Question:

### a) Find the value of the constant c To find the constant $c$, we use the property that the total probability over the entire support must be 1. The integral of the joint probability density function (PDF) over its region must equal 1. $$ \int \int_R f(y_1, y_2) \, dy_1 \, dy_2 = 1 $$ The region of support $R$ is defined by $0 \le y_1 \le 2$, $0 \le y_2 \le 1$, and $2y_2 \le y_1$. This region is a triangle with vertices at (0,0), (2,0), and (2,1). The area of this triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$. Since the PDF is a constant $c$ over this region, the integral is $c \times \text{Area}$. $$ c \times \text{Area} = 1 \implies c \times 1 = 1 \implies c = 1 $$ So, the value of the constant **c is 1**. --- ### b) Find $P(Y_1 \ge 3Y_2)$ We need to integrate the PDF, $f(y_1, y_2) = 1$, over the region where $y_1 \ge 3y_2$ within the original support. This new region is defined by the intersection of $2y_2 \le y_1 \le 2$ and $y_1 \ge 3y_2$, which simplifies to $3y_2 \le y_1 \le 2$. The integration is over a smaller triangular region with vertices (0,0), (2,0), and (2, 2/3). The probability is the area of this new region (since the PDF is 1). $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times \frac{2}{3} = \frac{2}{3} $$ Alternatively, we can calculate the integral: $$ P(Y_1 \ge 3Y_2) = \int_0^{2/3} \int_{3y_2}^2 1 \, dy_1 \, dy_2 = \int_0^{2/3} (2-3y_2) \, dy_2 = [2y_2 - \frac{3}{2}y_2^2]_0^{2/3} = \frac{4}{3} - \frac{3}{2}\left(\frac{4}{9}\right) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} $$ So, **$P(Y_1 \ge 3Y_2) = \frac{2}{3}$**. --- ### c) Given that $Y_2=0.5$, calculate the probability that $Y_1$ exceeds 1.5 This is a conditional probability, $P(Y_1 > 1.5 \mid Y_2 = 0.5)$. We first need the conditional PDF $f(y_1 \mid y_2) = \frac{f(y_1, y_2)}{f_2(y_2)}$. 1. **Find the marginal PDF of $Y_2$**: $$ f_2(y_2) = \int_{2y_2}^2 1 \, dy_1 = [y_1]_{2y_2}^2 = 2 - 2y_2, \quad \text{for } 0 \le y_2 \le 1 $$ 2. **Find the conditional PDF of $Y_1$ given $Y_2=y_2$**: $$ f(y_1 \mid y_2) = \frac{1}{2-2y_2}, \quad \text{for } 2y_2 \le y_1 \le 2 $$ 3. **Set $y_2 = 0.5$**: The conditional PDF is $f(y_1 \mid 0.5) = \frac{1}{2-2(0.5)} = 1$, for the interval $2(0.5) \le y_1 \le 2$, which is $1 \le y_1 \le 2$. This means $Y_1$ given $Y_2=0.5$ is uniformly distributed on \[1, 2]. 4. **Calculate the probability**: $$ P(Y_1 > 1.5 \mid Y_2 = 0.5) = \int_{1.5}^2 f(y_1 \mid 0.5) \, dy_1 = \int_{1.5}^2 1 \, dy_1 = [y_1]_{1.5}^2 = 2 - 1.5 = 0.5 $$ The probability that **$Y_1$ exceeds 1.5 given $Y_2=0.5$ is 0.5**. --- ### d) Are $Y_1$ and $Y_2$ independent? No, **$Y_1$ and $Y_2$ are not independent**. For two random variables to be independent, the support of their joint distribution must be a rectangle, and the joint PDF must be a product of their marginal PDFs, i.e., $f(y_1, y_2) = f_1(y_1)f_2(y_2)$. The support region $2y_2 \le y_1 \le 2$ is a