Let's solve the problem step by step: --- ### **Problem 2** Given: \[ f_{X}(x)= \begin{cases} 1 / 3 & 0 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases} \] \[ f_{Y}(y)= \begin{cases} c y & 0 \leq y \leq 5 \\ 0 & \text{otherwise} \end{cases} \] \(X\) and \(Y\) are independent. --- ## **(a) Determine \(c\).** Since \(f_Y(y)\) is a probability density function, it must integrate to 1 over its domain: \[ \int_{-\infty}^{\infty} f_Y(y) \, dy = 1 \] Since \(f_Y(y) = c y\) for \(0 \leq y \leq 5\): \[ \int_0^{5} c y \, dy = 1 \] \[ c \int_0^{5} y \, dy = 1 \] \[ c \left[ \frac{y^2}{2} \right]_0^5 = 1 \] \[ c \left( \frac{25}{2} - 0 \right) = 1 \] \[ c \cdot \frac{25}{2} = 1 \] \[ c = \frac{2}{25} \] **Final answer for (a):** \[ \boxed{c = \frac{2}{25}} \] --- ## **(b) Determine the joint PDF \(f_{X,Y}(x, y)\).** Since \(X\) and \(Y\) are independent: \[ f_{X,Y}(x, y) = f_X(x) f_Y(y) \] So, - \(f_X(x) = \frac{1}{3}\) for \(0 \leq x \leq 3\), 0 otherwise. - \(f_Y(y) = \frac{2}{25}y\) for \(0 \leq y \leq 5\), 0 otherwise. Thus, \[ f_{X,Y}(x, y) = \begin{cases} \frac{1}{3} \cdot \frac{2}{25} y = \frac{2}{75} y & 0 \leq x \leq 3,\; 0 \leq y \leq 5 \\ 0 & \text{otherwise} \end{cases} \] **Final answer for (b):** \[ \boxed{ f_{X,Y}(x, y) = \begin{cases} \frac{2}{75} y & 0 \leq x \leq 3,\; 0 \leq y \leq 5 \\ 0 & \text{otherwise} \end{cases} } \] --- ## **(c) Determine \(\operatorname{cov}(X, Y)\).** Recall: \[ \operatorname{cov}(X,Y) = E[XY] - E[X]E[Y] \] First, calculate \(E[X]\), \(E[Y]\), and \(E[XY]\): ### **Find \(E[X]\):** \[ E[X] = \int_{0}^{3} x \cdot \frac{1}{3} dx = \frac{1}{3} \int_0^3 x dx = \frac{1}{3} \cdot \left[\frac{x^2}{2}\right]_0^3 = \frac{1}{3} \cdot \frac{9}{2} = \frac{3}{2} \] ### **Find \(E[Y]\):** \[ E[Y] = \int_0^5 y \cdot \frac{2}{25}y dy = \frac{2}{25} \int_0^5 y^2 dy = \frac{2}{25} \cdot \left[ \frac{y^3}{3} \right]_0^5 = \frac{2}{25} \cdot \frac{125}{3} = \frac{2 \times 125}{25 \times 3} = \frac{250}{75} = \frac{10}{3} \] ### **Find \(E[XY]\):** Since they are independent: \[ E[XY] = E[X]E[Y] \] ### **Covariance:** \[ \operatorname{cov}(X, Y) = E[XY] - E[X]E[Y] = 0 \] **Final answer for (c):** \[ \boxed{0} \] --- ## **(d) Determine \(P(X < Y)\).** \[ P(X < Y) = \iint_{x < y} f_{X,Y}(x, y) dx dy \] Since \(0 \leq x \leq 3\), \(0 \leq y \leq 5\): For each \(y\), \(x\) goes from 0 to \(\min(y, 3)\): But \(x < y\) means \(x\) goes from 0 up to \(y\) (as long as \(y \leq 3\)), and from 0 to 3 (as long as \(y > 3\)), but in that case, \(x < y\) is always satisfied for \(x \in [0,3]\). Let's consider the two ranges for \(y\): - \(0 \leq y \leq 3\): \(x\) goes from 0 to \(y\). - \(3 < y \leq 5\): \(x\) goes from 0 to 3. So, \[ P(X < Y) = \int_{y=0}^{3} \int_{x=0}^{y} f_{X,Y}(x,y) dx dy + \int_{y=3}^{5} \int_{x=0}^{3} f_{X,Y}(x,y) dx dy \] Recall \(f_{X,Y}(x, y) = \frac{2}{75} y\): First term: \[ \int_{y=0}^{3} \int_{x=0}^{y} \frac{2}{75} y dx dy = \int_{y=0}^{3} \frac{2}{75} y \int_{x=0}^{y} dx dy = \int_{y=0}^{3} \frac{2}{75} y (y - 0) dy = \int_{y=0}^{3} \frac{2}{75} y^2 dy = \frac{2}{75} \int_{y=0}^{3} y^2 dy = \frac{2}{75} \left[ \frac{y^3}{3} \right]_0^3 = \frac{2}{75} \cdot \frac{27}{3} = \frac{2}{75} \cdot 9 = \frac{18}{75} = \frac{6}{25} \] Second term: \[ \int_{y=3}^{5} \int_{x=0}^{3} \frac{2}{75} y dx dy = \int_{y=3}^{5} \frac{2}{75} y \int_{x=0}^{3} dx dy = \int_{y=3}^{5} \frac{2}{75} y \cdot 3 dy = \int_{y=3}^{5} \frac{6}{75} y dy = \frac{2}{25} \int_{y=3}^{5} y dy = \frac{2}{25} \left[ \frac{y^2}{2} \right]_3^5 = \frac{1}{25} \left( 25 - 9 \right ) = \frac{1}{25} \cdot 16 = \frac{16}{25} \] Add the two results: \[ P(X < Y) = \frac{6}{25} + \frac{16}{25} = \frac{22}{25} \] **Final answer for (d):** \[ \boxed{\frac{22}{25}} \] --- ## **(e) Determine the joint CDF \(F_{X,Y}(x, y)\).** By definition, \[ F_{X,Y}(x, y) = P(X \leq x, Y \leq y) = \int_{-\infty}^x \int_{-\infty}^y f_{X,Y}(u, v) dv\, du \] But since \(f_{X,Y}(u,v) = 0\) unless \(0 \leq u \leq 3\) and \(0 \leq v \leq 5\), we can restrict the integration accordingly: For \(x < 0\) or \(y < 0\), \(F_{X,Y}(x, y) = 0\). For \(

Let's carefully derive the joint CDF \(F_{X,Y}(x,y) = P(X \leq x, Y \leq y)\). --- ### **Part (e): Find the joint CDF \(F_{X,Y}(x,y)\).** Recall: \[ F_{X,Y}(x,y) = P(X \leq x, Y \leq y) = \iint_{u \leq x, v \leq y} f_{X,Y}(u,v) \, du \, dv \] Given the joint PDF from part (b): \[ f_{X,Y}(u,v) = \begin{cases} \frac{2}{75} v & 0 \leq u \leq 3, \quad 0 \leq v \leq 5 \\ 0 & \text{otherwise} \end{cases} \] --- ### **Step-by-step derivation:** Since the support is limited to \(0 \leq u \leq 3\) and \(0 \leq v \leq 5\), the joint CDF depends on the values of \(x\) and \(y\): 1. **If \(x < 0\) or \(y < 0\):** \[ F_{X,Y}(x,y) = 0 \] 2. **If \(x \geq 0\) and \(y \geq 0\):** - The limits of integration are from \(u=0\) to \(u = \min(x, 3)\), - and for each \(u\), from \(v=0\) to \(v = \min(y, 5)\), - considering the support restrictions. So: \[ F_{X,Y}(x,y) = \int_{u=0}^{\min(x,3)} \int_{v=0}^{\min(y,5)} \frac{2}{75} v \, dv \, du \] --- ### **Evaluate the inner integral:** \[ \int_{v=0}^{\min(y,5)} \frac{2}{75} v \, dv = \frac{2}{75} \left[ \frac{v^2}{2} \right]_0^{\min(y,5)} = \frac{1}{75} \left[ v^2 \right]_0^{\min(y,5)} = \frac{1}{75} (\min(y,5))^2 \] --- ### **Evaluate the outer integral:** \[ F_{X,Y}(x,y) = \int_{u=0}^{\min(x,3)} \frac{1}{75} (\min(y,5))^2 \, du \] Since \(\frac{1}{75} (\min(y,5))^2\) is constant with respect to \(u\), the integral simplifies to: \[ F_{X,Y}(x,y) = \frac{1}{75} (\min(y,5))^2 \times \left[ u \right]_0^{\min(x,3)} = \frac{1}{75} (\min(y,5))^2 \times \min(x,3) \] --- ### **Final expression:** \[ \boxed{ F_{X,Y}(x,y) = \begin{cases} 0, & x < 0 \text{ or } y < 0 \\ \frac{\min(x,3) \cdot (\min(y,5))^2}{75}, & x \geq 0, y \geq 0 \end{cases} } \] --- ### **Summary:** - For \(x \geq 0\) and \(y \geq 0\): \[ \boxed{ F_{X,Y}(x,y) = \frac{\min(x,3) \cdot (\min(y,5))^2}{75} } \] - For \(x < 0\) or \(y < 0\): \[ F_{X,Y}(x,y) = 0 \] This completes part (e).