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.
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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 $\sigma(W)$-measureable, 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. |
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