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Give step-by-step solution with explanation and final answer: Of course! I can walk you through this problem on order statistics. It's a great exercise that combines concepts of probability distributions, calculus, and expectations. Let's solve it step-by-step. ### Given Information: * We have 4 iid (independent and identically distributed) random variables: $X_1, X_2, X_3, X_4$. So, the sample size is $n=4$. * The probability density function (PDF) is $f_X(x) = 2x$ for $0 < x < 1$. *** ### (a) What is the CDF of $X_1$? **Concept:** The Cumulative Distribution Function (CDF), denoted as $F_X(x)$, gives the probability that the random variable $X$ is less than or equal to a certain value $x$. We find it by integrating the PDF from the lower bound of its support up to $x$. **Calculation:** The CDF is $F_X(x) = P(X \le x) = \int_{-\infty}^{x} f_X(t) \,dt$. Since our PDF is defined for $0 < x < 1$, we'll integrate from 0 to $x$: $F_X(x) = \int_{0}^{x} 2t \,dt = [t^2]_0^x = x^2 - 0^2 = x^2$ For a complete definition, the CDF is: $F_X(x) = \begin{cases} 0 & \text{for } x \le 0 \\ x^2 & \text{for } 0 < x < 1 \\ 1 & \text{for } x \ge 1 \end{cases}$ **Answer (a):** The CDF of $X_1$ is $F_X(x) = x^2$ for $0 < x < 1$. *** ### (b) What is the expected value of the second order statistics, $E(X_{(2)})$? **Concept:** To find the expected value of an order statistic, we first need to find its PDF. The general formula for the PDF of the $k$-th order statistic, $X_{(k)}$, from a sample of size $n$ is: $f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F_X(x)]^{k-1} [1 - F_X(x)]^{n-k} f_X(x)$ For our problem, $n=4$ and we want the second order statistic, so $k=2$. **Step 1: Find the PDF of $X_{(2)}$, $f_{X_{(2)}}(x)$.** Using the formula with $n=4$ and $k=2$: $f_{X_{(2)}}(x) = \frac{4!}{(2-1)!(4-2)!} [F_X(x)]^{2-1} [1 - F_X(x)]^{4-2} f_X(x)$ $f_{X_{(2)}}(x) = \frac{24}{1!2!} [x^2]^1 [1 - x^2]^2 (2x)$ $f_{X_{(2)}}(x) = 12 (x^2) (1 - 2x^2 + x^4) (2x)$ $f_{X_{(2)}}(x) = 24x^3 (1 - 2x^2 + x^4)$ $f_{X_{(2)}}(x) = 24x^3 - 48x^5 + 24x^7$, for $0 < x < 1$. **Step 2: Calculate the expected value, $E(X_{(2)})$.** $E(X_{(2)}) = \int_{0}^{1} x \cdot f_{X_{(2)}}(x) \,dx$ $E(X_{(2)}) = \int_{0}^{1} x (24x^3 - 48x^5 + 24x^7) \,dx$ $E(X_{(2)}) = \int_{0}^{1} (24x^4 - 48x^6 + 24x^8) \,dx$ $E(X_{(2)}) = [24\frac{x^5}{5} - 48\frac{x^7}{7} + 24\frac{x^9}{9}]_0^1$ $E(X_{(2)}) = (\frac{24}{5} - \frac{48}{7} + \frac{24}{9}) - 0$ $E(X_{(2)}) = \frac{24}{5} - \frac{48}{7} + \frac{8}{3}$ To add these fractions, we find a common denominator, which is $5 \times 7 \times 3 = 105$. $E(X_{(2)}) = \frac{24 \times 21}{105} - \frac{48 \times 15}{105} + \frac{8 \times 35}{105}$ $E(X_{(2)}) = \frac{504 - 720 + 280}{105} = \frac{64}{105}$ **Answer (b):** $E(X_{(2)}) = \frac{64}{105}$ *** ### (c) What is the variance of the second order statistics, $Var(X_{(2)})$? **Concept:** The variance is calculated using the formula $Var(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2$. We already have $E(X_{(2)})$, so we just need to find $E(X_{(2)}^2)$. **Step 1: Calculate $E(X_{(2)}^2)$.** $E(X_{(2)}^2) = \int_{0}^{1} x^2 \cdot f_{X_{(2)}}(x) \,dx$ $E(X_{(2)}^2) = \int_{0}^{1} x^2 (24x^3 - 48x^5 + 24x^7) \,dx$ $E(X_{(2)}^2) = \int_{0}^{1} (24x^5 - 48x^7 + 24x^9) \,dx$ $E(X_{(2)}^2) = [24\frac{x^6}{6} - 48\frac{x^8}{8} + 24\frac{x^{10}}{10}]_0^1$ $E(X_{(2)}^2) = [4x^6 - 6x^8 + \frac{12}{5}x^{10}]_0^1$ $E(X_{(2)}^2) = (4 - 6 + \frac{12}{5}) - 0 = -2 + \frac{12}{5} = \frac{-10+12}{5} = \frac{2}{5}$ **Step 2: Calculate the variance.** $Var(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2$ $Var(X_{(2)}) = \frac{2}{5} - (\frac{64}{105})^2$ $Var(X_{(2)}) = \frac{2}{5} - \frac{4096}{11025}$ $Var(X_{(2)}) = \frac{2 \times 2205}{11025} - \frac{4096}{11025} = \frac{4410 - 4096}{11025} = \frac{314}{11025}$ **Answer (c):** $Var(X_{(2)}) = \frac{314}{11025}$ *** ### (d) What is the covariance between the second and third order statistics, $Cov(X_{(2)}, X_{(3)})$? **Concept:** The formula for covariance is $Cov(X_{(2)}, X_{(3)}) = E(X_{(2)}X_{(3)}) - E(X_{(2)})E(X_{(3)})$. This is the most involved part. We need to calculate $E(X_{(3)})$ and then $E(X_{(2)}X_{(3)})$. **Step 1: Calculate $E(X_{(3)})$.** First, find the PDF of $X_{(3)}$ using $n=4, k=3$. $f_{X_{(3)}}(x) = \frac{4!}{(3-1)!(4-3)!} [F_X(x)]^{3-1} [1-F_X(x)]^{4-3} f_X(x)$ $f_{X_{(3)}}(x) = \frac{24}{2!1!} [x^2]^2 [1-x^2]^1 (2x)$ $f_{X_{(3)}}(x) = 12 (x^4) (1-x^2) (2x) = 24x^5(1-x^2) = 24x^5 - 24x^7$ Now find the expectation: $E(X_{(3)}) = \int_0^1 x \cdot f_{X_{(3)}}(x) dx = \int_0^1 x(24x^5 - 24x^7) dx$ $E(X_{(3)}) = \int_0^1 (24x^6 - 24x^8) dx = [24\frac{x^7}{7} - 24\frac{x^9}{9}]_0^1$ $E(X_{(3)}) = \frac{24}{7} - \frac{24}{9} = \frac{24}{7} - \frac{8}{3} = \frac{72-56}{21} = \frac{16}{21}$ **Step 2: Calculate $E(X_{(2)}X_{(3)})$.** We need the joint PDF of $X_{(2)}$ and $X_{(3)}$. For $j

Question:

Give step-by-step solution with explanation and final answer: Of course! I can walk you through this problem on order statistics. It's a great exercise that combines concepts of probability distributions, calculus, and expectations. Let's solve it step-by-step. ### Given Information: * We have 4 iid (independent and identically distributed) random variables: $X_1, X_2, X_3, X_4$. So, the sample size is $n=4$. * The probability density function (PDF) is $f_X(x) = 2x$ for $0 < x < 