Solutions - 5260-02

Integrate the joint PDF and solve for $c$:

$$\begin{array}{c}{\int }_0^1{\int }_0^{y}cxydxdy=1\\ \\ \gg \gg {\int }_0^1\frac{c{y}^3}{2}dy\gg \gg \frac{c}{8}=1\gg \gg c=8\end{array}$$

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(a)

Integrate the joint PDF with respect to $y$ to obtain $f_X$:

$$\begin{array}{c}{f}_X(x)={\int }_x^{1}8xydy\\ \\ \gg \gg 8x{\left[\frac{y^2}{2}\right]|}_x^1\gg \gg 4x(1-x^2)\end{array}$$

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(b)

Apply the conditional density formula:

$$\begin{array}{c}{f}_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}\\ \\ \gg \gg \frac{8xy}{4x(1-x^2)}\gg \gg \frac{2y}{1-x^2},x<y\end{array}$$

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(c)

(1) Plug in $x=0.5$ into the conditional density from part (b):

$$\begin{array}{c}{f}_{Y|X}(y|0.5)=\frac{2y}{1-{0.5}^2}\gg \gg \frac{8y}{3}\end{array}$$

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(2) Compute the conditional expectation:

$$\begin{array}{c}E[Y|X=0.5]={\int }_{0.5}^1\frac{8y^2}{3}dy\\ \\ \gg \gg \frac{8}{3}{\left[\frac{y^3}{3}\right]|}_{0.5}^1\gg \gg {\displaystyle \frac{7}{9}}\end{array}$$

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(d)

Integrate the conditional density from part (b) to find $E[Y|X=x]$:

$$\begin{array}{c}E[Y|X=x]={\int }_x^1\frac{2y^2}{1-x^2}dy\\ \\ \gg \gg \frac{2}{1-x^2}{\left[\frac{y^3}{3}\right]|}_x^1\gg \gg {\displaystyle \frac{2(1-x^3)}{3(1-x^2)}}\end{array}$$