# What is the variance of a probability distribution function of the form: f(x)=ke^(-2x)?

E ( ${x}^{2}$ )= 1/2 $\implies$ V (x) = E ( ${x}^{2}$) - ${\left\{E \left(x\right)\right\}}^{2}$ - 1/2 - ${\left(\frac{1}{2}\right)}^{2}$ = 1/2 - 1/4 = 1/4.
The limit of the distribution is (0, $\infty$ ) To find k,
${\int}_{0}^{B}$ k ${e}^{-} \left(2 x\right)$ dx = k $\Gamma$ (1)/ 2 = 1 $\implies$ k/2 = 1 $\implies$ k = 2.