1$. *** ### (a) What is the CDF of $X_1$? **Concept:** The Cumulative Distribution Function (CDF), denoted as $F_X(x)$, gives the probability that the random variable $X$ is less than or equal to a certain value $x$. We find it by integrating the PDF from the lower bound of its support up to $x$. **Calculation:** The CDF is $F_X(x) = P(X \le x) = \int_{-\infty}^{x} f_X(t) \,dt$. Since our PDF is defined for $0 < x < 1$, we'll integrate from 0 to $x$: $F_X(x) = \int_{0}^{x} 2t \,dt = [t^2]_0^x = x^2 - 0^2 = x^2$ For a complete definition, the CDF is: $F_X(x) = \begin{cases} 0 & \text{for } x \le 0 \\ x^2 & \text{for } 0 < x < 1 \\ 1 & \text{for } x \ge 1 \end{cases}$ **Answer (a):** The CDF of $X_1$ is $F_X(x) = x^2$ for $0 < x < 1$. *** ### (b) What is the expected value of the second order statistics, $E(X_{(2)})$? **Concept:** To find the expected value of an order statistic, we first need to find its PDF. The general formula for the PDF of the $k$-th order statistic, $X_{(k)}$, from a sample of size $n$ is: $f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F_X(x)]^{k-1} [1 - F_X(x)]^{n-k} f_X(x)$ For our problem, $n=4$ and we want the second order statistic, so $k=2$. **Step 1: Find the PDF of $X_{(2)}$, $f_{X_{(2)}}(x)$.** Using the formula with $n=4$ and $k=2$: $f_{X_{(2)}}(x) = \frac{4!}{(2-1)!(4-2)!} [F_X(x)]^{2-1} [1 - F_X(x)]^{4-2} f_X(x)$ $f_{X_{(2)}}(x) = \frac{24}{1!2!} [x^2]^1 [1 - x^2]^2 (2x)$ $f_{X_{(2)}}(x) = 12 (x^2) (1 - 2x^2 + x^4) (2x)$ $f_{X_{(2)}}(x) = 24x^3 (1 - 2x^2 + x^4)$ $f_{X_{(2)}}(x) = 24x^3 - 48x^5 + 24x^7$, for $0 < x < 1$. **Step 2: Calculate the expected value, $E(X_{(2)})$.** $E(X_{(2)}) = \int_{0}^{1} x \cdot f_{X_{(2)}}(x) \,dx$ $E(X_{(2)}) = \int_{0}^{1} x (24x^3 - 48x^5 + 24x^7) \,dx$ $E(X_{(2)}) = \int_{0}^{1} (24x^4 - 48x^6 + 24x^8) \,dx$ $E(X_{(2)}) = [24\frac{x^5}{5} - 48\frac{x^7}{7} + 24\frac{x^9}{9}]_0^1$ $E(X_{(2)}) = (\frac{24}{5} - \frac{48}{7} + \frac{24}{9}) - 0$ $E(X_{(2)}) = \frac{24}{5} - \frac{48}{7} + \frac{8}{3}$ To add these fractions, we find a common denominator, which is $5 \times 7 \times 3 = 105$. $E(X_{(2)}) = \frac{24 \times 21}{105} - \frac{48 \times 15}{105} + \frac{8 \times 35}{105}$ $E(X_{(2)}) = \frac{504 - 720 + 280}{105} = \frac{64}{105}$ **Answer (b):** $E(X_{(2)}) = \frac{64}{105}$ *** ### (c) What is the variance of the second order statistics, $Var(X_{(2)})$? **Concept:** The variance is calculated using the formula $Var(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2$. We already have $E(X_{(2)})$, so we just need to find $E(X_{(2)}^2)$. **Step 1: Calculate $E(X_{(2)}^2)$.