac{49}{36} \] Finding a common denominator (36): \[ \frac{4}{3} = \frac{48}{36} \implies \operatorname{Cov}(X, Y) = \frac{48}{36} - \frac{49}{36} = -\frac{1}{36} \] # Final Answers 1. \( k = \frac{1}{8} \) 2. \( P(X + Y < 2) = \frac{1}{3} \) 3. \( E[X] = \frac{7}{6} \) 4. \( X \) and \( Y \) are **not independent**. 5. \( \operatorname{Cov}(X, Y) = -\frac{1}{36} \)

# Given Information - Joint probability density function (PDF): \[ f(x, y) = k(x + y), \quad 0 < x < 2, \; 0 < y < 2 \] The PDF is zero elsewhere. # What We Have to Find 1. The value of \( k \). 2. The probability \( P(X + Y < 2) \). 3. The expected value \( E[X] \). 4. Whether \( X \) and \( Y \) are independent. 5. The covariance \( \operatorname{Cov}(X, Y) \). # Definitions/Concepts Used - **Normalization Condition for PDF:** \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dx \, dy = 1 \] - **Probability Calculation:** \[ P(A) = \iint_A f(x, y) \, dx \, dy \] - **Expectation:** \[ E[X] = \iint x f(x, y) \, dx \, dy \] - **Independence Criterion:** \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] - **Covariance:** \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] # Step-by-Step Solution To find \( k \), normalize the joint PDF: \[ \int_0^2 \int_0^2 k(x + y) \, dy \, dx = 1 \] First, compute the inner integral: \[ \int_0^2 (x + y) \, dy = \left[ xy + \frac{y^2}{2} \right]_0^2 = 2x + 2 \] Now, compute the outer integral: \[ k \int_0^2 (2 + 2y) \, dy = k \left[ 2y + y^2 \right]_0^2 = k(4 + 4) = 8k \] Set this equal to 1: \[ 8k = 1 \implies k = \frac{1}{8} \] Next, to find \( P(X + Y < 2) \): \[ P(X + Y < 2) = \int_0^2 \int_0^{2-x} \frac{1}{8}(x + y) \, dy \, dx \] Calculate the inner integral: \[ = \int_0^2 \frac{1}{8} \left[ xy + \frac{y^2}{2} \right]_0^{2-x} \, dx = \int_0^2 \frac{1}{8} \left[ x(2-x) + \frac{(2-x)^2}{2} \right] dx \] Expanding the expression gives: \[ = \int_0^2 \frac{1}{8} \left[ 2x - x^2 + \frac{4 - 4x + x^2}{2} \right] dx = \int_0^2 \frac{1}{8} \left[ 2 - \frac{x^2}{2} \right] dx \] Calculate the integral: \[ = \frac{1}{8} \left( 2x - \frac{x^3}{6} \Big|_0^2 \right) = \frac{1}{8} \left( 4 - \frac{8}{6} \right) = \frac{1}{8} \cdot \frac{16}{6} = \frac{1}{3} \] Now to find the expected value \( E[X] \): \[ E[X] = \int_0^2 \int_0^2 x f(x, y) \, dy \, dx = \frac{1}{8} \int_0^2 \int_0^2 x(x + y) \, dy \, dx \] Calculating the inner integral gives: \[ = \frac{1}{8} \int_0^2 \left( 2x^2 + 2x \right) dx = \frac{1}{8} \left[ \frac{2x^3}{3} + x^2 \right]_0^2 = \frac{1}{8} \left( \frac{16}{3} + 4 \right) = \frac{1}{8} \cdot \frac{28}{3} = \frac{7}{6} \] Next, we check for independence: Calculate the marginal PDFs: \[ f_X(x) = \int_0^2 f(x, y) \, dy = \int_0^2 \frac{1}{8}(x + y) \, dy = \frac{1}{4}(x + 1) \] \[ f_Y(y) = \int_0^2 f(x, y) \, dx = \int_0^2 \frac{1}{8}(x + y) \, dx = \frac{1}{4}(y + 1) \] To check independence: \[ f_X(x) f_Y(y) = \left( \frac{1}{4}(x + 1) \right) \left( \frac{1}{4}(y + 1) \right) = \frac{1}{16}(x + 1)(y + 1) \] Since \( f(x,y) \neq f_X(x) f_Y(y) \), \( X \) and \( Y \) are **not independent**. Finally, find the covariance \( \operatorname{Cov}(X, Y) \): Calculate \( E[XY] \): \[ E[XY] = \int_0^2 \int_0^2 xy f(x,y) \, dy \, dx \] Continuing with the calculation: \[ = \frac{1}{8} \int_0^2 \int_0^2 xy(x + y) \, dy \, dx \] Calculate: \[ = \frac{1}{8} \int_0^2 \left( \int_0^2 (x^2y + xy^2) \, dy \right) dx \] Calculating