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This is a simple question but I'm not sure what to do when we're not summing the distributions from $i=1$ to $i=n$.

Let's assume $X_1, \dots, X_n$ are i.i.d observations sampled from a Poisson distribution with parameter $\lambda > 0$. Let $S = \sum_{i=1}^{n} X_i$ and $W = \sum_{i=2}^{n} X_i$.

What are the marginal distributions of $S$ and $W$?

I think the marginal distribution of $S$ is $e^{-\lambda} \sum_{i=1}^{n} \dfrac{\lambda^{x_i}}{{x_i}!}$. But what about W? Do I just take the marginal distribution of $S$ and subtract one Poisson random variable from it?

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Since $S$ is the sum of $n$ Poisson random variables with parameter $\lambda$, $S$ should be poisson with parameter $n\lambda$. By similar logic, $W$ is the sum of $n-1$ Poisson random variables with parameter $\lambda$, so $W$ must also be Poisson with parameter $(n-1)\lambda$.

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  • $\begingroup$ Ah right, so it's the parameter that changes, not the form of the distribution? Ok, that makes sense, thanks. I'll just wait for a few more upvotes for you and then I'll accept it. $\endgroup$ – meenaparam Jul 27 '18 at 17:32

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