# Distribution of $Z$ from Moment Generating Function

Suppose that $X_1, X_2, ..., X_n$ are independent and identically distributed Exp(λ) random variables and let $Z = X_1 + X_2 + · · · + X_n$.

Determine $M_Z(θ)$, the moment generating function of $Z$ and hence, or otherwise, determine the distribution of $Z$

Working:

I know that $M_z(\theta)=\left(\frac{\lambda}{\lambda - \theta}\right)^n$. But I don't even know where to start after that! Any help is appreciated!

## 1 Answer

$$f_{Z}(z) = \frac{z^{n-1} e^{-\lambda z} \lambda^n}{(n-1)!}$$

Pf 1: (I'm going to use t instead of $\theta$) Given that $M_{Z}(t) = (\frac{\lambda}{\lambda - t})^n$ for $t < \lambda$, compare that to the mgf of a gamma pdf: $(1-t\theta)^{-k}$ for $t < \frac{1}{\theta}$.

If $X$ ~ Gamma$(k, \theta)$, then

$f_X(x) =$

Now let $\lambda = \frac{1}{\theta}$ and $k = n \ \text{QED}$.

Pf 2 (using convolution and mathematical induction):

Let $Z = S_n = \sum_{k=1}^n X_k$.

Step 1: $n = 1$

$S_1 = X_1$ has pdf (for $z > 0$):

$$f_{S_1}(z) = f_{X_1}(z)=\frac{z^{n-1} e^{-\lambda z} \lambda^n}{(n-1)!}\mid_{n=1}$$

Step 2: $n = k \to n = k + 1$

We must show that if $Z = S_n$ has pdf

$$f_{Z}(z) = f_{S_n}(z) = \frac{z^{n-1} e^{-\lambda z} \lambda^n}{(n-1)!},$$

then $S_{n+1}$ has pdf

$$f_{S_{n+1}}(z) = \frac{z^{n} e^{-\lambda z} \lambda^{n+1}}{n!}:$$

Since $S_{n+1} = S_n + X_{n+1}$, we can use convolution to determine the pdf of $S_{n+1}$:

$$f_{S_{n+1}}(z) = \int_\mathbb R f_{S_n}(\tau)f_{X_1}(z - \tau)\ \mathsf d\tau$$

Note that just as $z > 0$ earlier, we have $\tau > 0$ and $z - \tau > 0 \to 0 < \tau < z$

$$= \int_{0}^{z} \frac{\tau^{n-1} e^{-\lambda \tau} \lambda^n}{(n-1)!} e^{-\lambda (z-\tau)} \lambda d\tau$$

$$= \frac{z^{n} e^{-\lambda z} \lambda^{n+1}}{n!} \ \text{QED}$$