An accepted formula that is applied successfully in many conditional probability problems is: $$E[f(X,Y)\mid Y=t]=E[f(X,t)\mid Y=t]$$ This formula says that the conditional expectations $E[f(X,Y)\mid Y]$ and $E[f(X,t)\mid Y]$ agree at $\quad$ "$Y=t$". But what sense does that equality make when we know that conditional expectation is a global concept, not a pointwise one? The formula would make sense if we could say, if at all, that the equality holds for almost all $Y=y$, not just $Y=t$, but this is not true. It sounds like an absurd equality, and yet it works. How do we then have to interpret that equality? When does it make sense to apply it? Why does it work?
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Formula states that the average value of $f(X, Y)$ when $Y$ is fixed at t is equal to the average value of $f(X, t)$ when $X$ varies according to its conditional distribution given $Y$ = t. The left-hand side of the formula, $E[f(X, Y) | Y = t]$, is the average value of $f(X, Y)$ when $Y$ is fixed at $t$, and $X$ varies according to its conditional distribution given $Y = t$. The right-hand side is the average value of $f(X, t)$ when $X$ varies according to its conditional distribution given $Y = t$, and t is a fixed constant. – rumathe Mar 17 '23 at 10:12
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The reason this formula works is that it is consistent with the definition of conditional expectation. It makes use of the fact that the conditional expectation is a function of $Y$, and that the average value of $f(X, Y)$ for a fixed value of $Y$ should be the same as the average value of $f(X, t)$, where $t$ is a fixed constant, and $X$ varies according to its conditional distribution given $Y = t$. – rumathe Mar 17 '23 at 10:12