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 $Z$


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!


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

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

$f_X(x) =$ enter image description here

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 $z - \tau > 0 \to 0 < \tau < z$

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

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


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.