1. **Problem statement:** We are given a joint probability density function (pdf) for random variables $X$ and $Y$: $$f(x,y) = \begin{cases} k(2x + 3y), & 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0, & \text{elsewhere} \end{cases}$$ (i) Find the constant $k$. (ii) Find $P[(X,Y) \in A]$ where $A = \{(x,y) | 0 < x < \frac{1}{2}, \frac{1}{4} < y < \frac{1}{2}\}$. (iii) Compute $P(X + Y \leq 1)$. (b) Show that $E(Y) = E[E(Y|X)]$. (c) Prove $E(X) = \int_{-\infty}^{\infty} E[X|Y=y] f_Y(y) dy$. (d) Given $X$ with pdf $$f(x) = \begin{cases} \frac{1}{100} e^{-\frac{x}{100}}, & x > 0 \\ 0, & \text{elsewhere} \end{cases}$$ Two independent components in series: find pdf of system life $Y$. --- ### (a)(i) Find $k$: 1. The total probability must be 1: $$\int_0^1 \int_0^1 k(2x + 3y) dy dx = 1$$ 2. Compute inner integral: $$\int_0^1 (2x + 3y) dy = 2x \int_0^1 dy + 3 \int_0^1 y dy = 2x(1) + 3 \cdot \frac{1}{2} = 2x + \frac{3}{2}$$ 3. Now outer integral: $$\int_0^1 k \left(2x + \frac{3}{2}\right) dx = k \left( \int_0^1 2x dx + \int_0^1 \frac{3}{2} dx \right) = k \left( 2 \cdot \frac{1}{2} + \frac{3}{2} \cdot 1 \right) = k (1 + \frac{3}{2}) = k \cdot \frac{5}{2}$$ 4. Set equal to 1: $$k \cdot \frac{5}{2} = 1 \implies k = \frac{2}{5}$$ --- ### (a)(ii) Find $P[(X,Y) \in A]$ where $A = \{0 < x < \frac{1}{2}, \frac{1}{4} < y < \frac{1}{2}\}$: 1. Probability is integral over $A$: $$P = \int_0^{\frac{1}{2}} \int_{\frac{1}{4}}^{\frac{1}{2}} k(2x + 3y) dy dx$$ 2. Substitute $k=\frac{2}{5}$: $$= \frac{2}{5} \int_0^{\frac{1}{2}} \int_{\frac{1}{4}}^{\frac{1}{2}} (2x + 3y) dy dx$$ 3. Inner integral: $$\int_{\frac{1}{4}}^{\frac{1}{2}} (2x + 3y) dy = 2x \left(y \Big|_{\frac{1}{4}}^{\frac{1}{2}}\right) + 3 \left( \frac{y^2}{2} \Big|_{\frac{1}{4}}^{\frac{1}{2}} \right) = 2x \left( \frac{1}{2} - \frac{1}{4} \right) + \frac{3}{2} \left( \left(\frac{1}{2}\right)^2 - \left(\frac{1}{4}\right)^2 \right)$$ 4. Calculate: $$2x \cdot \frac{1}{4} + \frac{3}{2} \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{x}{2} + \frac{3}{2} \cdot \frac{3}{16} = \frac{x}{2} + \frac{9}{32}$$ 5. Outer integral: $$\int_0^{\frac{1}{2}} \left( \frac{x}{2} + \frac{9}{32} \right) dx = \int_0^{\frac{1}{2}} \frac{x}{2} dx + \int_0^{\frac{1}{2}} \frac{9}{32} dx = \frac{1}{2} \cdot \frac{x^2}{2} \Big|_0^{\frac{1}{2}} + \frac{9}{32} \cdot \frac{1}{2}$$ 6. Calculate: $$\frac{1}{2} \cdot \frac{(\frac{1}{2})^2}{2} + \frac{9}{64} = \frac{1}{2} \cdot \frac{1}{8} + \frac{9}{64} = \frac{1}{16} + \frac{9}{64} = \frac{4}{64} + \frac{9}{64} = \frac{13}{64}$$ 7. Multiply by $k$: $$P = \frac{2}{5} \cdot \frac{13}{64} = \frac{26}{320} = \frac{13}{160}$$ --- ### (a)(iii) Compute $P(X + Y \leq 1)$: 1. The region is $\{(x,y) | 0 \leq x \leq 1, 0 \leq y \leq 1, x + y \leq 1\}$. 