TITLE: Probability that an integer number having Poisson distribution is even
QUESTION [10 upvotes]: The probability $P(X=n)$ that an event X takes place $n$ times in a fixed period of time follows the Poisson distribution with parameter $\lambda$ i.e.
$$ P(X = n) = e^{-\lambda} \frac{\lambda ^ n}{n!}$$
I have to evaluate the probability that the event $X$ takes place an even number of times.
I know that: 
$$ P(X \text{ is even} ) = e^{-\lambda} \cdot \sum_{k=0}^{+\infty} \frac{\lambda^{2k}}{(2k)!}$$
but I can't solve the series.
I guess that I have to use the fact that $e^\lambda = \sum_{n = 0}^{+\infty} \lambda^n/n!$, but I got stuck.
How can I evaluate $P(X \text{ is even})$ (alternative solutions appreciated).

REPLY [14 votes]: Hint:
$$
e^x + e^{-x} = \sum_{n=0}^\infty \frac{x^n}{n!} + \sum_{n=0}^\infty \frac{(-x)^n}{n!}
= 2\cdot\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}
$$

REPLY [7 votes]: Let $\displaystyle f(\lambda)= \sum_{k\ge0}\frac{\lambda^{2k}}{(2k)!}$.
Then $f''(\lambda) = f(\lambda)$.  Solving this differential equation and applying the initial conditions $f(0)=1$ and $f'(0)=0$, we get
$$
f(\lambda) = \frac{e^\lambda+e^{-\lambda}}{2} = \cosh(\lambda).
$$
So $\Pr(\text{even}) = \dfrac{1+e^{-2\lambda}}{2}$.