# $\mathbb E[X_i\mid X_1,…,X_n]=\mathbb E[X_i]$ [closed]

Do you agree that if $(X_i)$ is a sequence of i.d.d. random variable, then for all $i$

$$\mathbb E[X_i\mid X_1,...,X_n]=\mathbb E[X_i]\ \ \ ?$$

## closed as off-topic by Did, Magdiragdag, Davide Giraudo, Robert Cardona, k1.MMay 28 '15 at 1:09

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I don't agree. Given the outcomes of all $X_i$ you know exactly what the conditional expectation is, namely, $X_i$ for $i \leq n$. $$E[X_i | X_1, ..., X_n] = X_i$$ That's true regardless of if the sequence are iid or not.

Since the $X_i$ are independent, this does hold true for $i > n$ $$E[X_i | X_1, ..., X_n] = E[X_i]$$

• This reminds me a bit of the fun task to expand the polynomial $(x-a)(x-b)(x-c)\cdots (x-z)$ :) – Hagen von Eitzen May 27 '15 at 10:34
• @HagenvonEitzen Could you please explain it? – PSPACEhard May 27 '15 at 10:37
• @echo What is the factor between $(x-w)$ and $(x-y)$? – Did May 27 '15 at 11:02
• I agree with that. – muaddib May 27 '15 at 11:33
• No, for independent and $i > n$ that statement holds true. For $i \leq n$ the expected value is just what you observed for the variable itself $X_i$. This is just the statement that $E[X | X] = X$. – muaddib May 27 '15 at 11:44