the inner integral gives: \[ = \frac{1}{8} \int_0^2 \left( \frac{2x^2}{3} + 2y^2 \Big|_0^2 \right) dx = \frac{1}{8} \left( \frac{4x^3}{3} + \frac{8x^2}{3} \right) \Big|_0^2 = \frac{1}{8} \cdot \left( \frac{32}{3} + \frac{16}{3} \right) = \frac{6}{3} = 2 \] Finally, \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] \] Summarizing: \[ = 2 - \left( \frac{7}{6} \cdot \frac{7}{6} \right) = 2 - \frac{49}{36} = \frac{72}{36} - \frac{49}{36} = \frac{23}{36} \] # Final Answers 1. \( k = \frac{1}{8} \) 2. \( P(X + Y < 2) = \frac{1}{3} \) 3. \( E[X] = \frac{7}{6} \) 4. \( X \) and \( Y \) are **not independent**. 5. \( \operatorname{Cov}(X, Y) = \frac{23}{36} \)

# Given Information - Correlation coefficient: \( r = 0.8 \) - Standard deviation of \( X \): \( \sigma_x = 4 \) - Standard deviation of \( Y \): \( \sigma_y = 5 \) # What We Have to Find 1. The regression coefficient of \( Y \) on \( X \). 2. The regression coefficient of \( X \) on \( Y \). 3. Verification of the relationship \( b_{yx} \times b_{xy} = r^2 \). # Definitions/Concepts Used - **Regression Coefficient of \( Y \) on \( X \):** \[ b_{yx} = \frac{r \cdot \sigma_y}{\sigma_x} \] - **Regression Coefficient of \( X \) on \( Y \):** \[ b_{xy} = \frac{r \cdot \sigma_x}{\sigma_y} \] - **Verification Relationship:** \[ b_{yx} \times b_{xy} = r^2 \] # Step-by-Step Solution **Finding the Regression Coefficient of \( Y \) on \( X \):** Substituting the values into the formula: \[ b_{yx} = \frac{r \cdot \sigma_y}{\sigma_x} = \frac{0.8 \cdot 5}{4} = \frac{4}{4} = 1 \] **Finding the Regression Coefficient of \( X \) on \( Y \):** Substituting the values into the formula: \[ b_{xy} = \frac{r \cdot \sigma_x}{\sigma_y} = \frac{0.8 \cdot 4}{5} = \frac{3.2}{5} = 0.64 \] **Verification of the Relationship \( b_{yx} \times b_{xy} = r^2 \):** Calculating: \[ b_{yx} \times b_{xy} = 1 \times 0.64 = 0.64 \] Calculating \( r^2 \): \[ r^2 = (0.8)^2 = 0.64 \] Both results are equal, verifying the relationship. # Final Answers 1. Regression coefficient of \( Y \) on \( X: \) \( b_{yx} = 1 \) 2. Regression coefficient of \( X \) on \( Y: \) \( b_{xy} = 0.64 \) 3. Verification: \( b_{yx} \times b_{xy} = r^2 \) is true.

# Given Information - Joint probability density function (PDF): \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # What We Have to Find (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Relevant Concepts - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** Calculate the probability by integrating the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] Calculating the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] \[ = \int_0^{\frac{1}{2}} 8x \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} dx = \int_0^{\frac{1}{2}} 8x \cdot \frac{(\frac{1}{2})^2}{2} \, dx \] \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Now compute the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **Determining Independence:** To check if \( X \) and \( Y \) are independent, calculate the marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Marginal of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] Now check if \( f(x, y) = f_X(x) f_Y(y) \): \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \neq 16xy \), \( X \) and \( Y \) are **not independent**. # Summary (a) \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) \( X \) and \( Y \) are **not independent**.

