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Let $X$ and $Y$ be continuous random variables and $Z = f(X,Y)$ for some function $f$. Additionally, let $g(\cdot)$ be a different function of $X$ and $Y$. Is the following true? $$ E_Z[g(X,Y)] = E_{XY}[g(X,Y)] $$ or equivalently, if $\mathcal X, \mathcal Y,$ and $\mathcal Z$ are the ranges of $X,Y,$ and $Z$ respectively (I'm aware that the integration on the left-hand side of the equation below does not make much sense, as $g(x,y)$ can be taken out of the integration, but I'm not sure if there's a better way of stating what I mean), $$ \int_{\mathcal Z} g(x,y) p_Z(z) dz = \int_{\mathcal X} \int_{\mathcal Y} g(x,y) p_{XY}(x,y) dx dy $$ I think this statement is true, as \begin{align} E_Z[g(X,Y)] &= E_{XY}[E_Z[g(X,Y) \mid X,Y]] \\ &= E_{XY}[g(X,Y)] \end{align}

mhdadk
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    What do notations "$E_Z$" and "$E_{XY}$" mean? There is no need to attach a random variable to the expectation operator (though I do see this sometimes in the literature): this is because, if $X, Y, Z$ are all defined on the same probability space $(\Omega, \mathscr{F}, P)$, then by definition, $E[X] = \int_\Omega X(\omega) P(d\omega), E[Z] = E[g(X, Y)] = \int_\Omega g(X(\omega), Y(\omega))P(d\omega)$, either $X$ or $Z$ are simply integrands (functions), while "$E$" should be interpreted as a shorthand for the integration operator. – Zhanxiong Feb 07 '23 at 02:20
  • If you tried to use $E_Z[X]$ to mean the conditional expectation $E[X|Z]$, then this is non-standard. The standard notation $E[X|Z]$ is succinct and clear already. It is pointless to create a new system for conditional expectation. – Zhanxiong Feb 07 '23 at 02:23
  • @Zhanxiong sorry, I should've been clearer with my notation. I've edited my question with what $E_Z$ and $E_{XY}$ mean. – mhdadk Feb 07 '23 at 13:31
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    The left hand side of your expectation makes no sense because the arguments $x$ and $y$ of the integrand $g$ are undefined. – whuber Feb 07 '23 at 13:57

1 Answers1

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You probably want to check if (see my comments below your question) \begin{align} E[g(X, Y)|f(X, Y)] = E[g(X, Y)|X, Y] = g(X, Y) \end{align} is true.

This is not true. A simple counterexample is: let $X, Y \text{ i.i.d.} \sim N(0, 1)$, $f(X, Y) = X + Y$, $g(X, Y) = X - Y$. It is well known that $f(X, Y)$ and $g(X, Y)$ are independent, whence \begin{align} E[g(X, Y)|f(X, Y)] = E[g(X, Y)] = 0 \neq g(X, Y) = X - Y. \end{align}

Another way to disprove your conjecture is by noting that $E[g(X, Y)|f(X, Y)]$ has to be a function of $f(X, Y)$, say $h(f(X, Y))$, then you are requiring $(h \circ f)(X, Y) = g(X, Y)$, which is not true unless $g = h \circ f$. This is in general very unlikely.

It seems that your argument tries to apply the tower property of the conditional expectation: if $\mathscr{F}_1 \subset \mathscr{F}_2 \subset \mathscr{F}$, then \begin{align} E[X|\mathscr{F}_1] = E[E[X|\mathscr{F}_2]|\mathscr{F}_1] = E[E[X|\mathscr{F}_1]|\mathscr{F}_2]. \tag{1} \end{align} While $\sigma(f(X, Y))$ ("$\mathscr{F}_1$") is indeed a sub-$\sigma$-field of $\sigma(X, Y)$ ("$\mathscr{F}_2$"), the problem lies in that neither $E[g(X, Y)|X, Y]$ is $\sigma(f(X, Y))$-measurable (if you want to use the first equality in $(1)$) nor $E[g(X, Y)|f(X, Y)]$ is equal to $g(X, Y)$ (if you want to use the second equality in $(1)$).

Zhanxiong
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  • In your statement $$E[g(X, Y)|f(X, Y)] = E[g(X, Y)|X, Y] = g(X, Y)$$ I only conjecture that the second equality is true, $$E[g(X, Y)|X, Y] = g(X, Y)$$ and not the first equality, $$E[g(X, Y)|f(X, Y)] = E[g(X, Y)|X, Y]$$ I'm not familiar with the measure-theoretic foundations of probability, so could you please explain how the second equality in your statement is disproven in your answer? – mhdadk Feb 07 '23 at 13:36
  • @mhdadk The second equality is true. If you only conjectured the second equality, then $f(X, Y)$ plays no role and should be removed for clarity. – Zhanxiong Feb 07 '23 at 14:29
  • But what I wrote in my attempt at a proof is $$E_{XY}[E_Z[g(X,Y) \mid X,Y]] = E_{XY}[g(X,Y)]$$ I didn't write $$E_{XY}[E_Z[g(X,Y) \mid f(X,Y)]] = E_{XY}[g(X,Y)]$$ Is this what you meant? – mhdadk Feb 07 '23 at 14:33
  • @mhdadk Again, I couldn't understand your notations "$E_{XY}, E_Z$" (even with your updated question), they make no sense to me. My last comment responds to your first comment, where you asked me to explain why "$E[g(X, Y)|X, Y] = g(X, Y)$". – Zhanxiong Feb 07 '23 at 15:54
  • After your editing, it becomes more ambiguous to me. Can you simply clarify: is what you conjectured an equality of the unconditional expectation (i.e., the common-sense expectation) or the conditional expectation? – Zhanxiong Feb 07 '23 at 16:05
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    It looks like what you conjectured is similar to the law of unconscious statistician, but in this theorem, $f$ must be equal to $g$, in which case $\int zp_Z(z)dz = \int g(x, y)p_{X, Y}(x, y)dxdy$ indeed holds. – Zhanxiong Feb 07 '23 at 16:16
  • Then it seems that I’ve made a mistake somewhere in my question. To be honest, I originally asked this question as a simpler version of a more complicated one, but it seems I oversimplified. If you are interested, and have the time, I’ve asked the more complicated question here. – mhdadk Feb 07 '23 at 16:37