6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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 '17 at 2:04
  • $\begingroup$ Would you give details for "An explanation I read in a book "? What is the context? $\endgroup$ – Jack Jan 10 '17 at 2:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Hardy Jan 10 '17 at 2:21
  • 1
    $\begingroup$ @user321627 : Yes. They both generate the same sigma-algebra. $\endgroup$ – Michael Hardy Jan 10 '17 at 2:28
  • 3
    $\begingroup$ 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 '17 at 9:27
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Do you mean $abs(X_1) = 2$? $\endgroup$ – probs_not12 Jan 10 '17 at 10:31
  • $\begingroup$ Oops, yes. I've edited it, thanks. $\endgroup$ – timbo Jan 16 '17 at 0:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.