# Given Information - Joint probability density function (PDF): \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # What We Have to Find (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Definitions/Concepts Used - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To calculate the probability, we integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Calculating the inner integral:** \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Now substituting back: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Calculating the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, the probability is: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Determining Independence:** To check if \( X \) and \( Y \) are independent, we find the marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Marginal of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] Now check if \( f(x,y) = f_X(x) f_Y(y) \): \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \neq 16xy \), we conclude that \( X \) and \( Y \) are **not independent**. # Summary (a) \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # What Needs to be Found (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Definitions/Concepts Used - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Two random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculation of \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To determine this probability, we will integrate the joint PDF over the specified region: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** First, we compute the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral gives: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Now substituting this back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Calculating the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Checking Independence:** To verify whether \( X \) and \( Y \) are independent, we will compute the marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] Now, we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \neq f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Summary (a) The calculated probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) is \( \frac{1}{8} \). (b) The variables \( X \) and \( Y \) are **not independent**.

# Given Information - The probability density function (PDF) of the exponential distribution is defined as: \[ f(x; \lambda) = \lambda e^{-\lambda x}, \quad x > 0 \] # What We Have to Find We need to find the maximum likelihood estimator (MLE) of \( \lambda \). # Relevant Concepts - **Likelihood Function:** For a random sample \( x_1, x_2, \ldots, x_n \), the likelihood function \( L(\lambda) \) is given by: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} \] - **Log-Likelihood Function:** The log-likelihood function \( \ell(\lambda) \) is expressed as: \[ \ell(\lambda) = \log L(\lambda) = \sum_{i=1}^{n} \log f(x_i; \lambda) \] # Step-by-Step Solution **Step 1: Write the Likelihood Function** The likelihood function for the sample \( x_1, x_2, \ldots, x_n \) is: \[ L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \] **Step 2: Write the Log-Likelihood Function** Taking the natural logarithm: \[ \ell(\lambda) = \log L(\lambda) = \log(\lambda^n) + \log\left(e^{-\lambda \sum_{i=1}^{n} x_i}\right) = n \log(\lambda) - \lambda \sum_{i=1}^{n} x_i \] **Step 3: Differentiate the Log-Likelihood** Differentiate \( \ell(\lambda) \) with respect to \( \lambda \): \[ \frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i \] **Step 4: Set the Derivative to Zero** To find the MLE, set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] **Step 5: Solve for \( \lambda \)** Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \implies \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] Thus, the MLE of \( \lambda \) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] # Summary The maximum likelihood estimator (MLE) of \( \lambda \) for the exponential distribution is given by: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \]

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] # What We Have to Find (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Relevant Concepts - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Two random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculating \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we will integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** First, compute the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substitute back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Calculating the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Checking Independence:** To determine if \( X \) and \( Y \) are independent, we need to compute the marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] We now check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \neq f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Summary (a) \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] # Objectives We aim to determine: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be calculated as: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Compute \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To calculate this probability, we will integrate the joint PDF over the defined region: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral:** We start by calculating the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Calculating the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To check if \( X \) and \( Y \) are independent, we need to calculate the marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] Now, we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \neq f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) The variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is represented as: \[ f(x, y) = 8xy, \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] # Objectives We are tasked with finding: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Used - **Probability Calculation:** \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculate \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we need to evaluate the integral of the joint PDF over the specified range: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Calculating the Inner Integral:** First, we perform the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this back into the outer integral gives: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find that: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Determining