Solutions - 5150-03

(a)

Compute $E[X]$:

$$\begin{array}{c}E[X]={\int }_0^2\frac{3}{4}{x}^2(2-x)dx\\ \\ \gg \gg {[\frac{x^3}{2}-\frac{3}{16}{x}^4]|}_0^2\gg \gg 1\end{array}$$

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(b)

(1) Find the CDF of $X$:

$$\begin{array}{c}{F}_X(x)={\int }_{-\infty }^x{f}_X(t)dt\\ \\ \gg \gg {\int }_0^x\frac{3}{4}t(2-t)dt\gg \gg {[\frac{3}{4}{t}^2-\frac{t^3}{4}]|}_0^x\\ \\ \gg \gg \frac{3}{4}{x}^2-\frac{1}{4}{x}^3\end{array}$$

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(2) Find the CDF of $X^2$:

Since $X^2$ is monotone increasing, $F_{X^2}(x)=F_X\left(\sqrt{x}\right)$.

$$F_X\left(\sqrt{x}\right)=\{\begin{array}{cc}0& x<0\\ \frac{3}{4}x-\frac{1}{4}{x}^{\frac{3}{2}}& 0\le x<4\\ 1& 4\le x\end{array}$$

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(3) Find the PDF of $X^2$ by differentiating:

$$f_{X^2}(x)=\{\begin{array}{cc}0& x<0\\ \frac{3}{4}-\frac{3}{8}{x}^{\frac{1}{2}}& 0\le x<4\\ 0& 4\le x\end{array}$$

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(c)

Find $E\left[X^2\right]$:

$$\begin{array}{c}E\left[X^2\right]={\int }_0^{4}x(\frac{3}{4}-\frac{3}{8}{x}^{\frac{1}{2}})dx\\ \\ \gg \gg {[\frac{3}{8}{x}^2-\frac{3}{20}{x}^{\frac{5}{2}}]|}_0^4\gg \gg \frac{6}{5}\end{array}$$

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(d)

Compute $\text{Var}[X]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2\\ \\ \gg \gg \frac{6}{5}-1\gg \gg \frac{1}{5}\end{array}$$