equality of conditional expectation

Question

Let $X_1,...,X_k$ be independent exponentially distributed with parameter $\lambda$ random variables.

Is this equality holds?

$$E[e^{-X_1-...-X_k}|X_1+...+X_k]=e^{-X_1-...-X_k}$$

My textbook says it holds, but I don't know why.

I know that if random variable $X$ is measuable with respect to sigma algebra generated by random variable $Y$ then: $E[X|Y]=X$. But why is $e^{-X_1-...-X_k}$ measurable with respect to sigma algebra generated by $X_1+...+X_k$

I tried to use theroem which says: Image of borel set under continuous, injective function from $R$ to $R$ is borel set, but in my situation I would like to apply it to $-ln(x)$ but this function is from $(0,\infty)$ not $R$ and I think it matter because borel set can have some negative numbers and $ln(x)$ is undefined for negative numbers.

Answer 1

As a non-rigorous rule which "almost surely" works is that functions(random variables) which are measurable wrt $\sigma(Y)$ are precisely the one's which are determined if $Y$ is known.

So keeping with your notation, let $f:\Bbb{R}\to\Bbb{R}$ be a Borel Measurable function, then $f(Y)$ is measurable wrt $\sigma(Y)$ i.e. if $B\in\mathcal{B}(\Bbb{R})$ then $f^{-1}(B)$ is Borel and hence $Y^{-1}(f^{-1}(B))\in\sigma(Y)$ and that's it. $f(Y)$ is thus measurable wrt $\sigma(Y)$.

So in particular $e^{\pm Y}$ is also measurable and that's precisely what your book is using.

An advice would be to not overthink while doing Conditional Expectation as it can lead to a whole lot of unintended conceptual mistakes.

Answer 2

Is is a direct consequence of conditioning ("taking out what is known"). The rv's only enter through a sum, so, let's define $Z=\sum_{n=1}^k X_k$.

You are trying to evaluate $E[e^{-Z}|Z]$, which is a function of $Z$. But because of conditioning in this case, the value of $Z$ is "known", since you condition on it, hence there is nothing to integrate. So, it follows directly that $E[e^{-Z}|Z]=e^{-Z}E[1|Z]=e^{-Z}$.