odcinaknr

2022-10-08

Conditional Expectation Property Geometric

I need to calculate $\mathbb{E}(X|Z)$ where X, Y are Geometric r.v. and $Z=X+Y$. The conditional distribution of $\mathbb{E}(X|Z)$ is $\frac{1}{n-1}$

So now, $\mathbb{E}(X|Z)=\sum _{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}$. However, $\mathbb{E}(X)=\frac{1}{p}$. And by property of conditional Expectation $\mathbb{E}(\mathbb{E}(X|Z))=\mathbb{E}(X)$. So I get $\frac{n}{2}=\frac{1}{p}$. What am I doing wrong? How do I get $\mathbb{E}(\mathbb{E}(X|Z))=\frac{1}{p}$

I need to calculate $\mathbb{E}(X|Z)$ where X, Y are Geometric r.v. and $Z=X+Y$. The conditional distribution of $\mathbb{E}(X|Z)$ is $\frac{1}{n-1}$

So now, $\mathbb{E}(X|Z)=\sum _{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}$. However, $\mathbb{E}(X)=\frac{1}{p}$. And by property of conditional Expectation $\mathbb{E}(\mathbb{E}(X|Z))=\mathbb{E}(X)$. So I get $\frac{n}{2}=\frac{1}{p}$. What am I doing wrong? How do I get $\mathbb{E}(\mathbb{E}(X|Z))=\frac{1}{p}$

Demarion Thornton

Beginner2022-10-09Added 11 answers

Step 1

Your sin are sloppy notations. In fact, you (correctly) calculate $\mathbb{E}(X|Z=n)=\sum _{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}.$

Step 2

You could write that as $\mathbb{E}(X|Z)=\frac{1}{2}Z$, too. Then, $\mathbb{E}(\mathbb{E}(X|Z))=\mathbb{E}\left(\frac{1}{2}Z\right)=\frac{1}{2}(\mathbb{E}X+\mathbb{E}Y)=\frac{1}{p}.$

Your sin are sloppy notations. In fact, you (correctly) calculate $\mathbb{E}(X|Z=n)=\sum _{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}.$

Step 2

You could write that as $\mathbb{E}(X|Z)=\frac{1}{2}Z$, too. Then, $\mathbb{E}(\mathbb{E}(X|Z))=\mathbb{E}\left(\frac{1}{2}Z\right)=\frac{1}{2}(\mathbb{E}X+\mathbb{E}Y)=\frac{1}{p}.$

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