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A poisson random variable has a mean of x=6.25. A random sample of this variable is drawn. What is the probability function for sum S = $\sum_{i=1}^n X_i$, as specifically as possible.

I have no idea how to attack this.

Nick
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  • The sum of independent poisson distributions is a poisson distribution. have you proved this? – Tyler Dec 13 '13 at 16:09
  • I have not. the hint suggests I use the moment generating function somehow, is that involved in the proof? – Nick Dec 13 '13 at 16:16
  • As you know, the mgf of a sum of independent random variables is the product of the mgf. If you apply this to Poisson $X_i$ with parameters $\lambda_1$ to $\lambda_n$, you will get the mgf of a Poisson, parameter $\sum \lambda_i$. Yours is the special case $\lambda_i$ all equal to $6.25$. – André Nicolas Dec 13 '13 at 16:27
  • @AndréNicolas so the 0th moment of the mgf is the probability function for this sum? i don't understand. – Nick Dec 13 '13 at 16:38
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    If two random variables have the same mgf, they have the same distribution. So if you compute the mgf of your sum, and it looks like the mgf of a Poisson, then the distribution is Poisson, so you will have proved what you want to prove, namely that your sum is Poisson with parameter $6.25n$. The $0$-th moment has nothing to do with anything, it is always $1$. – André Nicolas Dec 13 '13 at 16:43

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Recall that a random variable $W$ is Poisson with parameter $\lambda$ if and only if the mgf of $W$ is equal to $\exp(\lambda(e^t-1))$.

Our $X_i$ therefore all have mgf equal to $\exp((6.25)(e^t-1))$.

The mgf of a sum $X_1+X_2+\cdots+X_n$ of independent random variables is the product of the individual mgf.

Thus in our case the mgf of $X_1+X_2+\cdots+X_n$ is $\left(\exp((6.25)(e^t-1))\right)^n$.

This is equal to $\exp((6.25n)(e^t-1))$. That has shape $\exp(\lambda(e^t-1))$, where $\lambda=6.25n$.

Thus by the first paragraph of this post, $X_1+X_2+\cdots+X_n$ has Poisson distribution with parameter $6.25n$.

If we want to be really specific, we can write $$\Pr(S=k)=e^{-6.25n}\frac{(6.25n)^k}{k!}.$$

Remark: The moment generating function, and its close relative the characteristic function, can be very useful tools in recognizing that a random variable has a certain distribution. To use these tools, we need to have a familiarity with the mgf (or characteristic functions) of commonly occurring distributions.

André Nicolas
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