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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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    $\begingroup$ Knowing $e^{X_1}$ is equivalent to knowing $X_1$. So the conditional expectation is based on the exact same "evidence". $\endgroup$
    – Alt
    Jan 10, 2017 at 2:04
  • $\begingroup$ Would you give details for "An explanation I read in a book "? What is the context? $\endgroup$
    – user9464
    Jan 10, 2017 at 2:07
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    $\begingroup$ 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. $\endgroup$ Jan 10, 2017 at 2:21
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    $\begingroup$ @user321627 : Yes. They both generate the same sigma-algebra. $\endgroup$ Jan 10, 2017 at 2:28
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    $\begingroup$ 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 $$\sigma(X_1)=\sigma(e^{X_1})$$ $\endgroup$
    – Did
    Jan 10, 2017 at 9:27

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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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    $\begingroup$ Do you mean $abs(X_1) = 2$? $\endgroup$ Jan 10, 2017 at 10:31
  • $\begingroup$ Oops, yes. I've edited it, thanks. $\endgroup$
    – timbo
    Jan 16, 2017 at 0:58

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