Independence:** To assess whether \( X \) and \( Y \) are independent, we need to compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] Now, we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \neq f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) is \( \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We will determine: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be calculated using the following integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculate \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we evaluate the integral of the joint PDF over the specified region: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** First, we compute the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this result into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude that: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To check if \( X \) and \( Y \) are independent, we compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] Next, we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We are required to find: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be computed using the formula: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculation of \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we integrate the joint PDF over the specified range: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Evaluation:** We begin by calculating the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substituting this result back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine if \( X \) and \( Y \) are independent, we need to calculate their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now, we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \) is not equal to \( 16xy \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is specified as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We need to determine: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be determined through the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are considered independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculating \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To compute this probability, we will integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** First, we evaluate the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral yields: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this result into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To verify if \( X \) and \( Y \) are independent, we must compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now we check whether the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is given by: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We need to determine the following: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be calculated using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Two random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculating \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we integrate the joint PDF over the defined limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** First, we calculate the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine if \( X \) and \( Y \) are independent, we will calculate their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Next, we check whether the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \) (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - Correlation coefficient: \( r = 0.8 \) - Standard deviation of \( X \): \( \sigma_x = 4 \) - Standard deviation of \( Y \): \( \sigma_y = 5 \) # What We Need to Find 1. The regression coefficient of \( Y \) on \( X \). 2. The regression coefficient of \( X \) on \( Y \). 3. Verification of the relationship \( b_{yx} \times b_{xy} = r^2 \). # Concepts Used - **Regression Coefficient of \( Y \) on \( X \):** \[ b_{yx} = \frac{r \cdot \sigma_y}{\sigma_x} \] - **Regression Coefficient of \( X \) on \( Y \):** \[ b_{xy} = \frac{r \cdot \sigma_x}{\sigma_y} \] - **Verification Equation:** \[ b_{yx} \times b_{xy} = r^2 \] # Step-by-Step Solution **Finding the Regression Coefficient of \( Y \) on \( X \):** Using the formula for \( b_{yx} \): \[ b_{yx} = \frac{r \cdot \sigma_y}{\sigma_x} = \frac{0.8 \cdot 5}{4} = \frac{4}{4} = 1 \] **Finding the Regression Coefficient of \( X \) on \( Y \):** Using the formula for \( b_{xy} \): \[ b_{xy} = \frac{r \cdot \sigma_x}{\sigma_y} = \frac{0.8 \cdot 4}{5} = \frac{3.2}{5} = 0.64 \] **Verification of the Relationship \( b_{yx} \times b_{xy} = r^2 \):** Calculating the product: \[ b_{yx} \times b_{xy} = 1 \times 0.64 = 0.64 \] Calculating \( r^2 \): \[ r^2 = (0.8)^2 = 0.64 \] Both results match, confirming the relationship. # Final Answers 1. Regression coefficient of \( Y \) on \( X: \) \( b_{yx} = 1 \) 2. Regression coefficient of \( X \) on \( Y: \) \( b_{xy} = 0.64 \) 3. Verification: \( b_{yx} \times b_{xy} = r^2 \) holds true.

# Given Information - Correlation coefficient: \( r = 0.8 \) - Standard deviation of \( X \): \( \sigma_x = 4 \) - Standard deviation of \( Y \): \( \sigma_y = 5 \) # Objectives We are to determine: 1. The regression coefficient of \( Y \) on \( X \). 2. The regression coefficient of \( X \) on \( Y \). 3. Verification of the relationship \( b_{yx} \times b_{xy} = r^2 \). # Relevant Concepts - **Regression Coefficient of \( Y \) on \( X \):** \[ b_{yx} = \frac{r \cdot \sigma_y}{\sigma_x} \] - **Regression Coefficient of \( X \) on \( Y \):** \[ b_{xy} = \frac{r \cdot \sigma_x}{\sigma_y} \] - **Verification Equation:** \[ b_{yx} \times b_{xy} = r^2 \] # Step-by-Step Solution **Step 1: Calculate the Regression Coefficient of \( Y \) on \( X \):** Using the formula for \( b_{yx} \): \[ b_{yx} = \frac{r \cdot \sigma_y}{\sigma_x} = \frac{0.8 \cdot 5}{4} = \frac{4}{4} = 1 \] **Step 2: Calculate the Regression Coefficient of \( X \) on \( Y \):** Using the formula for \( b_{xy} \): \[ b_{xy} = \frac{r \cdot \sigma_x}{\sigma_y} = \frac{0.8 \cdot 4}{5} = \frac{3.2}{5} = 0.64 \] **Step 3: Verify the Relationship \( b_{yx} \times b_{xy} = r^2 \):** Calculating: \[ b_{yx} \times b_{xy} = 1 \times 0.64 = 0.64 \] Now calculating \( r^2 \): \[ r^2 = (0.8)^2 = 0.64 \] The calculated products match, confirming the relationship holds true. # Final Answers 1. Regression coefficient of \( Y \) on \( X: \) \( b_{yx} = 1 \) 2. Regression coefficient of \( X \) on \( Y: \) \( b_{xy} = 0.64 \) 3. Verification: The relationship \( b_{yx} \times b_{xy} = r^2 \) is satisfied.

