Solutions - 5250-01

(a)

(1) Compute the marginal $f_Y(y)$:

$$\begin{array}{c}{f}_Y(y)=\frac{12}{5}{\int }_0^1(2x-x^2-xy)dx\\ \\ \gg \gg \frac{12}{5}{[x^2-\frac{x^3}{3}-\frac{x^{2}y}{2}]|}_0^1\\ \\ \gg \gg \frac{12}{5}(\frac{2}{3}-\frac{y}{2})\end{array}$$

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(2) Apply the conditional density formula:

$$\begin{array}{c}{f}_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\ \\ \gg \gg \frac{\frac{12}{5}x(2-x-y)}{\frac{12}{5}(\frac{2}{3}-\frac{y}{2})}\gg \gg \frac{6x(2-x-y)}{4-3y}\end{array}$$ $$f_{X|Y}(x|y)=\{\begin{array}{cc}{\displaystyle \frac{6x(2-x-y)}{4-3y}}& x,y\in [0,1]\\ 0& \text{otherwise}\end{array}$$

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(b)

Integrate the conditional density over $(\frac{1}{2},1)$:

$$\begin{array}{c}P[X>\frac{1}{2}|Y=y]={\int }_{1/2}^1\frac{6x(2-x-y)}{4-3y}dx\\ \\ \gg \gg \frac{6}{4-3y}{\int }_{1/2}^1(2x-x^2-xy)dx\\ \\ \gg \gg \frac{6}{4-3y}{[x^2-\frac{x^3}{3}-\frac{x^{2}y}{2}]|}_{1/2}^1\\ \\ \gg \gg {\displaystyle \frac{11-9y}{4(4-3y)}}\end{array}$$ $$P[X>\frac{1}{2}|Y=y]=\{\begin{array}{cc}{\displaystyle \frac{11-9y}{4(4-3y)}}& y\in [0,1]\\ 0& \text{otherwise}\end{array}$$