Question

Find the mean and variance for the probability density function : $f\left(x\right)=\{\begin{array}{c}\alpha e^{-\alpha x}\text{if}x>0\\ 0\text{otherwise}\end{array}$

Answer 1

$f\left(x\right)=\{\begin{array}{c}\alpha e^{-\alpha x}\text{if}x>0\\ 0\text{otherwise}\end{array}$
$E\left(X\right)={\int }_{-\infty }^{\infty }xf\left(x\right)dx={\int }_0^{\infty }x\left(\alpha \right)\left(e^{-\alpha x}\right)dx$
$=\alpha {\int }_0^{\infty }x{e}^{-\alpha x}dx=\frac{\alpha \left(1\right)}{{\alpha }^2}=\frac{1}{\alpha }$
[By Bernoulli's formula ${\int }_0^{\infty }{x}^n{e}^{-\alpha x}dx=\frac{n}{a^{n+1}}$]
$E\left(X^2\right)={\int }_0^{\infty }{x}^2\left(\alpha e^{-\alpha x}\right)dx$
$\left[{\int }_0^{\infty }{x}^n{e}^{-\alpha x}dx=\frac{n}{a^{n+1}}\right]=\alpha {\int }_0^{\infty }{x}^2{e}^{-\alpha x}dx=\frac{\alpha \left(2\right)}{{\alpha }^3}=\frac{2}{{\alpha }^2}$
[By Bernoulli's formula]
Mean$=E\left(X\right)=\frac{1}{\alpha }$
Variance $=E\left(X^2\right)-{\left[E\right(X\left)\right]}^2=\frac{2}{{\alpha }^2}-\frac{1}{{\alpha }^2}=\frac{1}{{\alpha }^2}$