Question

If $X$is an exponential random variable with a 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

Using the equation of expectation, determine $E\left(X^k\right)$and the Gamma function within the integral.

Step-by-step solution

Find the Exponential variable.

We are provided that$X$is an exponential random variable with a mean $\frac{1}{\lambda }$In other words, we have got that $X~$Expo$\left(\lambda \right)$By the idea about the mean of a function of a random variable quantity, we have that

localid="1649619285130" $$\begin{gathered} E\left(X^k\right)={\int }_{\mathrm{ℝ}}{x}^k{f}_X\left(x\right)dx \\ ={\int }_0^{\infty }{x}^k\lambda e^{-\lambda x}dx \end{gathered}$$

Calculate the integral.

Let's calculate the integral, we've that

${\int }_0^{\infty }{x}^k\lambda e^{-\lambda x}dx=\lambda {\int }_0^{\infty }{x}^k{e}^{-\lambda x}dx$

Making substitutions $s=\lambda x$we've that

localid="1649619305189" $$\begin{gathered} \lambda {\int }_0^{\infty }{x}^k{e}^{-\lambda x}dx=\lambda {\int }_0^{\infty }{\left(\frac{s}{\lambda }\right)}^k{e}^{-s}\frac{ds}{\lambda } \\ =\frac{1}{{\lambda }^k}{\int }_0^{\infty }{s}^k{e}^{-s}ds \\ =\frac{\Gamma (k+1)}{{\lambda }^k} \end{gathered}$$

Finally, we've obtained that

localid="1649619316547" $$\begin{gathered} E\left(X^k\right)=\frac{\Gamma (k+1)}{{\lambda }^k} \\ =\frac{k!}{{\lambda }^k} \end{gathered}$$

Which has been claimed.