Question

Find the mean and variance of distribution

$f\left(x\right)=\{\begin{array}{cc}3e^{-x}& 0<x<\infty \\ 0& \text{elsewhere}\end{array}$

Answer 1

$f\left(x\right)=\{\begin{array}{c}3e^{-x},0<x<\infty \\ 0,-\infty <x\le 0\end{array}$

$Mean={\int }_{-\infty }^{\infty }xf\left(x\right)dx$

$\mu =$ ${\int }_{-\infty }^{0}x.0dx+{\int }_0^{\infty }3x{e}^{-x}dx$

$\mu =$ $0-$ $3{\left|x{e}^{-x}\right|}_0^{\infty }-3{\int }_0^{\infty }-e^{-x}dx$

$\mu =0+(-3){\left|e^{-x}\right|}_0^{\infty }$

$\mu =3$

$Variance={\int }_{-\infty }^{\infty }{(x-\mu )}^{2}f\left(x\right)dx$

${\sigma }^2=$ $0+{\int }_0^{\infty }{(x-3)}^2.3e^{-x}dx$

${\sigma }^2=$ ${\int }_0^{\infty }3x^2{e}^{-x}dx+{\int }_0^{\infty }27e^{-x}dx-{\int }_0^{\infty }6.3x{e}^{-x}dx$

${\sigma }^2=$ $6-27{\left|e^{-x}\right|}_0^{\infty }-6\ast 3$

${\sigma }^2=15$