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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 $\frac{1}{n-1}$ (See here: Conditional probability distribution with geometric random variables)

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 $\mathbb{E}(\mathbb{E}(X|Z))= \frac{1}{p}$.

Dom
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1 Answers1

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Your sin are sloppy notations. In fact, you (correctly) calculate $$\mathbb{E}(X|Z=n) = \sum_{k=1}^{n-1}\frac{k}{n-1}=\frac{n}{2}.$$ You could write that as $\mathbb{E}(X|Z)=\frac12Z$, too. Then, $$\mathbb{E}(\mathbb{E}(X|Z))=\mathbb{E}\left(\frac12Z\right)=\frac12(\mathbb{E}X+\mathbb{E}Y)= \frac{1}{p}.$$