** $E(X_{(2)}^2) = \int_{0}^{1} x^2 \cdot f_{X_{(2)}}(x) \,dx$ $E(X_{(2)}^2) = \int_{0}^{1} x^2 (24x^3 - 48x^5 + 24x^7) \,dx$ $E(X_{(2)}^2) = \int_{0}^{1} (24x^5 - 48x^7 + 24x^9) \,dx$ $E(X_{(2)}^2) = [24\frac{x^6}{6} - 48\frac{x^8}{8} + 24\frac{x^{10}}{10}]_0^1$ $E(X_{(2)}^2) = [4x^6 - 6x^8 + \frac{12}{5}x^{10}]_0^1$ $E(X_{(2)}^2) = (4 - 6 + \frac{12}{5}) - 0 = -2 + \frac{12}{5} = \frac{-10+12}{5} = \frac{2}{5}$ **Step 2: Calculate the variance.** $Var(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2$ $Var(X_{(2)}) = \frac{2}{5} - (\frac{64}{105})^2$ $Var(X_{(2)}) = \frac{2}{5} - \frac{4096}{11025}$ $Var(X_{(2)}) = \frac{2 \times 2205}{11025} - \frac{4096}{11025} = \frac{4410 - 4096}{11025} = \frac{314}{11025}$ **Answer (c):** $Var(X_{(2)}) = \frac{314}{11025}$ *** ### (d) What is the covariance between the second and third order statistics, $Cov(X_{(2)}, X_{(3)})$? **Concept:** The formula for covariance is $Cov(X_{(2)}, X_{(3)}) = E(X_{(2)}X_{(3)}) - E(X_{(2)})E(X_{(3)})$. This is the most involved part. We need to calculate $E(X_{(3)})$ and then $E(X_{(2)}X_{(3)})$. **Step 1: Calculate $E(X_{(3)})$.** First, find the PDF of $X_{(3)}$ using $n=4, k=3$. $f_{X_{(3)}}(x) = \frac{4!}{(3-1)!(4-3)!} [F_X(x)]^{3-1} [1-F_X(x)]^{4-3} f_X(x)$ $f_{X_{(3)}}(x) = \frac{24}{2!1!} [x^2]^2 [1-x^2]^1 (2x)$ $f_{X_{(3)}}(x) = 12 (x^4) (1-x^2) (2x) = 24x^5(1-x^2) = 24x^5 - 24x^7$ Now find the expectation: $E(X_{(3)}) = \int_0^1 x \cdot f_{X_{(3)}}(x) dx = \int_0^1 x(24x^5 - 24x^7) dx$ $E(X_{(3)}) = \int_0^1 (24x^6 - 24x^8) dx = [24\frac{x^7}{7} - 24\frac{x^9}{9}]_0^1$ $E(X_{(3)}) = \frac{24}{7} - \frac{24}{9} = \frac{24}{7} - \frac{8}{3} = \frac{72-56}{21} = \frac{16}{21}$ **Step 2: Calculate $E(X_{(2)}X_{(3)})$.** We need the joint PDF of $X_{(2)}$ and $X_{(3)}$. For $j

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Asked by: Umesh
Created at: 2025-08-07 23:28:01

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Certainly! Here is your solution, rewritten with equations in **rendered LaTeX** (MathJax). The text remains otherwise unchanged. --- ### Given Information: - We have 4 iid (independent and identically distributed) random variables: \( X_1, X_2, X_3, X_4 \). So, the sample size is \( n=4 \). - The probability density function (PDF) is \( f_X(x) = 2x \) for \( 0 < x < 1 \). --- ### (a) What is the CDF of \( X_1 \)? **Concept:** The Cumulative Distribution Function (CDF), denoted as \( F_X(x) \), gives the probability that the random variable \( X \) is less than or equal to a certain value \( x \). We find it by integrating the PDF from the lower bound of its support up to \( x \). **Calculation:** The CDF is \( F_X(x) = P(X \le x) = \int_{-\infty}^{x} f_X(t) \,dt \). Since our PDF is defined for \( 0 < x < 1 \), we'll integrate from 0 to \( x \): \[ F_X(x) = \int_{0}^{x} 