1. **Problem Statement:** Given the joint probability mass function (pmf) of random variables $X$ and $Y$: $$(x,y) \quad f(x,y)$$ $$(1,1) \quad \frac{3}{8}$$ $$(2,1) \quad \frac{1}{8}$$ $$(1,2) \quad \frac{1}{8}$$ $$(2,2) \quad \frac{3}{8}$$ Find: (a) (i) Conditional pmf $f_{X|Y}(x|y)$, (ii) Conditional pmf $f_{Y|X}(y|x)$. (b) (i) Conditional mean $\mu_{X|y}$, (ii) Conditional mean $\mu_{Y|x}$. (c) Describe conditional variance $\sigma^2_{Y|X}$. (d) Determine if $X$ and $Y$ are independent. 2. **Formulas and Rules:** - Conditional pmf: $$f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)}$$ and $$f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}$$ - Marginal pmf: $$f_X(x) = \sum_y f(x,y)$$, $$f_Y(y) = \sum_x f(x,y)$$ - Conditional mean: $$\mu_{X|y} = E[X|Y=y] = \sum_x x f_{X|Y}(x|y)$$, $$\mu_{Y|x} = E[Y|X=x] = \sum_y y f_{Y|X}(y|x)$$ - Conditional variance: $$\sigma^2_{Y|X=x} = E[(Y - \mu_{Y|x})^2 | X=x] = \sum_y (y - \mu_{Y|x})^2 f_{Y|X}(y|x)$$ - Independence: $X$ and $Y$ are independent if and only if $$f(x,y) = f_X(x) f_Y(y)$$ for all $x,y$. 3. **Calculate Marginal pmfs:** - For $Y=1$: $$f_Y(1) = f(1,1) + f(2,1) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$$ - For $Y=2$: $$f_Y(2) = f(1,2) + f(2,2) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$$ - For $X=1$: $$f_X(1) = f(1,1) + f(1,2) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$$ - For $X=2$: $$f_X(2) = f(2,1) + f(2,2) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$$ 4. **(a) Conditional pmfs:** (i) For $Y=1$: $$f_{X|Y}(1|1) = \frac{f(1,1)}{f_Y(1)} = \frac{3/8}{1/2} = \frac{3}{4}$$ $$f_{X|Y}(2|1) = \frac{f(2,1)}{f_Y(1)} = \frac{1/8}{1/2} = \frac{1}{4}$$ For $Y=2$: $$f_{X|Y}(1|2) = \frac{f(1,2)}{f_Y(2)} = \frac{1/8}{1/2} = \frac{1}{4}$$ $$f_{X|Y}(2|2) = \frac{f(2,2)}{f_Y(2)} = \frac{3/8}{1/2} = \frac{3}{4}$$ (ii) For $X=1$: $$f_{Y|X}(1|1) = \frac{f(1,1)}{f_X(1)} = \frac{3/8}{1/2} = \frac{3}{4}$$ $$f_{Y|X}(2|1) = \frac{f(1,2)}{f_X(1)} = \frac{1/8}{1/2} = \frac{1}{4}$$ For $X=2$: $$f_{Y|X}(1|2) = \frac{f(2,1)}{f_X(2)} = \frac{1/8}{1/2} = \frac{1}{4}$$ $$f_{Y|X}(2|2) = \frac{f(2,2)}{f_X(2)} = \frac{3/8}{1/2} = \frac{3}{4}$$ 5. **(b) Conditional means:** (i) For $Y=1$: $$\mu_{X|1} = 1 \times \frac{3}{4} + 2 \times \frac{1}{4} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} = 1.25$$ For $Y=2$: $$\mu_{X|2} = 1 \times \frac{1}{4} + 2 \times \frac{3}{4} = \frac{1}{4} + \frac{6}{4} = \frac{7}{4} = 1.75$$ (ii) For $X=1$: $$\mu_{Y|1} = 1 \times \frac{3}{4} + 2 \times \frac{1}{4} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} = 1.25$$ For $X=2$: $$\mu_{Y|2} = 1 \times \frac{1}{4} + 2 \times \frac{3}{4} = \frac{1}{4} + \frac{6}{4} = \frac{7}{4} = 1.75$$ 6. **(c) Conditional variance $\sigma^2_{Y|X}$:** For $X=1$: $$\sigma^2_{Y|1} = (1 - 1.25)^2 \times \frac{3}{4} + (2 - 1.25)^2 \times \frac{1}{4} = ( -0.25)^2 \times \frac{3}{4} + (0.75)^2 \times \frac{1}{4} = 0.0625 \times 0.75 + 0.5625 \times 0.25 = 0.046875 + 0.140625 = 0.1875$$ For $X=2$: $$\sigma^2_{Y|2} = (1 - 1.75)^2 \times \frac{1}{4} + (2 - 1.75)^2 \times \frac{3}{4} = ( -0.75)^2 \times \frac{1}{4} + (0.25)^2 \times \frac{3}{4} = 0.5625 \times 0.25 + 0.0625 \times 0.75 = 0.140625 + 0.046875 = 0.1875$$ 7. **(d) Independence check:** Check if $$f(x,y) = f_X(x) f_Y(y)$$ for all $(x,y)$. For $(1,1)$: $$f_X(1) f_Y(1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ Given $$f(1,1) = \frac{3}{8} = 0.375 \neq 0.25$$ Since $$f(1,1) \neq f_X(1) f_Y(1)$$, $X$ and $Y$ are **not independent**. **Final answers:** - (a)(i) $f_{X|Y}(x|y)$ as above. - (a)(ii) $f_{Y|X}(y|x)$ as above. - (b)(i) $\mu_{X|1} = 1.25$, $\mu_{X|2} = 1.75$. - (b)(ii) $\mu_{Y|1} = 1.25$, $\mu_{Y|2} = 1.75$. - (c) $\sigma^2_{Y|1} = 0.1875$, $\sigma^2_{Y|2} = 0.1875$. - (d) $X$ and $Y$ are not independent.