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For two random variables $X_1, X_2$, is it always necessarily the case that $E \left(e^{X_2}\mid e^{X_1}\right) = E\left(e^{X_2}\mid X_1\right)$? If not, in what cases are they like so? An explanation I read in a book is that $X_1$ is increasing in $e^{X_1}$ but that doesn't make sense to me.

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    Knowing $e^{X_1}$ is equivalent to knowing $X_1$. So the conditional expectation is based on the exact same "evidence". – 2017-01-10
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    Would you give details for "An explanation I read in a book "? What is the context? – 2017-01-10
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    The context I found this in was that $X_1, X_2$ are multivariate normal. – 2017-01-10
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    @Alt Is there something about the exponential function that makes this work? I imagine that $E(X|f(Y)) \neq E(X|Y)$ for measurable $f$ in general, since $E(X||Y|) \neq E(X|Y)$ as the absolute value "kills" information out of $Y$. What is the formal reasoning? Is it that the sigma algebra of $e^{X_1}$ is contained in $X_1$? – 2017-01-10
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    If it said it's because $X_1$ is an increasing function of $e^{X_1}$, it probably meant that therefore it's a one-to-one function of $e^{X_1}$. That is why if you know either $X_1$ or $e^{X_1}$ you can find the other, and that is why $X_1$ and $e^{X_1}$ both give the same information. – 2017-01-10
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    @MichaelHardy From a theoretical standpoint, is there a way to view the one-to-one function in terms of sigma algebras? In other words, the sigma algebras of $e^{X_1}$ and $X_1$ are the same? – 2017-01-10
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    @user321627 : Yes. They both generate the same sigma-algebra. – 2017-01-10
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    The key fact here is that $$\sigma(u(X))\subseteq\sigma(X)$$ for every random variable $X$ and every measurable function $u$. (Can you prove this?) Then, applying the key fact to $X=X_1$, $u=\exp$, and to $X=e^{X_1}$ and $u=\log$, one gets $$\sigma(X_1)=\sigma(e^{X_1})$$ – 2017-01-10
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    @Did In this case, for the counter-example, $|X|$ is NOT a measurable function? – 2017-01-10
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    If you are referring to your comment above, considering $Z=|X|$, in general $E(Y\mid X)\ne E(Y\mid Z)$ because $\sigma(Z)\subset\sigma(X)$ but the reverse inclusion does not hold. And of course, the random variables $E(Y\mid X)$ and $E(Y\mid Z)$ are related by the identity $E(E(Y\mid X)\mid Z)=E(X\mid Z)$. – 2017-01-10

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I think this is much the same as a comment already given, but since $X_1$ is functionally dependant on $e^{X_1}$ and vice versa, if you have $e^{X_1}$, you have exactly the same information as having $X_1$. $X_1$ is just $ln(e^{X_1})$. Therefore they are the equivalent.

Note that this works in this case, but not all. It will work if the function is injective or one-to-one (that is, for a value of $F(X_1)$ there is only one $X_1$ that can lead to it).

It is not special to the exponential function exactly, but for a counter example, $E( X_2 \mid abs(X_1))$ does not necessarily equal $E( X_2 \mid X_1)$. This is because the absolute function is not injective; if we were told $abs(X_1)$ was $2$, $X_1$ could be either $-2$ or $2$, so we have less information than if we knew $X_1$ exactly.

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    Do you mean $abs(X_1) = 2$? – 2017-01-10
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    Oops, yes. I've edited it, thanks. – 2017-01-16