odcinaknr

2022-10-08

Conditional Expectation Property Geometric
I need to calculate $\mathbb{E}\left(X|Z\right)$ where X, Y are Geometric r.v. and $Z=X+Y$. The conditional distribution of $\mathbb{E}\left(X|Z\right)$ is $\frac{1}{n-1}$
So now, $\mathbb{E}\left(X|Z\right)=\sum _{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}$. However, $\mathbb{E}\left(X\right)=\frac{1}{p}$. And by property of conditional Expectation $\mathbb{E}\left(\mathbb{E}\left(X|Z\right)\right)=\mathbb{E}\left(X\right)$. So I get $\frac{n}{2}=\frac{1}{p}$. What am I doing wrong? How do I get $\mathbb{E}\left(\mathbb{E}\left(X|Z\right)\right)=\frac{1}{p}$

Demarion Thornton

Step 1
Your sin are sloppy notations. In fact, you (correctly) calculate $\mathbb{E}\left(X|Z=n\right)=\sum _{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}.$
Step 2
You could write that as $\mathbb{E}\left(X|Z\right)=\frac{1}{2}Z$, too. Then, $\mathbb{E}\left(\mathbb{E}\left(X|Z\right)\right)=\mathbb{E}\left(\frac{1}{2}Z\right)=\frac{1}{2}\left(\mathbb{E}X+\mathbb{E}Y\right)=\frac{1}{p}.$

Do you have a similar question?