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I have a question regarding the conditional expectation. Any hint or help is appreciated!

Consider $a,b,$c are random variables on probability space (X, $\Sigma$, $P$),all of them are integrable and furthermore, $a$ is independent of $c$ and $b$ is also independent of $c$. Then does the following claim hold?

$$E(a|b,c) =_{a.s} E(a|b)$$

I plan to use the fact that $E(E(a|b,c)|b) = E(a|b)$, since $\sigma(b) \subset\sigma(b,c) $, but then I get stuck at showing $E(E(a|b,c)|b) = E(a|b,c)$.

Any help or hint is appreciated!

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1 Answer 1

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The usual counterexample applies: assume that $a$ and $b$ are i.i.d. symmetric Bernoulli $\pm1$ random variables and that $c=ab$. Then $a$ and $b$ are independent hence $E[a\mid b]=E[a]=0$ while $a=bc$ is measurable with respect to $(b,c)$ hence $E[a\mid b,c]=a$.

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    $\begingroup$ +1 for "get(ting) stuck at showing" a result which does not hold. $\endgroup$
    – Did
    Commented May 4, 2013 at 9:31
  • $\begingroup$ Thank you so much for your help! Before closing out this question, I am trying to show that in your example, both a and b are independent of c = ab, which I don't see it immediately, could you please explain this part? Thank you! $\endgroup$
    – user7788
    Commented May 4, 2013 at 16:34
  • $\begingroup$ Indeed, $a$ is independent of $c$, and $b$ is independent of $c$ (but of course, $(a,b)$ is not independent of $c$). An easy way to check these independence properties is to prove, for example, that $(a,c)$ is uniformly distributed on $\{-1,+1\}^2$. $\endgroup$
    – Did
    Commented May 4, 2013 at 20:49
  • $\begingroup$ Thank you Did, but i am still confusing the fact that, by your construction, a is independent of both b and c, then isn't it trivailly that E(a|b,c)= a.s E(a|b) if it is the case? RHS = LHS = E(a)? $\endgroup$
    – user7788
    Commented May 4, 2013 at 21:27
  • $\begingroup$ No: a is independent of b, a is independent of c, but a is not independent of (b,c). $\endgroup$
    – Did
    Commented May 4, 2013 at 21:28

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