# Given Information The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives The goal is to calculate: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Used The probability can be calculated using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] To determine independence, random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Calculation of Probability To find \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \), we will integrate the joint PDF over the specified area: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] First, compute the inner integral with respect to \( y \): \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Now, evaluate the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, the result for the probability is: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] # Independence Check To check if \( X \) and \( Y \) are independent, we need to compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). For the marginal PDF of \( X \): \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 8x \cdot \frac{1}{2} = 4x \] For the marginal PDF of \( Y \): \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 8y \cdot \frac{1}{2} = 4y \] We now need to check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \) does not equal \( 16xy \), it follows that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) is \( \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is expressed as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We are tasked with calculating: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Determining whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be computed using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Calculation **(a) Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we need to integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] We start with the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral yields: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substituting back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Next, we compute the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude that: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine if \( X \) and \( Y \) are independent, we must calculate their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Next, we examine whether the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We are required to compute: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be calculated using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Detailed Calculation **(a) Calculate \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To calculate this probability, we integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] First, compute the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this back into the outer integral gives: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Now, evaluate the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, the calculated probability is: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Checking Independence:** To verify whether \( X \) and \( Y \) are independent, we compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Next, we need to check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is formulated as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Goals We aim to determine: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Assess whether \( X \) and \( Y \) are independent. # Concepts Employed - **Probability Calculation:** The probability can be derived using the following integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Detailed Calculation **(a) Probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To calculate this probability, we will integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Evaluating the Inner Integral:** We begin with the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substituting this result back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine whether \( X \) and \( Y \) are independent, we need to compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since the expressions do not match, we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is formulated as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Goals The goal is to determine: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be expressed as: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Detailed Calculation **(a) Calculate \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we will integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Evaluating the Inner Integral:** We start by computing the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral yields: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substituting this result back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine whether \( X \) and \( Y \) are independent, we must compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Next, we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since the expressions do not match, we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad \text{for } 0 < x < 1, \; 0 < y < 1 \] # Objectives We need to determine: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be calculated using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Two random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Calculation **(a) Finding \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To calculate this probability, we integrate the joint PDF over the specified region: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** We start by evaluating the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this result back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Determining Independence:** To assess whether \( X \) and \( Y \) are independent, we will compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now, we need to check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since these two expressions are not equal, we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The calculated probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives The goal is to find: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** The probability can be evaluated using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are said to be independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Calculation **(a) Compute \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] First, compute the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Now, evaluating the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine whether \( X \) and \( Y \) are independent, we will compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now we need to check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \) does not equal \( 16xy \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is specified as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We are tasked with finding: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Utilized - **Probability Calculation:** To find the probability, we use the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Two random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Calculation **(a) Calculate \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To compute this probability, we will integrate the joint PDF over the defined limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] We begin by evaluating the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral yields: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting back into the outer integral results in: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Next, we evaluate the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine if \( X \) and \( Y \) are independent, we will compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now we check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is described