triangle, not a rectangle. The range of $y_1$ depends on the value of $y_2$, which immediately implies dependence. --- ### e) Find $E(Y_1)$, $E(Y_2)$, $V(Y_1)$, and $V(Y_2)$ First, we find the marginal PDFs: * $f_1(y_1) = \int_0^{y_1/2} 1 \, dy_2 = \frac{y_1}{2}$, for $0 \le y_1 \le 2$ * $f_2(y_2) = 2 - 2y_2$, for $0 \le y_2 \le 1$ **Expected Values**: * $E(Y_1) = \int_0^2 y_1 f_1(y_1) \, dy_1 = \int_0^2 y_1 \left(\frac{y_1}{2}\right) \, dy_1 = \frac{1}{2} \left[\frac{y_1^3}{3}\right]_0^2 = \frac{1}{2} \left(\frac{8}{3}\right) = \frac{4}{3}$ * $E(Y_2) = \int_0^1 y_2 f_2(y_2) \, dy_2 = \int_0^1 y_2 (2 - 2y_2) \, dy_2 = \left[y_2^2 - \frac{2y_2^3}{3}\right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3}$ **Variances**: First, we find $E(Y_1^2)$ and $E(Y_2^2)$. * $E(Y_1^2) = \int_0^2 y_1^2 f_1(y_1) \, dy_1 = \int_0^2 y_1^2 \left(\frac{y_1}{2}\right) \, dy_1 = \frac{1}{2} \left[\frac{y_1^4}{4}\right]_0^2 = \frac{1}{2} \left(\frac{16}{4}\right) = 2$ * $E(Y_2^2) = \int_0^1 y_2^2 f_2(y_2) \, dy_2 = \int_0^1 y_2^2 (2 - 2y_2) \, dy_2 = \left[\frac{2y_2^3}{3} - \frac{2y_2^4}{4}\right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$ Now we can calculate the variances: * $V(Y_1) = E(Y_1^2) - [E(Y_1)]^2 = 2 - \left(\frac{4}{3}\right)^2 = 2 - \frac{16}{9} = \frac{18 - 16}{9} = \frac{2}{9}$ * $V(Y_2) = E(Y_2^2) - [E(Y_2)]^2 = \frac{1}{6} - \left(\frac{1}{3}\right)^2 = \frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18}$ The results are: * **$E(Y_1) = \frac{4}{3}$** * **$E(Y_2) = \frac{1}{3}$** * **$V(Y_1) = \frac{2}{9}$** * **$V(Y_2) = \frac{1}{18}$** --- ### f) Find $E(Y_1 - Y_2)$ and $V(Y_1 - Y_2)$ **Expected Value of the Difference**: Using the linearity of expectation: $$ E(Y_1 - Y_2) = E(Y_1) - E(Y_2) = \frac{4}{3} - \frac{1}{3} = 1 $$ **Variance of the Difference**: The formula for the variance of a difference is: $$ V(Y_1 - Y_2) = V(Y_1) + V(Y_2) - 2 \cdot \text{Cov}(Y_1, Y_2) $$ We need to find the covariance, $\text{Cov}(Y_1, Y_2) = E(Y_1 Y_2) - E(Y_1)E(Y_2)$ 1. **Calculate $E(Y_1 Y_2)$**: $$ E(Y_1 Y_2) = \int_0^1 \int_{2y_2}^2 (y_1 y_2)(1) \, dy_1 \, dy_2 = \int_0^1 y_2 \left[\frac{y_1^2}{2}\right]_{2y_2}^2 \, dy_2 $$ $$ = \int_0^1 y_2 \left(\frac{4}{2} - \frac{4y_2^2}{2}\right) \, dy_2 = \int_0^1 (2y_2 - 2y_2^3) \, dy_2 = \left[y_2^2 - \frac{y_2^4}{2}\right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} $$ 2. **Calculate Covariance**: $$ \text{Cov}(Y_1, Y_2) = E(Y_1 Y_2) - E(Y_1)E(Y_2) = \frac{1}{2} - \left(\frac{4}{3}\right)\left(\frac{1}{3}\right) = \frac{1}{2} - \frac{4}{9} = \frac{9 - 8}{18} = \frac{1}{18} $$ 3. **Calculate Variance of the Difference**: $$ V(Y_1 - Y_2) = V(Y_1) + V(Y_2) - 2\text{Cov}(Y_1, Y_2) = \frac{2}{9} + \frac{1}{18} - 2\left(\frac{1}{18}\right) $$ $$ = \frac{4}{18} + \frac{1}{18} - \frac{2}{18} = \frac{3}{18} = \frac{1}{6} $$ The results are: * **$E(Y_1 - Y_2) = 1$** * **$V(Y_1 - Y_2) = \frac{1}{6}$** rewrite it as it is /human