2t \,dt = [t^2]_0^x = x^2 - 0^2 = x^2 \] For a complete definition, the CDF is: \[ F_X(x) = \begin{cases} 0 & \text{for } x \le 0 \\ x^2 & \text{for } 0 < x < 1 \\ 1 & \text{for } x \ge 1 \end{cases} \] **Answer (a):** \[ \boxed{ F_X(x) = x^2 \text{ for } 0 < x < 1 } \] --- ### (b) What is the expected value of the second order statistic, \( E(X_{(2)}) \)? **Concept:** To find the expected value of an order statistic, we first need to find its PDF. The general formula for the PDF of the \( k \)-th order statistic, \( X_{(k)} \), from a sample of size \( n \) is: \[ f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F_X(x)]^{k-1} [1 - F_X(x)]^{n-k} f_X(x) \] For our problem, \( n=4 \) and we want the second order statistic, so \( k=2 \). **Step 1: Find the PDF of \( X_{(2)} \), \( f_{X_{(2)}}(x) \).** Using the formula with \( n=4 \) and \( k=2 \): \[ f_{X_{(2)}}(x) = \frac{4!}{(2-1)!(4-2)!} [x^2]^1 [1 - x^2]^2 (2x) \] \[ = \frac{24}{1!2!} x^2 (1 - x^2)^2 (2x) \] \[ = 12 x^2 (1 - 2x^2 + x^4) (2x) \] \[ = 24x^3 (1 - 2x^2 + x^4) \] \[ = 24x^3 - 48x^5 + 24x^7, \quad 0 < x < 1 \] **Step 2: Calculate the expected value, \( E(X_{(2)}) \).** \[ E(X_{(2)}) = \int_{0}^{1} x \cdot f_{X_{(2)}}(x) \,dx \] \[ = \int_{0}^{1} x (24x^3 - 48x^5 + 24x^7) \,dx \] \[ = \int_{0}^{1} (24x^4 - 48x^6 + 24x^8) \,dx \] \[ = \left[24\frac{x^5}{5} - 48\frac{x^7}{7} + 24\frac{x^9}{9}\right]_0^1 \] \[ = \frac{24}{5} - \frac{48}{7} + \frac{24}{9} \] To add these fractions, we use common denominator \( 5 \times 7 \times 3 = 105 \): \[ E(X_{(2)}) = \frac{24 \times 21}{105} - \frac{48 \times 15}{105} + \frac{8 \times 35}{105} \] \[ = \frac{504 - 720 + 280}{105} = \frac{64}{105} \] **Answer (b):** \[ \boxed{ E(X_{(2)}) = \frac{64}{105} } \] --- ### (c) What is the variance of the second order statistic, \( \text{Var}(X_{(2)}) \)? **Concept:** The variance is calculated using the formula \( \text{Var}(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2 \). We already have \( E(X_{(2)}) \), so we just need to find \( E(X_{(2)}^2) \). **Step 1: Calculate \( E(X_{(2)}^2) \).** \[ E(X_{(2)}^2) = \int_{0}^{1} x^2 \cdot f_{X_{(2)}}(x) \,dx \] \[ = \int_{0}^{1} x^2 (24x^3 - 48x^5 + 24x^7) \,dx \] \[ = \int_{0}^{1} (24x^5 - 48x^7 + 24x^9) \,dx \] \[ = \left[24\frac{x^6}{6} - 48\frac{x^8}{8} + 24\frac{x^{10}}{10}\right]_0^1 \] \[ = \left[4x^6 - 6x^8 + \frac{12}{5}x^{10}\right]_0^1 \] \[ = 4 - 6 + \frac{12}{5} = -2 + \frac{12}{5} = \frac{-10+12}{5} = \frac{2}{5} \] **Step 2: Calculate the variance.