1. **Problem Statement:** (a) Given joint pdf $f_{X,Y}(x,y) = \frac{15}{2}(2 - x - y)$ for $0 < x < 1$, $0 < y < 1$, find the conditional density $f_{X|Y}(x|y)$. (b) Two independent uniform arrivals $X,Y \sim \text{Uniform}(0,60)$ minutes. Find $P(|X - Y| > 10)$, i.e., the probability the first to arrive waits more than 10 minutes. (c) Given joint pdf $f_{X,Y}(x,y) = \frac{e^{-(x+y)}}{y}$ for $x,y > 0$, find: (i) $P(X > \frac{1}{Y} | Y = y)$ (ii) $E[X|Y=y]$ (d) Given joint pdf $f_{X,Y}(x,y) = \frac{x}{5} + cy$ for $0 < x < 1$, $1 < y < 5$, find: (i) constant $c$ (ii) independence of $X$ and $Y$ (iii) $P(X + Y > 3)$ --- 2. **Part (a) Conditional Density:** - Formula: $f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$ - Find marginal $f_Y(y) = \int_0^1 f_{X,Y}(x,y) dx = \int_0^1 \frac{15}{2}(2 - x - y) dx$ Calculate: $$f_Y(y) = \frac{15}{2} \int_0^1 (2 - x - y) dx = \frac{15}{2} \left[(2 - y)\cdot 1 - \frac{1^2}{2}\right] = \frac{15}{2} \left(2 - y - \frac{1}{2}\right) = \frac{15}{2} \left(\frac{3}{2} - y\right) = \frac{15}{2} \cdot \frac{3 - 2y}{2} = \frac{15(3 - 2y)}{4}$$ Thus, $$f_{X|Y}(x|y) = \frac{\frac{15}{2}(2 - x - y)}{\frac{15(3 - 2y)}{4}} = \frac{(2 - x - y)}{\frac{3 - 2y}{2}} = \frac{2(2 - x - y)}{3 - 2y}$$ Domain: $0 < x < 1$, $0 < y < 1$. --- 3. **Part (b) Waiting Time Probability:** - $X,Y$ independent uniform on $[0,60]$ minutes. - Probability first to arrive waits more than 10 minutes is $P(|X - Y| > 10)$. Area of square: $60 \times 60 = 3600$. Region where $|X - Y| \leq 10$ is a band around the diagonal with width 20. Area where $|X - Y| \leq 10$ is $3600 - 2 \times$ area of two triangles each with base and height 50. Calculate area where $|X - Y| \leq 10$: $$= 3600 - 2 \times \frac{1}{2} \times 50 \times 50 = 3600 - 2500 = 1100$$ So, $$P(|X - Y| > 10) = 1 - \frac{1100}{3600} = \frac{2500}{3600} = \frac{25}{36} \approx 0.6944$$ --- 4. **Part (c) Joint pdf $f_{X,Y}(x,y) = \frac{e^{-(x+y)}}{y}$:** (i) Find $P\left(X > \frac{1}{Y} | Y = y\right) = P\left(X > \frac{1}{y}\right)$ given $Y=y$. Conditional pdf of $X$ given $Y=y$: $$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$ Marginal $f_Y(y) = \int_0^\infty \frac{e^{-(x+y)}}{y} dx = \frac{e^{-y}}{y} \int_0^\infty e^{-x} dx = \frac{e^{-y}}{y} \cdot 1 = \frac{e^{-y}}{y}$$ So, $$f_{X|Y}(x|y) = \frac{\frac{e^{-(x+y)}}{y}}{\frac{e^{-y}}{y}} = e^{-x}, \quad x > 0$$ Therefore, $$P\left(X > \frac{1}{y} | Y = y\right) = \int_{1/y}^\infty e^{-x} dx = e^{-1/y}$$ (ii) Conditional expectation: $$E[X|Y=y] = \int_0^\infty x e^{-x} dx = 1$$ Because $X|Y=y$ is exponential with parameter 1. --- 5. **Part (d) Joint pdf $f_{X,Y}(x,y) = \frac{x}{5} + cy$ for $0 < x < 1$, $1 < y < 5$:** (i) Find $c$ by normalization: $$\int_1^5 \int_0^1 \left(\frac{x}{5} + cy\right) dx dy = 1$$ Calculate inner integral: $$\int_0^1 \frac{x}{5} dx = \frac{1}{5} \cdot \frac{1^2}{2} = \frac{1}{10}$$ $$\int_0^1 cy dx = cy \cdot 1 = cy$$ So, $$\int_1^5 \left(\frac{1}{10} + cy\right) dy = \int_1^5 \frac{1}{10} dy + c \int_1^5 y dy = \frac{1}{10} (5 - 1) + c \left(\frac{5^2}{2} - \frac{1^2}{2}\right) = \frac{4}{10} + c \cdot \frac{25 - 1}{2} = 0.4 + 12c$$ Set equal to 1: $$0.4 + 12c = 1 \implies 12c = 0.6 \implies c = 0.05$$ (ii) Check independence: Marginal $f_X(x) = \int_1^5 \left(\frac{x}{5} + 0.05 y\right) dy = \frac{x}{5} (5 - 1) + 0.05 \cdot \frac{5^2 - 1^2}{2} = \frac{4x}{5} + 0.05 \cdot 12 = \frac{4x}{5} + 0.6$ Marginal $f_Y(y) = \int_0^1 \left(\frac{x}{5} + 0.05 y\right) dx = \frac{1}{5} \cdot \frac{1^2}{2} + 0.05 y = 0.1 + 0.05 y$ Check if $f_{X,Y}(x,y) = f_X(x) f_Y(y)$: $$\left(\frac{x}{5} + 0.05 y\right) \stackrel{?}{=} \left(\frac{4x}{5} + 0.6\right)(0.1 + 0.05 y)$$ Right side expands to terms not matching left side exactly, so $X$ and $Y$ are not independent. (iii) Find $P(X + Y > 3)$: Domain: $0 < x < 1$, $1 < y < 5$. Rewrite event: $$X + Y > 3 \implies X > 3 - Y$$ For $y$ in $(1,5)$, $3 - y$ varies: - If $y > 3$, then $3 - y < 0$, so $X >$ negative number always true since $X > 0$. - If $1 < y \leq 3$, then $3 - y$ in $(0,2)$. So, $$P = \int_1^3 \int_{3 - y}^1 \left(\frac{x}{5} + 0.05 y\right) dx dy + \int_3^5 \int_0^1 \left(\frac{x}{5} + 0.05 y\right) dx dy$$ Calculate inner integrals: For $1 < y \leq 3$: $$\int_{3 - y}^1 \frac{x}{5} dx = \frac{1}{5} \left[\frac{x^2}{2}\right]_{3 - y}^1 = \frac{1}{10} (1 - (3 - y)^2)$$ $$\int_{3 - y}^1 0.05 y dx = 0.05 y (1 - (3 - y)) = 0.05 y (y - 2)$$ Sum: $$\frac{1}{10} (1 - (3 - y)^2) + 0.05 y (y - 2)$$ For $3 < y < 5$: $$\int_0^1 \frac{x}{5} dx = \frac{1}{10}$$ $$\int_0^1 0.05 y dx = 0.05 y$$ Sum: $$\frac{1}{10} + 0.05 y$$ Now integrate over $y$: $$P = \int_1^3 \left[\frac{1}{10} (1 - (3 - y)^2) + 0.05 y (y - 2)\right] dy + \int_3^5 \left(\frac{1}{10} + 0.05 y\right) dy$$ Calculate: First integral: $$\int_1^3 \left[\frac{1}{10} (1 - (3 - y)^2) + 0.05 y (y - 2)\right] dy = \int_1^3 \left[\frac{1}{10} (1 - (9 - 6y + y^2)) + 0.05 (y^2 - 2y)\right] dy$$ $$= \int_1^3 \left[\frac{1}{10} (1 - 9 + 6y - y^2) + 0.05 y^2 - 0.1 y\right] dy$$ $$= \int_1^3 \left[-\frac{8}{10} + \frac{6}{10} y - \frac{1}{10} y^2 + 0.05 y^2 - 0.1 y\right] dy$$ $$= \int_1^3 \left[-0.8 + 0.6 y - 0.1 y^2 + 0.05 y^2 - 0.1 y\right] dy$$ $$= \int_1^3 \left[-0.8 + (0.6 - 0.1) y + (-0.1 + 0.05) y^2\right] dy = \int_1^3 \left[-0.8 + 0.5 y - 0.05 y^2\right] dy$$ Integrate termwise: $$= \left[-0.8 y + 0.25 y^2 - \frac{0.05}{3} y^3\right]_1^3 = \left[-0.8 y + 0.25 y^2 - 0.0167 y^3\right]_1^3$$ Calculate at 3: $$-0.8 \times 3 + 0.25 \times 9 - 0.0167 \times 27 = -2.4 + 2.25 - 0.45 = -0.6$$ Calculate at 1: $$-0.8 + 0.25 - 0.0167 = -0.5667$$ Difference: $$-0.6 - (-0.5667) = -0.0333$$ Second integral: $$\int_3^5 \left(0.1 + 0.05 y\right) dy = \left[0.1 y + 0.025 y^2\right]_3^5 = (0.5 + 0.625) - (0.3 + 0.225) = 1.125 - 0.525 = 0.6$$ Sum total probability: $$P = -0.0333 + 0.6 = 0.5667$$ --- **Final answers:** (a) $f_{X|Y}(x|y) = \frac{2(2 - x - y)}{3 - 2y}$ for $0 < x < 1$, $0 < y < 1$ (b) $P(|X - Y| > 10) = \frac{25}{36} \approx 0.6944$ (c)(i) $P\left(X > \frac{1}{Y} | Y = y\right) = e^{-1/y}$ (c)(ii) $E[X|Y=y] = 1$ (d)(i) $c = 0.05$ (d)(ii) $X$ and $Y$ are not independent (d)(iii) $P(X + Y > 3) \approx 0.5667$