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I am trying to find the probability of getting an odd/even number from the Poisson distribution and I found this answer.

I understand the reasoning, but my problem is that I can'r grasp how this expression $\sum_{n = 0}^{\infty}{e^{-\lambda}\lambda^{2n} \over (2n)!}$ is equal to ${1 \over 2}\,e^{-\lambda}\sum_{n = 0}^{\infty}{\lambda^{n} \over n!} + {1 \over 2}\,e^{-\lambda}\sum_{n = 0}^{\infty}{(-\lambda)^{n} \over n!}$.

I am pretty sure there is a simple explanation, but I am not being able to see it.

Vinícius Lopes Simões
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1 Answers1

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You are right, this does have a simple explanation. Every real function $x(\lambda)$ can be written in terms of its even and odd components as follows:

\begin{align*} x(\lambda) = x_e(\lambda) + x_o(\lambda)\\ \end{align*}

The definition for these two new functions are given as follows:

\begin{align*} x_e(\lambda) = \frac{1}{2}x(\lambda) + \frac{1}{2}x(-\lambda)\\ x_o(\lambda) = \frac{1}{2}x(\lambda) - \frac{1}{2}x(-\lambda)\\ \end{align*}

Now, the expansion that you see is simply the even part of your expression. This would give you the cumulative probability of outcomes that are part of an even cumulative distribution. This, however, does not seem like the answer that you are actually looking for, which is: "What is the probability that the individual outcome will have an even value?". At least this is what I am assuming from reading your question.

Gustav Streicher
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