Poisson Process Conditional Expectation Given $X_t$ a Poisson process such that $\lambda = 1$ find $E[X_1\mid X_2]$ and $E[X_2\mid X_1]$. 
The first one is pretty straight forward since we have $E[X_2 - X_1] = E[X_1] = 1$ so then we get $E[X_2 \mid X_1] = E[X_2 - X_1 + X_1 \mid X_1] = E[X_2 - X_1 \mid X_1] + E[X_1\mid X_1] = 1+ X_1$ as $X_2-X_1$ is independent of $X_1$. 
As for $E[X_1\mid X_2]$ I'm not entrirely sure, I already know the answer should be $\frac{X_2}{2}$ but that is not what I am getting. It relates to binomial if I understand correctly. 
I believe I should have $$P(X_1\mid X_2) = \frac{P(X_1 = x, X_2 = y)}{P(X_2 = y)} = \frac{P(X_1 = x, X_2 - X_1 = y-x)}{P(X_2 = y)}  = \frac{P(X_1= x)P(X_2 - X_1 = y-x)}{P(X_2 = y) } = \frac{e^{-1}(x!)^{-1}e^{-1}((y-x)!)^{-1}}{e^{-2}(2^{y}/y!)} = \frac{y!}{x!(y-x)!}2^{-y}$$ but Im missing the $2^{-(y-x)}$.
Maybe I made a mistake in my conditional probability. 

By what you calculated, the PMF of $X_1$ conditional on $X_2 =y$ is that of a $\mathrm{Bin}(y,1/2)$ random variable. Therefore, $\mathbb{E}[X_1\mid X_2 = y] = y/2$ and $\mathbb{E}[X_1\mid X_2] = X_2/2$.
