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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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    $\begingroup$ May be start with 2, then n. What is the sum of two independent poisson random variables? $\endgroup$ – vinux Apr 26 '13 at 6:36
  • $\begingroup$ I still don't get it.. Isn't the sum of the two independent poisson random variables the product of its moment generating functions? $\endgroup$ – Sue Apr 26 '13 at 6:42
  • $\begingroup$ No, the MGF of the distribution of the sum is the product of the MGFs. The MGF of the Poisson is simple. If you have trouble, you might like to let $a = (e^t - 1)$ and see how you go, then put $e^t-1$ back in for $a$ at the end. $\endgroup$ – Glen_b Apr 26 '13 at 6:50
  • $\begingroup$ @Sue, Through valid MGF (moment generating function) one can uniquely identify the random variable. follow Glen's correction. $\endgroup$ – vinux Apr 26 '13 at 6:57
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    $\begingroup$ The issue here is that the Answer given to Sue is not an answer to the question she has been asked: What is the distribution of their sum: $\sum_{i=1}^n X_i$? The correct answer to the question that she has been asked is that the sum has a Poisson distribution with parameter $\lambda=\sum_{i=1}^n \lambda_i$ and not The moment generating function of Poisson(sum of $\lambda$ ); that is the answer to a different question: What is the MGF of $\sum_{i=1}^n X_i$? $\endgroup$ – Dilip Sarwate Apr 26 '13 at 11:19
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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.

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