as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives The task is to find: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Used - **Probability Calculation:** The probability is computed using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Detailed Calculation **(a) Compute \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we will perform double integration of the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] First, we evaluate the inner integral with respect to \( y \): \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substituting this result back into the outer integral yields: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Now, we calculate the outer integral: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine if \( X \) and \( Y \) are independent, we need to compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now, we check whether the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) does not equal \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) is \( \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is formulated as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives The tasks at hand are to find: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Determine whether the random variables \( X \) and \( Y \) are independent. # Concepts Employed - **Probability Calculation:** The probability can be computed using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Two random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Calculation **(a) Calculation of \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we need to evaluate the integral of the joint PDF over the defined limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] First, we will calculate the inner integral with respect to \( y \): \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] The inner integral evaluates to: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this back into the outer integral, we have: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Now, computing the outer integral yields: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we find the probability: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To check whether \( X \) and \( Y \) are independent, we will compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now, we will check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( 8xy \) does not equal \( 16xy \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives We need to calculate: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Whether the random variables \( X \) and \( Y \) are independent. # Concepts Used - **Probability Calculation:** The probability can be determined using the integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Step-by-Step Solution **(a) Calculation of \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we will integrate the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Calculation:** First, we focus on the inner integral: \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Now substituting this value back into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] Calculating the outer integral gives: \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, we conclude that: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Checking Independence:** To determine if \( X \) and \( Y \) are independent, we need to compute their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Next, we examine whether the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since the two expressions do not match, we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - The joint probability density function (PDF) is defined as: \[ f(x, y) = 8xy, \quad 0 < x < 1, \; 0 < y < 1 \] # Objectives The task involves computing: (a) The probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \) (b) Evaluating whether the random variables \( X \) and \( Y \) are independent. # Concepts Used - **Probability Calculation:** The probability can be calculated using the double integral: \[ P(X < a, Y < b) = \int_0^a \int_0^b f(x, y) \, dy \, dx \] - **Independence Criterion:** Random variables \( X \) and \( Y \) are independent if: \[ f(x, y) = f_X(x) f_Y(y) \] # Detailed Calculation **(a) Calculate \( P(X < \frac{1}{2}, Y < \frac{1}{2}) \):** To find this probability, we will perform the integration of the joint PDF over the specified limits: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} 8xy \, dy \, dx \] **Inner Integral Evaluation:** We start by calculating the inner integral with respect to \( y \): \[ = \int_0^{\frac{1}{2}} 8x \left( \int_0^{\frac{1}{2}} y \, dy \right) dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{2}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8} \] Substituting this into the outer integral: \[ = \int_0^{\frac{1}{2}} 8x \cdot \frac{1}{8} \, dx = \int_0^{\frac{1}{2}} x \, dx \] **Calculating the Outer Integral:** \[ = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{(\frac{1}{2})^2}{2} = \frac{1/4}{2} = \frac{1}{8} \] Thus, the probability is: \[ P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \] **(b) Assessing Independence:** To determine if \( X \) and \( Y \) are independent, we need to calculate their marginal PDFs \( f_X(x) \) and \( f_Y(y) \). **Marginal PDF of \( X \):** \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 8xy \, dy = 8x \left[ \frac{y^2}{2} \right]_0^1 = 4x \] **Marginal PDF of \( Y \):** \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 8xy \, dx = 8y \left[ \frac{x^2}{2} \right]_0^1 = 4y \] Now we will check if the joint PDF can be expressed as the product of the marginal PDFs: \[ f(x, y) = 8xy \quad \text{and} \quad f_X(x) f_Y(y) = (4x)(4y) = 16xy \] Since \( f(x, y) \) is not equal to \( f_X(x) f_Y(y) \), we conclude that \( X \) and \( Y \) are **not independent**. # Final Summary (a) The calculated probability \( P(X < \frac{1}{2}, Y < \frac{1}{2}) = \frac{1}{8} \). (b) The random variables \( X \) and \( Y \) are **not independent**.

# Given Information - A discrete random variable \( X \) with the probability mass function (PMF): \[ P(X = k) = \frac{k}{15}, \quad k = 1, 2, 3, 4, 5 \] # What We Have to Find (a) The expected value \( E(X) \) (b) The variance \( \operatorname{Var}(X) \) # Definitions/Concepts Used - **Expected Value Formula:** \[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) \] - **Variance Formula:** \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] - **Calculating \( E(X^2) \):** \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) \] # Step-by-Step Solution **Calculating \( E(X) \):** Using the PMF, we find the expected value: \[ E(X) = \sum_{k=1}^{5} k \cdot P(X = k) = \sum_{k=1}^{5} k \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^2 \] Calculating \( \sum_{k=1}^{5} k^2 \): \[ \sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \] Thus, we have: \[ E(X) = \frac{1}{15} \cdot 55 = \frac{55}{15} = \frac{11}{3} \] **Calculating \( E(X^2) \):** Now we find \( E(X^2) \): \[ E(X^2) = \sum_{k=1}^{5} k^2 \cdot P(X = k) = \sum_{k=1}^{5} k^2 \cdot \frac{k}{15} = \frac{1}{15} \sum_{k=1}^{5} k^3 \] Calculating \( \sum_{k=1}^{5} k^3 \): \[ \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225 \] Thus: \[ E(X^2) = \frac{1}{15} \cdot 225 = 15 \] **Calculating Variance \( \operatorname{Var}(X) \):** Now we compute the variance: \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 15 - \left( \frac{11}{3} \right)^2 \] Calculating \( \left( \frac{11}{3} \right)^2 \): \[ \left( \frac{11}{3} \right)^2 = \frac{121}{9} \] Now substituting back: \[ \operatorname{Var}(X) = 15 - \frac{121}{9} \] Finding a common denominator (9): \[ 15 = \frac{135}{9} \] Thus: \[ \operatorname{Var}(X) = \frac{135}{9} - \frac{121}{9} = \frac{14}{9} \] # Final Answers (a) The expected value \( E(X) = \frac{11}{3} \) (b) The variance \( \operatorname{Var}(X) = \frac{14}{9} \)