Let \( X \) and \( Y \) be two jointly continuous random variables with pdf \[ f_{x, Y}(x, y)=\left\{\begin{array}{ll}6 x y & \text { if } 0 \leq x \leq 1, \quad 0 \leq y \leq \sqrt{x} \\ 0 & \text { otherwise }\end{array}\right. \] a. Find \( f_{x}(x) \) and \( f_{y}(y) \) b. Find the conditional pdf of \( X \) given \( Y=y, f_{x \mid Y}(x \mid y) \)

To solve the problem, we will first derive the marginal probability density functions (pdfs) \( f_{X}(x) \) and \( f_{Y}(y) \), and then we will find the conditional pdf \( f_{X|Y}(x|y) \). ### a. Finding \( f_{X}(x) \) and \( f_{Y}(y) \) **Step 1: Finding \( f_{X}(x) \)** The marginal pdf of \( X \) can be found by integrating the joint pdf \( f_{X,Y}(x,y) \) over all possible values of \( y \): \[ f_{X}(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy = \int_{0}^{\sqrt{x}} 6xy \, dy \] for \( 0 \leq x \leq 1 \). We compute the integral: \[ f_{X}(x) = \int_{0}^{\sqrt{x}} 6xy \, dy = 6x \int_{0}^{\sqrt{x}} y \, dy = 6x \left[ \frac{y^2}{2} \right]_{0}^{\sqrt{x}} = 6x \cdot \frac{x}{2} = 3x^2 \] Thus, \[ f_{X}(x) = \begin{cases} 3x^2 & \text{if } 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases} \] **Step 2: Finding \( f_{Y}(y) \)** Next, we find the marginal pdf of \( Y \) by integrating the joint pdf over \( x \): \[ f_{Y}(y) = \int_{0}^{1} f_{X,Y}(x,y) \, dx = \int_{y^2}^{1} 6xy \, dx \] The limits of integration for \( x \) are from \( y^2 \) to \( 1 \) because for \( f_{X,Y}(x,y) \) to be non-zero, we need \( 0 \leq y \leq \sqrt{x} \Rightarrow y^2 \leq x \leq 1 \). We compute the integral as follows: \[ f_{Y}(y) = 6y \int_{y^2}^{1} x \, dx = 6y \left[ \frac{x^2}{2} \right]_{y^2}^{1} = 6y \left( \frac{1}{2} - \frac{(y^2)^2}{2} \right) = 6y \left( \frac{1}{2} - \frac{y^4}{2} \right) = 3y(1 - y^4) \] Thus, \[ f_{Y}(y) = \begin{cases} 3y(1 - y^4) & \text{if } 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases} \] ### b. Finding the conditional pdf of \( X \) given \( Y=y \) The conditional pdf \( f_{X|Y}(x|y) \) is calculated using the formula: \[ f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_{Y}(y)} \] Substituting \( f_{X,Y}(x,y) \) and \( f_{Y}(y) \): \[ f_{X|Y}(x|y) = \frac{6xy}{3y(1 - y^4)} = \frac{2x}{1 - y^4} \] for \( y^2 \leq x \leq 1 \) (assuming \( 0 \leq y \leq 1 \)). Thus: \[ f_{X|Y}(x|y) = \begin{cases} \frac{2x}{1 - y^4} & \text{if } y^2 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases} \] And that completes the solution!