1
$\begingroup$

If $X\sim\text{Exp}(\lambda)$, then $\textbf{E}(X^{n}) = \displaystyle n\textbf{E}(X^{n-1})/\lambda$

MY SOLUTION

According to definition of $k$-th moment, we have \begin{align*} \textbf{E}(X^{n}) & = \int_{0}^{+\infty}x^{n}\lambda e^{-\lambda x}\mathrm{d}x = -x^{n}e^{-\lambda x}\biggr|_{0}^{+\infty} + \int_{0}^{+\infty}nx^{n-1}e^{-\lambda x}\mathrm{d}x\\\\ & = 0 + \frac{n}{\lambda}\int_{0}^{+\infty}x^{n-1}\lambda e^{-\lambda x}\mathrm{d}x = \frac{n}{\lambda}\textbf{E}(X^{n-1}) \end{align*}

Could someone provide an alternative proof based on moment generating functions or any other technique? Thanks in advance!

$\endgroup$
1
$\begingroup$

The MGF of the exponential distribution is $$ M(t) = \mathbb E[e^{tX}] = \frac{1}{1-t/\lambda},\ t < \lambda$$ (which, if you don't want to take as "known", you can get by an easy integration). This has Maclaurin series $$\sum_{j=0}^\infty \mathbb t^j E[X^j]/j! = M(t) = \sum_{j=0}^\infty (t/\lambda)^j $$ which says $$\mathbb E[X^j] = j!/\lambda^j$$ so that for $n \ge 1$ $$ \frac{\mathbb E[X^{n}]}{\mathbb E[X^{n-1}]} = \frac{n! \lambda^{n-1}}{(n-1)! \lambda^{n}} = \frac{n}{\lambda}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.