Conditional expectation and sums of independent exponential random variables Consider this simple case: Let $X$ and $Z$ be i.i.d. exponential RVs of parameter $1$ and let $Y=X+Z$. I obtain that
$$E[X|Y] = 1$$
Can you help me get an intuitive understanding of why this is true?
Brief derivation in case I made a mistake somewhere:


*

*$F_{Y|X}(y|x) = 1-e^{-(y-x)}$ and thus $f_{Y|X}(y|x) = e^{-(y-x)}$

*$f_{X,Y}(x,y) = f_X(x)\cdot f_{Y|X}(y|x) = e^{-y}$

*$f_Y(y) = y\ e^{-y}$

*$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{1}{y}$

*$E[X|Y=y] = \int_{R_x} x\cdot f_{X|Y}(x|y)\ dx = 1$


We have: $ X,Z\mathop{\sim}^\textsf{iid}\mathcal{Exp}(1)\\ Y=X+Z\\ \mathsf E(X+Z\mid Y) = \mathsf E(Y\mid Y) \\ \mathsf E(X\mid Y) = \mathsf E(Z\mid Y) \quad\textsf{due to symmetry, and identical distribution} \\ \therefore \mathsf E(X\mid Y) = Y/2$ 
