Given $Y \sim Pois(\mu)$ a random variable with mean $\mu > 0$.
$\Rightarrow \ P(Y=y) = e^{-\mu} \ \dfrac{\mu^{-y}}{y!}$
Prove that $Y$ is even?
I started my proof like this
$Y$ or $Y_n$ could be the number of phone calls received in a Mumbai call center with a mean of $\mu$ per hour. Let $n$ be the number of hours. Then $Y_{n-1}$ will be the number of calls among the first $n-1$ hours.
In $n-1$ hours we either have an even or an odd number of calls. Then in the $n\text{th}$ hour, we may get either have even or odd number of calls.
If we have an even number of calls among the first $n-1$ hours, then we need to get an even number in the $n\text{th}$ hour to have an even total.
On the other hand, if we have an odd number among the first $n-1$ hours, we need odd number of calls received in the last hour. To have an even total number of calls throughout our $n$ hours.
So my probabilities should look something like:
$P(Y_n\ \text{even}) = P(Y_{n-1}\ \text{odd}) P(\text{hour n has odd calls}) + P(Y_{n-1}\ \text{even}) P(\text{hour n has odd calls})$
Is this analysis correct? If so how do I proceed please?
Thank you
Are you trying to compute the probability of an even result?
If so, this is how you can do it: $$ P(Y=``even")=\sum_{n=0}^\infty P(Y=2n)=\frac{1}{2}\left(\sum_{n=0}^\infty P(Y=n)+\sum_{n=0}^\infty (-1)^n P(Y=n)\right). $$
The last step is correct because we counted both the even probabilities twice and the odd probabilities are canceled out. Now you can proceed to inserting the PMF and computing the sums. The final result should be: $$ P(Y=``even")=\frac{1}{2}\left(1+e^{-2\mu}\right). $$