$\mathbb{E}\big(e^{X}\big|$ $e^{Y}\big)\overset{?}{=}\mathbb{E}\big(e^{X}\big|$ $Y\big)$

Question

I have a quick question. Is the following reasoning correct? If not, why?

I know that $X|Y\in N(\rho Y, 1-\rho^2)$.

I want to deduce an expression for $\mathbb{E}\big(e^{X}\big|$ $e^{Y}\big)$. My idea was as following:

$\mathbb{E}\big(e^{X}\big|$ $e^{Y}\big)\overset{?}{=}\mathbb{E}\big(e^{X}\big|$ $Y\big)=M_{X|Y}(1)=e^{\rho Y+\frac{1}{2}(1-\rho^2)}$

where $M_{X|Y}$ is the moment generating function for $X|Y$.

I know that the expression that I got at the end is correct. I guess that what I'm really asking is whether my reasoning is correct and if a continuous function of a random variable $X$ is $X$-measurable?

Answer 1

Yes, provided that $e^{X}$ is integrable. It is because $\sigma(Y)=\sigma(e^{Y})$.

Proof: Clearly $e^{Y}$ is $\sigma(Y)$-measurable, so $\sigma(e^{Y})\subseteq\sigma(Y)$. On the other hand, $Y=\ln\left(e^{Y}\right)$ which is $\sigma(e^{Y})$-measurable, so $\sigma(Y)\subseteq\sigma(e^{Y})$.