1. **Problem statement:** Given the joint pmf of discrete random variables $X$ and $Y$ as $$\begin{array}{c|ccc|c} X \backslash Y & 0 & 1 & 3 & f_1(x) \\ \hline 0 & \frac{1}{8} & \frac{2}{8} & \frac{1}{8} & \frac{4}{8} \\ 1 & \frac{2}{8} & \frac{1}{8} & \frac{1}{8} & \frac{4}{8} \\ \hline f_2(y) & \frac{3}{8} & \frac{3}{8} & \frac{2}{8} & 1 \end{array}$$ Find: (i) $E(Y|X=x)$ (ii) $\mathrm{Var}(Y|X=x)$ (iii) $\mathrm{Cov}(X,Y)$ (iv) Correlation coefficient $\rho_{XY}$ (v) Are $X$ and $Y$ independent? --- 2. **Recall formulas:** - Conditional expectation: $E(Y|X=x) = \sum_y y P(Y=y|X=x)$ - Conditional variance: $\mathrm{Var}(Y|X=x) = E(Y^2|X=x) - [E(Y|X=x)]^2$ - Covariance: $\mathrm{Cov}(X,Y) = E(XY) - E(X)E(Y)$ - Correlation coefficient: $\rho_{XY} = \frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}}$ - Independence: $X$ and $Y$ are independent if $P(X=x,Y=y) = P(X=x)P(Y=y)$ for all $x,y$. --- 3. **Calculate conditional pmfs:** - For $X=0$, $P(Y=y|X=0) = \frac{P(X=0,Y=y)}{P(X=0)}$: - $P(Y=0|0) = \frac{1/8}{4/8} = \frac{1}{4}$ - $P(Y=1|0) = \frac{2/8}{4/8} = \frac{1}{2}$ - $P(Y=3|0) = \frac{1/8}{4/8} = \frac{1}{4}$ - For $X=1$, similarly: - $P(Y=0|1) = \frac{2/8}{4/8} = \frac{1}{2}$ - $P(Y=1|1) = \frac{1/8}{4/8} = \frac{1}{4}$ - $P(Y=3|1) = \frac{1/8}{4/8} = \frac{1}{4}$ --- 4. **Compute $E(Y|X=x)$:** - $E(Y|0) = 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 3 \times \frac{1}{4} = 0 + \frac{1}{2} + \frac{3}{4} = \frac{5}{4} = 1.25$ - $E(Y|1) = 0 \times \frac{1}{2} + 1 \times \frac{1}{4} + 3 \times \frac{1}{4} = 0 + \frac{1}{4} + \frac{3}{4} = 1$ --- 5. **Compute $E(Y^2|X=x)$ for variance:** - For $X=0$: $$E(Y^2|0) = 0^2 \times \frac{1}{4} + 1^2 \times \frac{1}{2} + 3^2 \times \frac{1}{4} = 0 + \frac{1}{2} + \frac{9}{4} = \frac{11}{4} = 2.75$$ - For $X=1$: $$E(Y^2|1) = 0^2 \times \frac{1}{2} + 1^2 \times \frac{1}{4} + 3^2 \times \frac{1}{4} = 0 + \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = 2.5$$ --- 6. **Calculate conditional variances:** - $\mathrm{Var}(Y|0) = E(Y^2|0) - [E(Y|0)]^2 = 2.75 - (1.25)^2 = 2.75 - 1.5625 = 1.1875$ - $\mathrm{Var}(Y|1) = 2.5 - (1)^2 = 2.5 - 1 = 1.5$ --- 7. **Calculate marginal expectations:** - $E(X) = 0 \times \frac{4}{8} + 1 \times \frac{4}{8} = 0 + 0.5 = 0.5$ - $E(Y) = 0 \times \frac{3}{8} + 1 \times \frac{3}{8} + 3 \times \frac{2}{8} = 0 + \frac{3}{8} + \frac{6}{8} = \frac{9}{8} = 1.125$ --- 8. **Calculate $E(XY)$:** Sum over all $x,y$ of $x y P(X=x,Y=y)$: - For $x=0$: $0 \times (0 \times \frac{1}{8} + 1 \times \frac{2}{8} + 3 \times \frac{1}{8}) = 0$ - For $x=1$: $1 \times (0 \times \frac{2}{8} + 1 \times \frac{1}{8} + 3 \times \frac{1}{8}) = 1 \times (0 + \frac{1}{8} + \frac{3}{8}) = \frac{4}{8} = 0.5$ So, $E(XY) = 0 + 0.5 = 0.5$ --- 9. **Calculate covariance:** $$\mathrm{Cov}(X,Y) = E(XY) - E(X)E(Y) = 0.5 - 0.5 \times 1.125 = 0.5 - 0.5625 = -0.0625$$ --- 10. **Calculate variances of $X$ and $Y$:** - $E(X^2) = 0^2 \times \frac{4}{8} + 1^2 \times \frac{4}{8} = 0 + 0.5 = 0.5$ - $\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 0.5 - (0.5)^2 = 0.5 - 0.25 = 0.25$ - $E(Y^2) = 0^2 \times \frac{3}{8} + 1^2 \times \frac{3}{8} + 3^2 \times \frac{2}{8} = 0 + \frac{3}{8} + \frac{18}{8} = \frac{21}{8} = 2.625$ - $\mathrm{Var}(Y) = 2.625 - (1.125)^2 = 2.625 - 1.265625 = 1.359375$ --- 11. **Calculate correlation coefficient:** $$\rho_{XY} = \frac{-0.0625}{\sqrt{0.25 \times 1.359375}} = \frac{-0.0625}{\sqrt{0.33984375}} = \frac{-0.0625}{0.583} \approx -0.107$$ --- 12. **Check independence:** Check if $P(X=x,Y=y) = P(X=x)P(Y=y)$ for all $x,y$. - For example, $P(0,0) = \frac{1}{8} = 0.125$ but $P(0)P(0) = \frac{4}{8} \times \frac{3}{8} = 0.1875 \neq 0.125$ Since this fails, $X$ and $Y$ are **not independent**. --- **Final answers:** - (i) $E(Y|X=0) = 1.25$, $E(Y|X=1) = 1$ - (ii) $\mathrm{Var}(Y|X=0) = 1.1875$, $\mathrm{Var}(Y|X=1) = 1.5$ - (iii) $\mathrm{Cov}(X,Y) = -0.0625$ - (iv) $\rho_{XY} \approx -0.107$ - (v) $X$ and $Y$ are not independent because $P(X,Y) \neq P(X)P(Y)$ for some $(x,y)$.