From Williams' Probability w/ Martingales:
Re $E[f(S_n)]$, how do I obtain $f_{S_n}(s)$? It seems that
$$f_{S_n}(s) = \frac{s^{n-1} e^{-\lambda s} \lambda^n}{(n-1)!}$$
I tried computing $M_{S_n}(t) = (\frac{\lambda}{\lambda - t})^n$ for $t < \lambda$ and then compared it to the mgf of a gamma function: $(1-t\theta)^{-k}$ for $t < \frac{1}{\theta}$ by letting $\lambda = \frac{1}{\theta}$ and $k = n$ and that seems to work, but is there a way to do it from using only CDFs? I was thinking:
$$F_{S_n}(s) = P(S_n \le s)$$
$$ = P(X_1 + ... + X_n \le s)$$
$$ = \int_{0}^{a_1} \int_{0}^{a_2} \cdots \int_{0}^{a_n} \lambda^n e^{-\lambda(x_1 + \cdots + x_n)} dx_1 dx_2 \cdots dx_n$$
where $a_1 + a_2 + ... + a_n = s$
$$ = \int_{0}^{a_1} \int_{0}^{a_2} \cdots \int_{0}^{a_n} \lambda^n e^{-\lambda(s_n)} dx_1 dx_2 \cdots dx_n$$
or something like that and then differentiate $F_{S_n}(s)$.
No need for moment generating functions or Laplace transforms; this follows from a simple induction argument.
Let $X_k\stackrel{\mathrm{i.i.d.}}\sim\operatorname{Exp}(\lambda)$ and $S_n=\sum_{k=1}^n X_k$. Then $X_1$ has density $$ f_{S_1}(t) = f_{X_1}(t)=\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t) = \frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t).$$
Now, assuming that $S_n$ has density
$$f_{S_n}(t) = \lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t) = \frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t)$$
for a positive integer $n$ and since we have $S_{n+1}=S_n+X_{n+1}$, by convolution the density of $S_{n+1}$ is, \begin{align} f_{S_{n+1}}(t) &= (f_{S_n}\star f_{X_1})(t)\\ &= \int_\mathbb R f_{S_n}(\tau)f_{X_1}(t-\tau)\ \mathsf d\tau\\ &= \int_\mathbb R \frac{(\lambda \tau)^{n-1}}{(n-1)!}\lambda e^{-\lambda \tau}\mathsf 1_{(0,\infty)}(\tau)\lambda e^{-\lambda(t-\tau)}\mathsf 1_{(0,\infty)}(t-\tau)\ \mathsf d\tau\\ &= \frac{\lambda^{n+1}}{(n-1)!}e^{-\lambda t}\mathsf 1_{(0,\infty)}(t)\int_0^t \tau^{n-1}\ \mathsf d\tau\\ &= \frac{(\lambda t)^n}{n!}\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t). \end{align}