Show that, if $X_{k}\sim\text{Poisson}(\lambda_{k})$ and they are independent for $1 \leq k \leq n$, then \begin{align*} Y = \displaystyle\sum_{k=1}^{n}X_{k}\sim\text{Poisson}\left(\sum_{k=1}^{n}\lambda_{k}\right) \end{align*}


To begin with, I am going determine the moment generating function of $X\sim\text{Poisson}(\lambda)$: \begin{align*} M_{X}(t) = \textbf{E}[e^{tX}] = \sum_{k=0}^{\infty}e^{tk}e^{-\lambda}\frac{\lambda^{k}}{k!} = e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^{t})^{k}}{k!} = e^{-\lambda}e^{\lambda e^{t}} = \exp(-\lambda(1 - e^{t})) \end{align*}

Once these random variables are independent, the moment generating function of its sum is given by the product of each moment generating function: \begin{align*} M_{Y}(t) = \prod_{k=1}^{n}\exp(-\lambda_{k}(1-e^{t})) = \exp\left(-\sum_{k=1}^{n}\lambda_{k}(1-e^{t})\right) \end{align*}

which means that $\displaystyle Y\sim\text{Poisson}\left(\sum_{k=1}^{n}\lambda_{k}\right)$, because random variables that have the same generating function have the same distribution.

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    $\begingroup$ Looks good to me. $\endgroup$ – kimchi lover Feb 5 at 22:51
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    $\begingroup$ Can't see anything wrong with this. Is there anything in particular that you are not sure about? $\endgroup$ – Mars Plastic Feb 5 at 22:57
  • $\begingroup$ I am particularly interested in other solutions not involving moment generating functions. I know that the convolution theorem may be applied, but are there other approaches? $\endgroup$ – APC89 Feb 5 at 22:59
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    $\begingroup$ Besides MGF's (and by extension, characteristic functions) and convolutions, there really isn't any other elementary, slick way to prove this. $\endgroup$ – Tom Chen Feb 5 at 23:09
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    $\begingroup$ Another method would be to use induction and convolutions. This is equally easy in this case. $\endgroup$ – Kavi Rama Murthy Feb 5 at 23:31

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