Solutions - 5120-05

(a)

$$f_X(x)=\{\begin{array}{cc}\frac{1}{2}& -1\le x<0\\ \frac{1}{2}& 2\le x<3\\ 0& \text{otherwise}\end{array}$$

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(b)

First find expectation of $X$:

$$\begin{array}{c}E[X]={\int }_{-\infty }^{\infty }xf(x)dx\\ \\ \gg \gg {\int }_{-1}^0\frac{1}{2}xdx+{\int }_2^3\frac{1}{2}xdx\\ \\ \gg \gg {\frac{1}{4}{x}^2|}_{-1}^0+{\frac{1}{4}{x}^2|}_2^3\\ \\ \gg \gg 0-\frac{1}{4}+\frac{1}{4}(9-4)\gg \gg 1\end{array}$$

To use the variance formula $\mathrm{Var}[X]=E[X^2]-E{[X]}^2$, we also need to find $E[X^2]$. For this we use $g(X)=X^2$ and the formula for $E[g(X)]$:

$$\begin{array}{c}E[X^2]={\int }_{-\infty }^{+\infty }{x}^{2}f(x)dx\\ \\ \gg \gg {\int }_{-1}^0\frac{1}{2}{x}^{2}dx+{\int }_2^3\frac{1}{2}{x}^{2}dx\\ \\ \gg \gg \frac{1}{6}{x}^3{|}_{-1}^0+\frac{1}{6}{x}^3{|}_2^3\\ \\ \gg \gg (0-\frac{1}{6}(-1{)}^3)+(\frac{1}{6}(3{)}^3-\frac{1}{6}(2{)}^3)\\ \\ \gg \gg \frac{1}{6}+\frac{19}{6}\gg \gg \frac{20}{6}\end{array}$$

Therefore:

$$\begin{array}{c}\mathrm{Var}[X]=E[X^2]-E{[X]}^2\\ \\ \gg \gg \frac{20}{6}-{(1)}^2\gg \gg \frac{7}{3}\end{array}$$