I know that this question has been asked several times, and to my knowledge I have read through all the versions. However, I can't find a complete proof, and I am having trouble finishing the algebraic manipulation. If anyone could help, I'd really appreciate it.
We have a r.v. $X$ that is poisson distributed with $\lambda$. So:
$P[X=k]=e^{-\lambda}\frac{\lambda^k}{k!}$ for $k=0,1,\dots$.
We want to find that the probability of $X$ taking on an even value.
So far, I have:
$$e^{t}+e^{-t}=\sum_{k=0}^{\infty} \frac{t^k}{k!}+\sum_{k=0}^{\infty}\frac{(-t)^k}{k!}$$
$$=2\sum_{k=0}^{\infty}\frac{t^{2k}}{(2k)!}$$
I know that the final answer is $$\frac{1+e^{-2\lambda}}{2}$$, but I am struggling with getting there.
Best Answer
The probability you want is
$$ P=\sum_{k \,{\rm even}}P(X=k)= \sum_{j=0}^{\infty} e^{-\lambda}\frac{\lambda^{2j}}{(2j)!}$$
But you know that, for any $\lambda$
$$ \sum_{j=0}^{\infty} \frac{\lambda^{2j}}{(2j)!}= \frac{e^{\lambda}+e^{-\lambda}}{2}$$
Hence... (what's your problem?)...