Question

If $X$is an exponential random variable with mean $1/\lambda$, show that

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

Hint: Make use of the gamma density function to evaluate the preceding.

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

Step 1. Defining X

The probability density function of random variable X is

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

Step 2. Calculations

Transforming the above variable to Gamma distribution and finding the k^th raw moment, we get-

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

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

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

which proves the required expression.