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I'm working on some statistics project and am not getting further because of some stupid prediction that doesn't want to be 0. That's why I was wondering if maybe the following holds: Suppose we have a random variable X and an unknown constant m. Is then $$E\left(X|X+m\right)=E(X)?$$ That would help a lot. I thought it might hold because m is unknown and thus knowing X+m doesn't say anything about X, so one could consider X and X+m as being independent?

Thanks a lot!

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I think you answered your own question. – Memming Jul 1 '12 at 15:42
Is $m$ a random variable? If not, $E[X|X+m] = X$, not $E[X]$. If $m$ is a random variable, then the answer depends on the distribution of $(X,m)$. – madprob Jul 1 '12 at 16:06
@Memming: No. $ $ – Did Jul 1 '12 at 20:24
If m is a constant and conditioning on X+m knowing both X and m would lead to E[X|X+m]=X But if you only know X+m and not X and m individually then it is not clear what E[X|X+m] would be but it would not be E{X] because X and X+m are highly dependent. – Michael Chernick Jul 1 '12 at 21:17
Pity. Thank you guys! – user34853 Jul 2 '12 at 11:10

one could consider $X$ and $X+m$ as being independent?

Certainly not. If $m$ is independent of $X$, then $X$ and $X+m$ are positively correlated.

If $m$ is constant, then $E(X|X+m)=X$ because we are conditioning on the $\sigma$-algebra generated by $X+m$, which is the same as $\sigma$-algebra generated by $X$. If $m$ is a random variable, then one should look into the joint distribuition of $X$ and $m$ to say anything about $E(X|X+m)$.

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