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Suppose $x_1$ and $x_2$ are independent Poisson random variables with parameters equal to $\lambda_1$ and $\lambda_2$ respectively. Show the sum of $x_1$ and $x_2$ is also a Poisson random variable using the moment generating function.

I know the function for Poisson is $\frac{\lambda^x e^{-\lambda}}{x!}$ and moment for it is $M_x(t) = e^{\lambda(e^t - 1)}$.

So I have to show $M_{x_1 + x_2}(t)$ is of the form $e^{k(e^t - 1)}$. But I don't know where to plug in $\frac{\lambda^{x_1} e^{-\lambda}}{x_1!} + \frac{\lambda^{x_2} e^{-\lambda}}{x_2!}$ into the moment function.

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2 Answers 2

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By a basic "law of exponentiation" we have $$e^{\lambda_1(e^t-1)}e^{\lambda_2(e^t-1)}=e^{(\lambda_1+\lambda_2)(e^t-1)}.$$

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  • $\begingroup$ I specifically used the mgf. $\endgroup$ May 17, 2014 at 1:16
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You have to use the fact that if the random variables are independent, then the MGF of the sum is the product of the MGFS.

$$M_{X_1+X_2}(t) = M_{X_1}(t) M_{X_2}(t).$$

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  • $\begingroup$ Do you know where I can find a proof for this? $\endgroup$ May 17, 2014 at 1:30
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    $\begingroup$ It follows from the definition. Since they're independent, $$E(e^{t(X+Y)}) = E(e^{tX})E(e^{tY})$$ $\endgroup$
    – Fred
    May 17, 2014 at 1:34

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