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

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!

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 $$n \ge 1$$ $$\frac{\mathbb E[X^{n}]}{\mathbb E[X^{n-1}]} = \frac{n! \lambda^{n-1}}{(n-1)! \lambda^{n}} = \frac{n}{\lambda}$$