Moment generating functions question 
Suppose that $X_1, X_2, ..., X_n$ are independent, where each $X_i$ has probability (mass) function $p_i(x_i)$ given as $p_i(x_i) = \frac{e^{-\lambda} \lambda_i^{x_i}}{x_i!}$ (only the parameter $\lambda_i$ differs int he distribution of each $X_i$ for $x_i = 0, 1, ...$. What is the distribution of their sum: $\sum\limits_{i=1}^n X_i$ ? Prove it (perhaps with moment generating functions). 

Answer: The moment generating function of Poisson (sum of $\lambda$)
I don't get how to do this question and I don't really understand the question. Can someone please help me? Thanks in advance.
 A: The MGF is $\psi_{X_i}(t)=e^{\lambda_i(e^t-1)}$. The MGF of $S_n$, $S_n=\sum_{i=1}^n X_i$ is $\psi_{S_n}(t)=\prod_{i=1}^n \psi_{X_i}(t)$. So what we get is
$$
\prod_{i=1}^n \psi_{X_i}(t)=\prod_{i=1}^ne^{\lambda_i(e^t-1)}=e^{\sum_{i=1}^n\lambda_i(e^t-1)}=e^{\lambda^*(e^t-1)}
$$
where $\lambda^*=\sum_{i=1}^n\lambda_i$. Since the result is the MGF of a Poisson with parameter $\lambda^*$, the conclusion is that the sum of independent Poisson random variables with different parameters is a Poisson random variable with the sum of their individual parameters as its parameter.
Edit: If it wasn't clear enough, the conclusion is this:
$$
\text{If }X_i \text{ are independently distributed, } X_i\sim Po(\lambda_i), \, i=1, \dots, n\\
\text{then } \sum_{i=1}^n X_i=S_n \sim Po\left(\sum_{i=1}^n\lambda_i\right).
$$
A special case of this relation is very often used, namely when the $X_i$ are not only independently but also identically distributed. In that case, $S_n$ is $Po(n\lambda)$. As initially shown, this is quite easily shown using moment generating functions.
