# Replacing a random variable in a conditional expectation

Suppose that $$X$$,$$Y$$, and $$Y'$$ are independent and identically distributed random variables defined on some probability space $$(\Omega,\mathcal F,P)$$. Assume that $$f:\mathbb R^2\to\mathbb R$$ is a measurable function so that $$f(X,Y)$$ is a random variable. Consider the conditional expectation $$\operatorname E[f(X,Y)\mid\sigma(X)],$$ where $$\sigma(X)$$ is the $$\sigma$$-algebra generated by $$X$$.

Is it true that $$\operatorname E[f(X,Y)\mid\sigma(X)]=\operatorname E[f(X,Y')\mid\sigma(X)]$$ almost surely? In other words, does the replacement of $$Y$$ with an independent and identically distributed copy $$Y'$$ change the conditional expectation? If it does not, how can we make sure that the conditional expectation remains unchanged?

Since the conditional expectation is with respect to the $$\sigma$$-algebra generated only by $$X$$, it seems that it should indeed be the case. Roughly speaking, the conditional expectation only depends on $$X$$ and we change $$Y$$ with an independent and identically distributed copy $$Y'$$. But I am trying to come up with some sort of a more rigorous argument.

Any help is much appreciated!

Yes, it is true. For any Borel set $$A$$ in $$\mathbb R$$, $$\int_{X^{-1}(A)} f(X,Y)dP=\int_{X^{-1}(A)} f(X,Y')dP$$ because $$(X,Y)$$ has the same two dimensional distribution as $$(X,Y')$$.

$$\int_{X^{-1}(A)} f(X,Y)dP$$ $$=\int_{A\times \mathbb R} f(x,y)dF_X(x)dF_Y(y)$$$$=\int_{A\times \mathbb R} f(x,y)dF_X(x)dF_Y'(y)$$ $$=\int_{X^{-1}(A)} f(X,Y')dP$$

• Hi @geetha290krm could you formlize the reasoning rewriting those integrals in terms of the bivariate distribution ? (I had some issues in the domains doing that was confusing me) Commented Nov 28, 2022 at 20:27
• @Thomas I have done that. Commented Nov 28, 2022 at 23:12
• Thanks. Trying to get to it but I am a bit confused about the notation in the second step. $\Omega$ is the underlying probability space, whereas shouldn't the domain of $y$ be something like the real line? Commented Nov 29, 2022 at 8:16
• @Thomas Yes, I made a silly mistake. Commented Nov 29, 2022 at 8:19
• I see (+1 for me :)). I think now you are considering a function $g:\Omega \rightarrow \Omega'=\mathbf{R}^2, \omega \rightarrow (X(\omega),Y(\omega))$ and rewriting the integral over $\Omega$ as an integral over $\Omega'$. The image measure should be the two dimensional density, that you factor according to independence. If this is a correct, could you please write some reference on the change of variables in measure spaces to check better these things? I had some old notes from university many years ago that I lost and cannot find a good reference again. Commented Nov 29, 2022 at 8:37

Under those circumstances $$E[f(X,Y)|X]=\int f(X,y)dP_Y$$.
Equality follows from being $$P_Y=P_{{Y}'}$$.

Maybe this answer is not very formal, but I think it may be convincing that the result generally holds.

$$E[f(X,Y)|X=\overline{x}]=\int p_{Y|X}(Y=y|X=\overline{x}) f(\overline{x},y) dy \ (1)$$

Now:

$$p_{Y|X}(Y=y|X=\overline{x})=(\text{independence of Y and X})=$$ $$=p_Y(y)=(\text{Y and Y' are identically dstributed})=p_{Y'}(y)=$$ $$=(\text{reverse previous steps})=p_{Y'|X}(Y'=y|X=\overline{x})$$

, so that substituting into (1):

$$E[f(X,Y)|X=\overline{x}]=\int p_{Y'|X}(Y'=y|X=\overline{x}) f(\overline{x},y) dy=E[f(X,Y')|X=\overline{x}]$$

from this we can conclude/be convinced that:

$$E[f(X,Y)|X]=E[f(X,Y')|X]$$

which is just a different way of writing:

$$E[f(X,Y)|\sigma(X)]=E[f(X,Y')|\sigma(X)]$$

• It is not given that densities exist. Commented Nov 29, 2022 at 8:54
• Well sure that is why I wrote that the answer is not formal. But these are manipulations to develop intuition that almost always lead to correct results I think. I am aware also that the notation is not very formal but one finds it commonly used, as it is suggestive. Commented Nov 29, 2022 at 9:09
• ( do not think this answer requires a downvote but that is my opinion so I would keep it there ) Commented Nov 29, 2022 at 9:10
• I don't know anything about downvoting. Commented Nov 29, 2022 at 9:12
• Eheh yes thanks for the clarification :) but I was not claiming that you downvoted, my message was just about the downvote itself ! Anyway thanks for helping me clarify that these manipulations are far from formal (for this there is your solution). Commented Nov 29, 2022 at 9:14