# Conditional Expectation of a Poisson Random Variable

Hello I'm having difficulties with the following problem:

Let X be a Poisson random variable with parameter $\lambda$. Find the conditional mean of X given X is odd.

What I tried this:
A = [X is odd.]

Back to the basic idea

E[X|A] = $\displaystyle\sum\limits_{x=0}^\infty xP_{X|A}(X=x|A=a) = \displaystyle\sum\limits_{x=0}^\infty x \frac{P_{X,A}(X=x,A=a)}{P_{A}(A=a)}$

I actually don't understand how the probability of X and X is odd is different from the probability of X is odd. I suppose the random variables X and A = [X is odd] are different, but if they are intersecting events how are they different in the end? My main confusion is that I don't know how to differentiate between the joint probability and the probability of X being odd.

Thank you.

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The expression E[X=k|Y=y] is misleading, you might want to replace it by E[X|A] with A=[X is odd]. – Did Feb 1 '13 at 8:12
The probability of "X" and "X is odd" is the same as "X is odd". ($P(\{X \in \mathbb{N}\}\cap\{X \text{ is odd}\})=P(\Omega\cap\{X \text{ is odd}\})=P(\{X \text{ is odd}\})$). Then A is not a random variable but a set - and random variables are not events... – AndreasS Feb 1 '13 at 21:42
Thanks for confirming that. I had thought they were the same, but then it seems like the expression I wrote would cancel and not give the right answer. – rhl Feb 2 '13 at 0:55

The idea is good. The conditional probabilities are a little bit off: recall that $\Pr(A|B)=\frac{\Pr(A\cap B}{\Pr(B)}$.

So we need to divide each of your terms, and hence your expression, by the probability that $X$ is odd. It is clear that you know how to compute that.

Added: The probability that $X$ is odd is $$\sum_{i=0}^\infty e^{-\lambda} \frac{\lambda^{2i+1}}{(2i+1)!}.$$ This can be simplified a lot. Write down the power series for $e^\lambda$, also for $e^{-\lambda}$ Subtract. We get twice our sum. So the probability that $X$ is odd is $e^{-\lambda}\left(\frac{e^\lambda-e^{-\lambda}}{2} \right)$. Note this is $e^{-\lambda}\sinh \lambda$.

Now the probability that $X=2k+1$ given that $X$ is odd is, by the usual conditiona probability formula, equal to $$\frac{e^{-\lambda}\frac{\lambda^{2k+1}}{(2k+1)!}}{e^{-\lambda}\sinh \lambda}.$$

For the conditional expectation, multiply the above expression by $2k+1$, and add up from $k=0$ to $\infty$. Since $\frac{2k+1}{(2k+1)!}=\frac{1}{(2k)!}$, we get, after a little manipulation, that the conditional expectation is $$\frac{\lambda}{\sinh \lambda}\sum_{k=0}^\infty \frac{\lambda^{2k}}{(2k)!}.$$ By a calculation with the expansions of $e^\lambda$ and $e^{-\lambda}$ of the type we did before, or in another way, we can show that the inner sum is $\cosh \lambda$.

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So what is the final answer? I am stuck :( – Natalie Mar 25 '13 at 19:45
I did the calculation. Don't believe every word of it, I could have lost a constant along the way. But it is either right or almost right. Have added to the answer, it would have been too hard in a comment. – André Nicolas Mar 25 '13 at 20:57
In addition, for whatever it's worth the book answer to this question is: $\lambda \frac{e^{\lambda}+e^{- \lambda}}{e^{\lambda} - e^{- \lambda}} = \lambda \frac{cosh( \lambda)}{sinh( \lambda)}$ – rhl Mar 28 '13 at 19:48
That's good to know, so my calculation was right not only in principle but in detail. – André Nicolas Mar 28 '13 at 19:53