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I'm trying to prove the "tower property" of conditional expectations, $$ E[V\mid W] = E[\ E[V\mid U,W]\ \mid W\ ], $$ where $U$, $V$ and $W$ are any random variables. $E[X \mid Y]$ is itself a random variable $f(Y)$ where $$f(y) = E[X \mid Y = y) = \sum_x x\cdot Pr[X=x\mid Y=y].$$ Keeping this observation in mind, I still don't see why $U$ is "averaged out" when moving from the right hand side to the left side.

somebody
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    The tower property is more simply/generally expressed as $E[E[V | U]] = E[V]$. Why do you add the conditioning on $W$? Nothing wrong with that, but you'd first try to prove/understand the more elementary formulation. – leonbloy Jun 04 '12 at 13:55
  • Related: http://math.stackexchange.com/questions/41536/intuitive-explanation-of-the-tower-property-of-conditional-expectation – leonbloy Jun 04 '12 at 13:56

1 Answers1

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The last equality in your observation does not apply in general (i.e. if $X$ is not discrete). Let $U,V,W$ be random variables such that $V\in \mathcal{L}^1(P)$. In order to show that $$ E[V\mid W]=E[E[V\mid U,W]\mid W] $$ we note that the right hand side is indeed $\sigma(W)$-measurable, so we only need to check the defining equation, i.e. check that $$ \int_A V\,\mathrm{d}P=\int_A E[V\mid U,W]\,\mathrm{d}P $$ for all $A\in\sigma(W)$. Let such an $A$ be given. Then $A\in\sigma(W)\subseteq \sigma(U,W)$ and therefore $$ \int_A E[V\mid U,W]\,\mathrm{d}P=\int_A V\,\mathrm{d}P $$ and we are done.

Did
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Stefan Hansen
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  • Could you explain why $\sigma(W) \subseteq \sigma(U,W)$ implies the last equation? – somebody Oct 09 '12 at 04:19
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    Because if $A\in\sigma(U,W)$, then by definition of the conditional expectation $E[V\mid U,W]$ we have the last equality. This is the defining equality of conditional expectations. – Stefan Hansen Oct 09 '12 at 05:16
  • Thanks for your answer, it's formally clear to me. I'm still having trouble with the intuition: As you said, $A \in \sigma(W)$, e.g., $A$ could be an event ${ W=w }$. We know that $E[V \mid U,W]$ depends on both, $U$ and $W$, but becomes a constant for this event $A$ (which only depends on $W$), isn't this counterintuitive? – somebody Oct 09 '12 at 15:01
  • It doesn't become a constant since it is still conditioned on $U$. – Stefan Hansen Oct 09 '12 at 15:11