Question

If $\mathit{X}$is an exponential random variable with a mean $\frac{\mathbf{1}}{\mathbf{\lambda }}$, show that

$\mathit{E}\mathbf{\left[}{\mathbf{X}}^{\mathbf{k}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{k}\mathbf{!}}{{\mathbf{\lambda }}^{\mathbf{k}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{k}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots }$

Short answer

The statement has been proved true, i.e.

$E\left(X^k\right)=\frac{k!}{{\lambda }^k}\text{;}k=1,2,\mathrm{...}$

Step-by-step solution

Given information.

$X$is an exponentially distributed random variable with a mean$\left(\frac{1}{\lambda }\right)$

Step 2. Defining X.

The probability density function of a random variable $X$ is

$f\left(x\right)=\frac{1}{\lambda }{e}^{-x/\lambda }\text{;}x>0$

Step 3. Calculation.

Transforming the above variable to Gamma distribution and finding the ${\text{k}}^{\text{th}}$raw moment, we get-

$\begin{array}{l}E\left(X^k\right)=\frac{1}{\Gamma \left(t\right)}\underset{0}{\overset{\infty }{\int }}{x}^k\lambda e^{-\lambda x}{\left(\lambda x\right)}^{t-1}dx\\ =\frac{{\lambda }^{-k}}{\Gamma \left(t\right)}\underset{0}{\overset{\infty }{\int }}\lambda e^{-\lambda x}{\left(\lambda x\right)}^{t+k-1}dx\\ =\frac{{\lambda }^{-k}}{\Gamma \left(t\right)}\Gamma (t+k)\text{;}k=1,2,\mathrm{...}\end{array}$

Putting $t=1$yields an exponential distribution, therefore, the final expression becomes

$\begin{array}{l}=\frac{{\lambda }^{-k}}{\Gamma \left(1\right)}\Gamma (t+1)\\ E\left(X^k\right)=\frac{k!}{{\lambda }^k}\text{;}k=1,2,\mathrm{...}\end{array}$

which proves the required expression.