I am preparing for a test and am unsure of my workings on this exercise:
Calculate $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
Where $A^+ = \max (A,0)$
My approach was to calculate the transformation $U = X-Y, V = Y$ and get the expected value $$E[Y^2|X-Y] = E[V^2| U]$$ Which was equal to $\frac {(1-U)^3}{3} I_{(0,1)}(U) + (\frac 13 + \frac {U^3}2) I_{(-1,0)}(U)$. I used this in the following manner $$E[V^2| U^+] = E[E[V^2|U]|U^+]$$ Which led to two cases $$U\in (0,1):\; \frac 13 E[(1-U)^3| U^+] = \frac 13 E[(1-U)^3| U] = \frac 13 (1-U)^3$$ $$U \in (-1,0): \; \frac 12 E[U^3|U^+] + \frac 13 = \frac 12 E[U^3] + \frac 13$$
I am frankly quite confused at this point as to what I'm actually doing. Questions:
- $E[(1-U)^3| U^+] = E[(1-U)^3| U]$ is fair since $[U \in (0,1)] = [U^+ \in (0,1)]$ is that correct?
- $E[U^3|U^+ ] = E[U^3]$ because $U^3 \bot U^+$, given $U\in (-1,0)$
- More importantly, this all seems very dodgy. The whole two-cases business is quite unjustified in my mind and I am not sure, for example, what $E[U^3]$ is now supposed to be if $U\in (-1,0)$. My guess is it'd be zero as we're dealing with the triangular (symmetric) distribution and we're only saying the expression equals for $U\in (-1,0)$. But this would render 2. untrue, if I apply the knowledge of what values I am currently considering only after I do the calculation - as I said, I am confused.
I'd appreciate an attempt at clearing this up in my head.
Thanks!
edit: it is entirely possible I made a calculation error, however, that should not matter - it is the general confusion with the approach that needs clarifying, not the result itself.