Moment generating function of Erlang Distribution

Question

I was looking to show the result for the MGF of the Erlang Distribution, and got stuck whilst computing the integral:

$$\\ \int e^{-st} \frac{\lambda^k t^{k-1}e^{-\lambda t}}{(k-1)!} dt$$

I know the result should evaluate to

$$\\ \left( \frac{\lambda}{\lambda+s}\right)^k $$ but I am unsure as to how to go about tackling the initial integral. I initially started by integrating by parts, but whilst the (k-1)! term appeared nicely I didn't see how the other terms would cancel/rearrange as required. Any help would be appreciated. Thanks.

Answer 1

Let $I_k$ be the integral in question, that is, $$ I_k = \int_0^\infty e^{-st} \frac{\lambda^k t^{k-1} e^{-\lambda t}}{(k-1)!} \, dt $$ and use induction on $k$: For $k=1$, we have $$ I_1 = \lambda \int_0^\infty e^{-(s+\lambda)t} \, dt = \lambda \left[-\frac{e^{-(s+\lambda) t}}{s+\lambda}\right]_0^\infty = \frac{\lambda}{s+\lambda} $$ assuming that $s+\lambda>0$. Then, using integration by parts with $u = t^k$ and $dv = e^{-(s+\lambda) t} dt$, we obtain \begin{align} I_{k+1} &= \frac{\lambda^{k+1}}{k!} \int_0^\infty t^k e^{-(s+\lambda)t} \, dt \\ &= \frac{\lambda^{k+1}}{k!} \left[ \left. t^{k-1} e^{-(s+\lambda) t} \right|_0^\infty + \frac{k}{s+\lambda} \int_0^\infty t^{k-1} e^{-(s+\lambda)t} \, dt \right] \\ &= \frac{\lambda}{s+\lambda} \int_0^\infty e^{-st} \frac{\lambda^k t^{k-1} e^{-\lambda t}}{(k-1)!} \, dt \\ &= \frac{\lambda}{s+\lambda} I_k \end{align} which completes the inductive step.

Answer 2

If $Y\sim\mathrm{Erlang}(k,\lambda)$ then $Y=\sum_{i=1}^k X_i$ where $X_1,\ldots,X_k\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Expo}(\lambda)$. We compute: \begin{align} M_{X_1}(t) :&= \mathbb E[e^{tX_1}]\\ &= \int_0^\infty e^{st}\lambda e^{-\lambda s}\ \mathsf ds\\ &= \frac\lambda{\lambda-t}, \ t<\lambda. \end{align} Since the moment-generating function of the sum of independent random variables is the product of their moment-generating functions, it follows readily that \begin{align} M_Y(t) :&= \mathbb E[e^{tY}]\\ &= \mathbb E[e^{t\sum_{i=1}^k X_i}]\\ &= \prod_{i=1}^k \mathbb E[e^{tX_i}]\\ &= M_{X_1}(t)^k\\ &= \left(\frac\lambda{\lambda-t}\right)^k,\ t<\lambda. \end{align}

Now, in your integral you used $e^{-st}$ as opposed to $e^{st}$, which would make this the Laplace-Stieltjes transform as opposed to the moment-generating function. That would make the result $$ \mathbb E[e^{-tY}] = \left(\frac\lambda{\lambda+t}\right)^k,\ t>-\lambda. $$