Calculate conditional expectation with exponential distributed random variables

Question

X, Y are independent random variables, and X, Y~Exp(λ) with λ>1.

Problem is to calculate $$E[e^X + e^Y| X^3]$$

Part of my solution: Since X,Y are independent, and first part uses "law of the unconscious statistician.", I have
$$E[e^X + e^Y| X^3]=E[e^Y]+E[e^X|X^3]=λ/(λ-1)+ E[e^X|X^3]$$

I get stuck on how to compute this part $$E[e^X|X^3]$$ Can I use $f(x^3)=e^x , E[f(X^3)|X^3]=f(X^3)?$

I have this formula

E[f(Z)|Z]=f(Z), where f is a Borel-measurable function

Is there such a formula, and how can I calculate this conditional expectation?

Thanks in advance!

Answer 1

The formula is correct. $e^{X}=f(X^{3})$ where $f(t)=e^{t^{1/3}}$. $f$ is a continuous, hence measuarble function. This makes $e^{X}$ a measurable function of $X^{3}$ so $E(e^{X}|X^{3})=e^{X}$.

In general, $E(X|Y)=X$ if $X=g(Y)$ for some measurable function $g: \mathbb R \to \mathbb R$.