Poisson distribution; P(X= n | X is odd) If a random variable $X$ has a Poisson distribution with mean $a$ then one can show that 
$$
P(X \text{ is odd } | X>0) = \frac{1-e^{-2a}}2.
$$
This is worked out by taking the sum of the $P_1 +P_3 + P_5$.... and using $e^a$ - $e^{-a}$.
The Poisson distribution density is given as 
$$P(X=n)=\frac{a^n\ e^{-a}}{n!}.$$
What I am trying to figure out is $P(X=n| X \text{ is odd})$ for an odd integer $n$.
Thank you.
 A: Assuming that $X$ is poisson distributed, the probability mass function (pmf) of $X$ is given as
$$P(X=n)=\frac{e^{-a}a^n}{n!}=P_n ~\mbox{(let)}~$$
The probability that $X$ takes an odd value will be
$$\sum_{n=0}^{\infty}P_{2n+1}=e^{-a} \times \frac{1}{2}\bigg(\sum_{n=0}^{\infty}\frac{a^n}{n!}-\sum_{n=0}^{\infty}\frac{(-a)^n}{n!}\bigg)=\frac{e^{-a}(e^a-e^{-a})}{2}=\frac{1-e^{-2a}}{2}$$
Now, we require the quantity
$$P(X=n|X~\mbox{ is odd}~)=\frac{P(X=n ~\cap~ X ~\mbox{ is odd}~)}{P(X ~\mbox{ is odd}~)}=P_{n|odd} ~\mbox{(let)}~$$
The numerator clearly vanishes when $n$ is even. Thus $P_{n|odd}=0$ if $n$ is even. On the other hand, if $n$ is odd, the numerator is simply $P(X=n)$. Thus if $n$ is odd, 
$$P_{n|odd}=\frac{P_n}{\big(\frac{1-e^{-2a}}{2}\big)}=\frac{\frac{e^{-a}a^n}{n!}}{\big(\frac{1-e^{-2a}}{2}\big)}=\frac{2a^n}{n!(e^a-e^{-a})}$$
In closed form:
$$
P_{n|odd}=
\begin{cases}
0, ~\mbox{if $n$ is even}\\
\frac{2a^n}{n!(e^a-e^{-a})}, ~\mbox{if $n$ is odd}
\end{cases}
$$
For your question in the comment section:
We further wish to calculate $E(n|n ~\mbox{ is odd})$.
\begin{align}
E(n|n ~\mbox{ is odd})&=\sum_{n ~\mbox{is odd}} n \times \frac{2a^n}{n!(e^a-e^{-a})}=\frac{2a}{(e^a-e^{-a})}\sum_{n ~\mbox{is odd}}\frac{a^{n-1}}{(n-1)!}\\
&=\frac{2a}{(e^a-e^{-a})}\sum_{n ~\mbox{is even}}\frac{a^n}{n!}=\frac{2a}{(e^a-e^{-a})} \times \frac{(e^a+a^{-a})}{2}\\
&=a\bigg( \frac{e^a+e^{-a}}{e^a - e^{-a}} \bigg)=a\bigg( \frac{1+e^{-2a}}{1-e^{-2a}} \bigg)
\end{align}
