For two random variables $X_1, X_2$, is it always necessarily the case that $E(e^{X_2}\mid e^{X_1}) = E(e^{X_2}\mid X_1)$?

Question

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.

Answer 1

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.