2. Probability: $$P = \int_0^1 \int_0^{1-x} k(2x + 3y) dy dx$$ 3. Substitute $k=\frac{2}{5}$: $$= \frac{2}{5} \int_0^1 \int_0^{1-x} (2x + 3y) dy dx$$ 4. Inner integral: $$\int_0^{1-x} (2x + 3y) dy = 2x (1-x) + \frac{3}{2} (1-x)^2$$ 5. So: $$P = \frac{2}{5} \int_0^1 \left[ 2x(1-x) + \frac{3}{2} (1-x)^2 \right] dx$$ 6. Expand terms: $$2x(1-x) = 2x - 2x^2$$ $$(1-x)^2 = 1 - 2x + x^2$$ 7. Substitute: $$P = \frac{2}{5} \int_0^1 \left( 2x - 2x^2 + \frac{3}{2} - 3x + \frac{3}{2} x^2 \right) dx = \frac{2}{5} \int_0^1 \left( \frac{3}{2} - x - \frac{1}{2} x^2 \right) dx$$ 8. Integrate term by term: $$\int_0^1 \frac{3}{2} dx = \frac{3}{2}$$ $$\int_0^1 x dx = \frac{1}{2}$$ $$\int_0^1 x^2 dx = \frac{1}{3}$$ 9. So: $$P = \frac{2}{5} \left( \frac{3}{2} - \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{3} \right) = \frac{2}{5} \left( 1 - \frac{1}{6} \right) = \frac{2}{5} \cdot \frac{5}{6} = \frac{1}{3}$$ --- ### (b) Show $E(Y) = E[E(Y|X)]$: 1. By definition: $$E(Y) = \int \int y f(x,y) dx dy$$ 2. Conditional expectation: $$E(Y|X=x) = \int y f_{Y|X}(y|x) dy$$ 3. Marginal pdf of $X$: $$f_X(x) = \int f(x,y) dy$$ 4. Then: $$E[E(Y|X)] = \int E(Y|X=x) f_X(x) dx = \int \left( \int y f_{Y|X}(y|x) dy \right) f_X(x) dx$$ 5. Since $f(x,y) = f_{Y|X}(y|x) f_X(x)$, $$E[E(Y|X)] = \int \int y f(x,y) dy dx = E(Y)$$ --- ### (c) Prove $E(X) = \int_{-\infty}^\infty E[X|Y=y] f_Y(y) dy$: 1. By definition: $$E(X) = \int \int x f(x,y) dx dy$$ 2. Conditional expectation: $$E[X|Y=y] = \int x f_{X|Y}(x|y) dx$$ 3. Marginal pdf of $Y$: $$f_Y(y) = \int f(x,y) dx$$ 4. Then: $$\int E[X|Y=y] f_Y(y) dy = \int \left( \int x f_{X|Y}(x|y) dx \right) f_Y(y) dy = \int \int x f(x,y) dx dy = E(X)$$ --- ### (d) Two independent components with exponential lifetimes $X_1, X_2$ with pdf: $$f(x) = \frac{1}{100} e^{-\frac{x}{100}}, x > 0$$ 1. System fails when either component fails, so system life $Y = \min(X_1, X_2)$. 2. CDF of $Y$: $$F_Y(y) = P(Y \leq y) = 1 - P(Y > y) = 1 - P(X_1 > y, X_2 > y)$$ 3. Independence: $$= 1 - P(X_1 > y) P(X_2 > y)$$ 4. For exponential: $$P(X_i > y) = e^{-\frac{y}{100}}$$ 5. So: $$F_Y(y) = 1 - e^{-\frac{y}{100}} e^{-\frac{y}{100}} = 1 - e^{-\frac{2y}{100}} = 1 - e^{-\frac{y}{50}}$$ 6. Differentiate to get pdf: $$f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{1}{50} e^{-\frac{y}{50}}, y > 0$$ --- **Final answers:** (i) $k = \frac{2}{5}$ (ii) $P = \frac{13}{160}$ (iii) $P = \frac{1}{3}$ (b) $E(Y) = E[E(Y|X)]$ holds by law of total expectation. (c) $E(X) = \int_{-\infty}^\infty E[X|Y=y] f_Y(y) dy$ holds by law of total expectation. (d) System life pdf: $$f_Y(y) = \frac{1}{50} e^{-\frac{y}{50}}, y > 0$$