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).


2 Answers 2


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

  • $\begingroup$ @Christoph.. thanks.. now I feel quite stupid :p $\endgroup$
    – lucacerone
    Aug 20, 2013 at 23:54

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 $\Pr(\text{even}) = \dfrac{1+e^{-2\lambda}}{2}$.

  • $\begingroup$ thanks Michael, I think that the hint from Christoph is slightly simpler, that's why I voted his answer, but yours is a nice explanation as well! $\endgroup$
    – lucacerone
    Aug 20, 2013 at 23:59
  • $\begingroup$ just a note.. for this to work one should also show that the initial series converges uniformly so that the sum of the derivatives also converges to $f'$ right? $\endgroup$
    – lucacerone
    Aug 21, 2013 at 0:07
  • $\begingroup$ That depends on what you've already proved. I think this series converges uniformly on bounded sets but doesn't converge uniformly on its whole domain, but uniform convergence on bounded sets is enough. In lots of books you can find a theorem stating that power series can be differentiated term-by-term. Definitely there is more to proving that, than there is to proving the same thing about finite sums. $\endgroup$ Aug 21, 2013 at 0:10
  • $\begingroup$ Thanks @Michael, I was just asking because I haven't "worked" on this for a while (a few years) and like to see what I can remember. $\endgroup$
    – lucacerone
    Aug 21, 2013 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.