We have a random variable $X$ that is poisson distributed with $\lambda$.
We wish to show:
$$E[X\mid X \text{ is even}]=\lambda \frac{1-e^{-2\lambda}}{1+e^{-2\lambda}}$$
So far, I have that
- $P(X \text{ is even})=\frac{1+e^{-2\lambda}}{2}$
- $P(X\text{ is odd})=\frac{1-e^{-2\lambda}}{2}$ and that
- $E[X]=\lambda$ for the Poisson distribution.
I am struggling because I'm not completely sure how these pieces fit together. Conditional expectation is still a little fuzzy to me. Any hints/solutions are appreciated.
Best Answer
Treat $X\mid X \text{ is even}$ as a random variable. What is its pmf? $$P(X=k\mid X \text{ is even})=0$$ if $k$ is odd (obviously) and for $k$ even $$P(X=k \mid X\text{ is even})=\frac{P(X=k, X \text{ is even})}{P(X \text{ is even})}=\frac{P(X=k)}{P(X \text{ is even})}=\frac{2e^{-λ}λ^k}{k!(1+e^{-2λ})}$$ So
\begin{align}E[X \mid X \text{ is even}]&=\sum_{k=0}^{+\infty}kP(X=k \mid X \text{ is even})=\sum_{k\text{ is even}}^{+\infty}k\frac{2e^{-λ}λ^k}{k!(1+e^{-2λ})}\\[0.2cm]&=\frac{2}{1+e^{-2λ}}\sum_{k\text{ is even}}^{+\infty}\frac{e^{-λ}λ^k}{(k-1)!}\\[0.2cm]&=\frac{2λ}{1+e^{-2λ}}\sum_{k\text{ is even}}^{+\infty}\frac{e^{-λ}λ^{k-1}}{(k-1)!}\\[0.2cm]&=\frac{2λ}{1+e^{-2λ}}\sum_{k\text{ is odd}}^{+\infty}\frac{e^{-λ}λ^{k}}{k!}=\frac{2λ}{1+e^{-2λ}}\cdot P(X \text{ is odd})\\[0.2cm]&=\frac{2λ}{1+e^{-2λ}}\cdot \frac{1-e^{-2λ}}{2}=λ\frac{1-e^{-2λ}}{1+e^{-2λ}}\end{align}