Question

5.6. Compute$E\left[X\right]$if $X$has a density function given by

$\left(a\right)f\left(x\right)=\left\{\begin{array}{ll}\frac{1}{4}x{e}^{\frac{-x}{2}}& x>0\\ 0& otherwise\end{array}\right.$;

$\left(b\right)f\left(x\right)=\left\{\begin{array}{ll}c(1-x^2)& -1<x<1\\ 0& otherwise\end{array}\right.$;

$\left(c\right)f\left(x\right)=\left\{\begin{array}{ll}\frac{5}{x^2}& x>5\\ 0& x\le 5\end{array}\right.$.

Short answer

Thus the answer of given question is

$\left(a\right),E\left(X\right)=4$

$\left(b\right),E\left(X\right)=0$

$\left(c\right),E\left(X\right)=\infty$

Step-by-step solution

Given Information.

$f\left(x\right)=\left\{\begin{array}{ll}\frac{1}{4}x{e}^{\frac{-x}{2}}& x>0\\ 0& otherwise\end{array}\right.$

Explanation.

$E\left(x\right)={\int }_{-\infty }^{\infty }x.\frac{1}{4}x{e}^{\frac{-x}{2}}dx$

$=\frac{1}{4}{\int }_0^{\infty }{x}^9{e}^{\frac{-x}{2}}dx$

$\frac{x}{2}=4,x=2y$

$E\left(X\right)={\int }_0^{\infty }2.y^2{e}^{-y}dy=2{\int }_0^{\infty }{y}^{3-1}{e}^{-1}dy$

$=2\times 2=4$

Explanation.

$f\left(x\right)=\left\{\begin{array}{ll}c(1-x^2)& -1<x<1\\ 0& otherwise\end{array}\right.$

${\int }_{-\infty }^{\infty }(1-x^2)dx=1,{\int }_{-\infty }^{\infty }c(1-x^2)dx=1$

$c.\frac{4}{3}=1\Rightarrow c=\frac{3}{4}$

$E\left(X\right)={\int }_{-\infty }^{\infty }x.\frac{3}{4}(1-x^2)dx=\frac{3}{4}{\int }_{-1}^1(x-\frac{3}{4})dx$

$=\frac{3}{4}{{\left[\frac{x^2}{2}-\frac{x^4}{4}\right]}^1}_{-1}$

$=\frac{3}{4}\times 0=0$

Explanation.

$E\left(X\right)={\int }_{-\infty }^{\infty }\frac{5}{x^0}dx={\int }_{}^{}\frac{5}{0x}dx=5$

$=5{{\left|\frac{u}{x^1}\right|}_5}^{\infty }$

$=\infty$