Solutions - 5220-01

(1) Write the CDF of $W=X+Y$:

Since $0\le y\le x\le 3$, we have $W\in [0,6]$.

$$F_W(w)=P[X+Y\le w]=\underset{\begin{array}{c}0\le y\le x\le 3\\ x+y\le w\end{array}}{\iint }\frac{8}{81}xydydx$$

For fixed $x\in [\mathrm{0,3}]$, $y$ ranges from $0$ to $-x+w$ when $x<w/2$ and from $0$ to $3$ when $x>w/2$.

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(2) Evaluate $F_W$ for $0\le w\le 3$:

$$\begin{array}{c}{F}_W(w)={\int }_0^{w/2}{\int }_0^x\frac{8}{81}xydydx+{\int }_{w/2}^w{\int }_0^{-x+w}\frac{8}{81}xydydx\\ \\ \gg \gg \frac{4}{81}{\int }_0^{w/2}{x}^{3}dx+\frac{4}{81}{\int }_{w/2}^{w}x{(-x+w)}^{2}dx\\ \\ \gg \gg \frac{4}{81}\cdot \frac{w^4}{64}+\frac{4}{81}\cdot \frac{5w^4}{192}\gg \gg \frac{w^4}{486}\end{array}$$

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(3) Evaluate $F_W$ for $3<w\le 6$:

$$\begin{array}{c}{F}_W(w)={\int }_0^{w/2}{\int }_0^x\frac{8}{81}xydydx+{\int }_{w/2}^3{\int }_0^{-x+w}\frac{8}{81}xydydx\\ \\ \gg \gg \frac{4}{81}{\int }_0^{w/2}{x}^{3}dx+\frac{4}{81}{\int }_{w/2}^{3}x{(-x+w)}^{2}dx\\ \\ \gg \gg \frac{4}{81}[\frac{w^4}{64}+\frac{9w^2}{2}-18w+\frac{81}{4}-\frac{11w^4}{192}]\\ \\ \gg \gg -\frac{w^4}{486}+\frac{2w^2}{9}-\frac{8w}{9}+1\end{array}$$

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(4) Differentiate for the PDF:

$$f_W(w)=\{\begin{array}{cc}{\displaystyle \frac{2}{243}}{w}^3& 0\le w\le 3\\ -{\displaystyle \frac{8}{9}}+{\displaystyle \frac{4}{9}}w-{\displaystyle \frac{2}{243}}{w}^3& 3<w\le 6\\ 0& \text{else}\end{array}$$