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Let $X, Y$ be continuous, real valued random variables, and let $f$ be a measurable function such that $f(X)$ is again a random variable.

EDITED: How would the conditional expectation $\mathbb{E}[Y|f(X)=f(x)]=\int y d\mathbb{P}_{Y|f(X)}(f(x))$ relate to $\mathbb{E}[Y|X=x]=\int y d\mathbb{P}_{Y|X}(x)$? Which are the minimum conditions that they are equal? How can I get from the one integral representation to the other and which properties must $f$ fullfil to do so?

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    $\begingroup$ What do you mean by this notation? $\endgroup$
    – Tim ♦
    Oct 11, 2021 at 9:03
  • $\begingroup$ Hmm thinking about it I might actually mean $E[Y|f(X) = f(x)]$ and how this relates to $E[Y|X=x]$. What I mean by that is $E[Y|f(X) = f(x)] = \int y dP_{Y|f(X)}(f(x))$ and $E[Y|X=x] = \int y dP_{Y|X}(x)$. And now how I can get from the one integral representation to the other and which properties f must fullfil to do so. $\endgroup$
    – guest1
    Oct 11, 2021 at 14:26
  • $\begingroup$ Please correct your question to reflect the change. $\endgroup$
    – Xi'an
    Oct 11, 2021 at 20:12

1 Answer 1

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The conditional expectation $\mathbb{E}[Y|X]$ is $\sigma (X)$-measurable, so there exists a measurable function $\varphi$ such that $\mathbb{E}[Y|X]=\varphi(X)$. Likewise, there exists a measurable function $\psi$ such that $\mathbb{E}[Y|f(X)]=\psi(f(X))$.

The sigma algebra $\sigma(f(X))$ generated by $f(X)$ is contained in the sigma algebra $\sigma(X)$ generated by $X$ (because $f(X)^{-1}(C)=X^{-1}(f^{-1}(C))\in\sigma(X)$ for every Borel set $C$). Hence, by the tower property for conditional expectations, $$\mathbb{E}[\varphi(X)|f(X)]=\psi(f(X))$$ This means that $$\mathbb{E}[Y|f(X)=f(x)]=\mathbb{E}[\mathbb{E}[Y|X]|f(X)=f(x)]$$

So, if $f$ is injective then $\mathbb{E}[Y|f(X)=f(x)]=\mathbb{E}[Y|X=x]$. In general, $\mathbb{E}[Y|f(X)=f(x)]$ is an integral of $\mathbb{E}[Y|X]$ over all $x'$ such that $f(x')=f(x)$.

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  • $\begingroup$ The quantity $\mathbb{E}[Y|X=x]=g(x)$ is defined uniquely almost everywhere so the question does make sense in this regard. $\endgroup$
    – Xi'an
    Oct 11, 2021 at 20:14
  • $\begingroup$ @Xi'an if a function is not uniquely defined everywhere then I'm not sure how much sense it makes to compare its value at two different points, the answer might depend on which version of $\mathbb{E}[Y|X]$ you're using.. $\endgroup$
    – S. Catterall
    Oct 12, 2021 at 7:34
  • $\begingroup$ hi thanks for your answer. Why can we know that $\sigma(f(X))$ is contained in $\sigma(X)$? $\endgroup$
    – guest1
    Oct 19, 2021 at 7:32
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    $\begingroup$ @guest1 see edit above. $\endgroup$
    – S. Catterall
    Oct 19, 2021 at 16:29
  • $\begingroup$ @S.Catterall Hi sorry for coming back to this after such a long time. I am currently wondering how I can understand you last sentence. So if I understand it correctly, it means $\mathbb{E}[Y|f(X)=f(x)] = \int_{\{x': f(x')=f(x)\}}\mathbb{E}[Y|X=x'] \mathbb{P}_X(dx')$ Is this correct? If so, what mathematical principle is going on that I can substitute the conditioning on $f(X)$ by a conditioning on $X$ and multiplying it with an indicator function over the set $\{x': f(x')=f(x)\}$? $\endgroup$
    – guest1
    Mar 25 at 14:49

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