# A question on Conditional Expectation from Breiman

Breiman Proposition 4.36 Let $X$ & $Y$ be random variables. Let $f(X,Y)$ be a real valued, random variable such that the expectation of the absolute value of $f(X,Y)$ is finite. If $Q(\cdot\mid Y=y)$ is a regular conditional distribution for $X$ given $Y=y$, then,

$E(f(X,Y)\mid Y=y)$=$\int_\ f(x,y) \, dQ(\cdot \mid Y=y)$ a.s. with respect to the Law of $Y$.

Quoting Brieman (Setion 4.3, page 80), "It is tempting to replace the right-hand of (4.37) (equation above) by $E(f(X,y)\mid Y=y)$. But this object cannot be defined through the standard definition of conditional expectation (4.18) (defined below)."

Definition 4.18. $E(X\mid Y=y)$ is any random variable on $R$, where $Q(B)=P(Y \in B)$, satisfying

$\displaystyle \int_B E(X\mid Y=y) dQ = \displaystyle \int_{Y \in B} X dP$ for all Borel subsets $B$.

That is essentially the relevant part of Breiman. (I have made small changes as my ability to write in scientific notation is limited.)

This should help clarify my questions below

Is $E(f(X,Y)\mid Y=y)$ = $E(f(X,y)\mid Y=y)$?

By my understanding, Brieman's Probability text (Setion 4.3, page 80) seems to suggest that$E(f(X,y)\mid Y=y)$ is not well defined, so the question is ill posed.

I was looking for an explanation of Breiman's statement, if my understanding of it is correct of course.

The intuitive answer from introductory probability would suggest that the two expressions are the same, as $Y$ can be treated as a constant when conditioning on $Y=y$.

Thanks

• What is "Breiman's statement"? That E(f(X,y)∣Y=y] is not well defined, or that it is well defined and equal to what you say?
– Did
Commented Sep 10, 2013 at 5:29
• @Did I have attempted to clarify my question by filling in more details. Of course limited by my understanding of the text, as well as a limitation of typing in scientific notation. Thanks for the help Commented Sep 10, 2013 at 12:22
• There was a mishmash in your use of X and Y, I corrected it.
– Did
Commented Sep 10, 2013 at 14:06
• @ Did Thanks for the edit Commented Sep 10, 2013 at 15:14

The expression $E[Z\mid Y=y]$ is not well defined when $P[Y=y]=0$. One approach to define it would be to come back to $E[Z\mid Y]$, a random variable well defined up to null events and which is $E[Z\mid Y]=u(Y)$ for some measurable function $u$, and to consider that $E[Z\mid Y=y]=u(y)$.
Unfortunately, changing the function $u$ at $y$ only yields a second measurable function $\bar u$ such that, when $P[Y=y]=0$, $u(Y)=\bar u(Y)$ almost surely, hence $E[Z\mid Y]=u(Y)$ almost surely and $E[Z\mid Y]=\bar u(Y)$ almost surely. Thus, there is no reason to consider that $E[Z\mid Y=y]=u(y)$ rather than $E[Z\mid Y=y]=\bar u(y)$.
• Thanks. I have added some more details to my question. In that light, is the difficulty with identifying $E[Z\mid Y=y]$ with a unique function $u$ what Breiman is alluding to when he states that "this object (namely $E(f(X,y)\mid Y=y)$) cannot be defined through the standard definition of conditional expectation (4.18)". Commented Sep 10, 2013 at 13:24
• Probably something like, for every $y$ let $E[f(X,y)\mid Y]=u(y,Y)$ then $E[f(X,Y)\mid Y]=u(Y,Y)$ is not guaranteed. Anyway, one way or another it has something to do with negligible sets not "gluing" together correctly.