# Moment generating function using probability function?

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.

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Here's one way to think about it without using moment generating functions:

For simplicity let's take the case where $n = 2$. So, for example, let $X$ and $Y$ be independent, $X$ be Poisson with parameter $\lambda_1$ and $Y$ be Poisson with parameter $\lambda_2$. And let $W = X+Y$. Then we want $p_W(w)$. We have:

$P(W = w) = P(X+Y = w) = \sum_{k=0}^w P(X = k, Y = w-k), \quad w = 0, 1, 2, \dots$.

That is, we sum up all of the possible ways that $x+y$ can equal $w$.

Since $X$ and $Y$ are independent we then have

\begin{align*} P(W = w) &= \sum_{k=0}^w P(X = k)P(Y = w-k) \\[5pt] &=\sum_{k=0}^w \frac{e^{-\lambda_1} \lambda_1^k }{k!} \cdot \frac{e^{-\lambda_2} \lambda_2^{w-k} }{(w-k)!} \\[5pt] &=e^{-\lambda_1}e^{-\lambda_2}\sum_{k=0}^w \frac{ \lambda_1^k\lambda_2^{w-k} }{k!(w-k)!} \\[5pt] &=\frac{e^{-(\lambda_1+\lambda_2)}}{w!}\sum_{k=0}^w \binom{w}{k} \lambda_1^k\lambda_2^{w-k} \quad \text{(we multiply and divide by $w!$)}\\[5pt] &=\frac{e^{-(\lambda_1+\lambda_2)}}{w!}\cdot(\lambda_1 + \lambda_2)^w \quad \text{(by the Binomial thm)} \\[5pt] &=\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1 + \lambda_2)^w}{w!} \end{align*}

which is the mass function of a Poisson random variable with parameter $\lambda_1 + \lambda_2$.

Hope that helps.

EDIT: For the sum of many independent random variables, it's actually often easier to use moment generating functions. Here's a hint:

The moment generating function of the sum of independent random variables is the product of their moment generating functions (which is one of the reasons moment generating functions are nice).

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