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 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 $\cosh \lambda$.