** \[ \text{Var}(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2 \] \[ = \frac{2}{5} - \left(\frac{64}{105}\right)^2 \] \[ = \frac{2}{5} - \frac{4096}{11025} \] \[ = \frac{2 \times 2205}{11025} - \frac{4096}{11025} = \frac{4410 - 4096}{11025} = \frac{314}{11025} \] **Answer (c):** \[ \boxed{ \text{Var}(X_{(2)}) = \frac{314}{11025} } \] --- ### (d) What is the covariance between the second and third order statistics, \( \text{Cov}(X_{(2)}, X_{(3)}) \)? **Concept:** The formula for covariance is \( \text{Cov}(X_{(2)}, X_{(3)}) = E(X_{(2)}X_{(3)}) - E(X_{(2)})E(X_{(3)}) \). This is the most involved part. We need to calculate \( E(X_{(3)}) \) and then \( E(X_{(2)}X_{(3)}) \). **Step 1: Calculate \( E(X_{(3)}) \).** First, find the PDF of \( X_{(3)} \) using \( n=4, k=3 \): \[ f_{X_{(3)}}(x) = \frac{4!}{(3-1)!(4-3)!} [F_X(x)]^{3-1} [1-F_X(x)]^{4-3} f_X(x) \] \[ = \frac{24}{2!1!} [x^2]^2 [1-x^2]^1 (2x) \] \[ = 12 (x^4) (1-x^2) (2x) = 24x^5(1-x^2) = 24x^5 - 24x^7 \] Now find the expectation: \[ E(X_{(3)}) = \int_0^1 x \cdot f_{X_{(3)}}(x) dx = \int_0^1 x(24x^5 - 24x^7) dx \] \[ = \int_0^1 (24x^6 - 24x^8) dx = [24\frac{x^7}{7} - 24\frac{x^9}{9}]_0^1 \] \[ = \frac{24}{7} - \frac{24}{9} = \frac{24}{7} - \frac{8}{3} \] \[ = \frac{72-56}{21} = \frac{16}{21} \] **Step 2: Calculate \( E(X_{(2)}X_{(3)}) \).** We need the joint PDF of \( X_{(2)} \) and \( X_{(3)} \). For \( j<k \), the formula is: \[ f_{X_{(j)},X_{(k)}}(x, y) = \frac{n!}{(j-1)!(k-j-1)!(n-k)!} [F(x)]^{j-1}[F(y)-F(x)]^{k-j-1}[1-F(y)]^{n-k} f(x)f(y) \] for \( x<y \). With \( n=4, j=2, k=3 \): \[ f_{X_{(2)},X_{(3)}}(x, y) = \frac{4!}{(1)!(0)!(1)!} [x^2]^1 [y^2-x^2]^0 [1-y^2]^1 (2x)(2y) \] \[ = 24 (x^2) (1) (1-y^2) (4xy) = 96x^3y(1-y^2), \quad 0<x<y<1 \] Now, calculate the expectation with a double integral: \[ E(X_{(2)}X_{(3)}) = \int_0^1 \int_x^1 xy \cdot f_{X_{(2)},X_{(3)}}(x, y) \,dy \,dx \] \[ = \int_0^1 \int_x^1 xy \cdot [96x^3y(1-y^2)] \,dy \,dx \] \[ = 96 \int_0^1 x^4 \left[ \int_x^1 y^2(1-y^2) \,dy \right] \,dx \] \[ = 96 \int_0^1 x^4 \left[ \int_x^1 (y^2-y^4) \,dy \right] \,dx \] Inner integral (w.r.t. \( y \)): \[ \int_x^1 (y^2-y^4) \,dy = \left[\frac{y^3}{3} - \frac{y^5}{5}\right]_x^1 = \left(\frac{1}{3} - \frac{1}{5}\right) - \left(\frac{x^3}{3} - \frac{x^5}{5}\right) = \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} \] Outer integral (w.r.t. \( x \)): \[ E(X_{(2)}X_{(3)}) = 96 \int_0^1 x^4 \left( \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} \right) \,dx \] \[ = 96 \int_0^1 \left( \frac{2}{15}x^4 - \frac{1}{3}x^7 + \frac{1}{5}x^9 \right) dx \] \[ = 96\left[ \frac{2}{15}\frac{x^5}{5} - \frac{1}{3}\frac{x^8}{8} + \frac{1}{5}\frac{x^{10}}{10} \right]_0^1 \] \[ = 96\left[ \frac{2}{75} - \frac{1}{24} + \frac{1}{50} \right] \] The common denominator for 75, 24, and 50 is 600. \[ = 96 \left[ \frac{2 \times 8}{600} - \frac{1 \times 25}{600}

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