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I have a question about the conditional expectation with some independence conditions for random variables and $\sigma$-fields.

For a random variable $X$ with $E|X| < \infty $, if $Y_1$ and $ Y_2 $ are random variables such that $\sigma(X,Y_1) $ and $\sigma (Y_2) $ independent, then I want to prove the following. $$ E\left(X | Y_{1}, Y_{2}\right)=E\left(X | Y_{1}\right) \quad \text { a.s. } $$

It seems very intuitive since $Y_2$ information is useless for $X$. But I don't know how to prove it. Can anyone help me?

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Since RHS is measurable w.r.t $\sigma (Y_1,Y_2)$ we only have to show that $E I_EX=EI_E E(X|Y_1)$ for every set $E \in \sigma (Y_1,Y_2)$. By a standard argument using the $\pi -\lambda$ theorem we can reduce this to the case when $E$ has the form $Y_1^{-1} (A) \cap Y_2^{-1}(B)$ where $A$ and $B$ are Borel sets in $\mathbb R$. Now $EXI_{Y_1^{-1}(A)}I_{Y_2^{-1}(B)}=EXI_{Y_1^{-1}(A)} EI_{Y_2^{-1}(B)}$ by the independence assumption. Also $$E E(X|Y_1)I_{Y_1^{-1}(A)} I_{Y_2^{-1}(B)}$$ $$=E(E(XI_{Y_1^{-1}(A)}|Y_1) I_{Y_2^{-1}(B)})$$ $$=E(E(XI_{Y_1^{-1}(A)}|Y_1)) EI_{Y_2^{-1}(B)}$$ $$=EXI_{Y_1^{-1}(A)} EI_{Y_2^{-1}(B)}. $$ This completes the proof.

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  • $\begingroup$ I think you have a right parenthesis out of place on line 2 of your displayed equation. Should the second right parenthesis not be on the far right? $\endgroup$ Commented May 5, 2020 at 7:46
  • $\begingroup$ @SpiritLevel Of course. I was careless in typing Thanks for pointing out. $\endgroup$ Commented May 5, 2020 at 7:51
  • $\begingroup$ In a hurry to be first! $\endgroup$ Commented May 5, 2020 at 8:06

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