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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!

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3 Answers 3

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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 $(X,Y')$.

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

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  • $\begingroup$ 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) $\endgroup$
    – Thomas
    Commented Nov 28, 2022 at 20:27
  • $\begingroup$ @Thomas I have done that. $\endgroup$ Commented Nov 28, 2022 at 23:12
  • $\begingroup$ 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? $\endgroup$
    – Thomas
    Commented Nov 29, 2022 at 8:16
  • $\begingroup$ @Thomas Yes, I made a silly mistake. $\endgroup$ Commented Nov 29, 2022 at 8:19
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Commented Nov 29, 2022 at 8:37
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Under those circumstances $E[f(X,Y)|X]=\int f(X,y)dP_Y$.
Equality follows from being $P_Y=P_{{Y}'}$.

See also this.

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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:

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

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    $\begingroup$ It is not given that densities exist. $\endgroup$ Commented Nov 29, 2022 at 8:54
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Commented Nov 29, 2022 at 9:09
  • $\begingroup$ ( do not think this answer requires a downvote but that is my opinion so I would keep it there ) $\endgroup$
    – Thomas
    Commented Nov 29, 2022 at 9:10
  • $\begingroup$ I don't know anything about downvoting. $\endgroup$ Commented Nov 29, 2022 at 9:12
  • $\begingroup$ 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). $\endgroup$
    – Thomas
    Commented Nov 29, 2022 at 9:14

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