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I am trying to prove $\Bbb E[X\Bbb1_A]=\Bbb E[X\mid A]\Bbb P(A)$ using the Tower rule. But I am not sure if I can take $\Bbb E[X\mid A]$ out of the expectation when I have $\Bbb E[\Bbb E[X\Bbb1_A\mid A]]=\Bbb E[\Bbb1_A\Bbb E[X\mid A]]$, because it seems that $\Bbb E[X\mid A]$ itself is a random variable.

Anyhow, if the Tower rule cannot be applied here, can anyone explain why it is wrong and offer a correct proof?

Thanks!

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No Tower Rule involved here, just the definition of conditional expectation conditionally on an event $A$ with positive probability, stating that $E[X\mid A]=E[X\mathbf 1_A]/P[A]$.

The confusion might come from the fact that, when $A$ is an event with positive probability, $E[X\mid A]$ is a number. By contrast, for every sigma-algebra $\mathcal G$, $E[X\mid\mathcal G]$ is a random variable. For example, if $P[A]\ne0$ and $P[A]\ne1$ then $E[X\mid\sigma(A)]=E[X\mid A]\mathbf 1_A+E[X\mid A^c]\mathbf 1_{A^c}$, with the obvious extension if $P[A]=0$ or $P[A]=1$.

To sum up, indeed $E[E[X\mid A]\mathbf 1_B]=E[X\mid A]P[B]$, for every events $A$ and $B$, and this has nothing to do with properties of conditional expectation, just with the fact that $E[X\mid A]$ is a number.

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  • $\begingroup$ Then in what situations can I use the Tower rule? $\endgroup$ – user99015 Oct 13 '13 at 16:26
  • $\begingroup$ When conditioning on a sigma-algebra. $\endgroup$ – Did Oct 13 '13 at 17:12

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