Q. 5.19.TE - A First Course in Probability

Question

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

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

Step-by-Step Solution

Answer

The statement has been proved true, i.e.

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

Step 1: 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.