1
$\begingroup$

Problem: Let $U,V$ be two i.i.d. standard uniform RV on $(0,1)$, compute $E((U-V)^+ \mid U)$


My solution:

We choose $Z \in L^\infty (\Omega, \sigma(U), \mathbb{P}) \iff Z=f(U)$ for some bounded borel measurable function $f$ and want to compute $$E(\max(U-V,0)f(U))= \int_{(0,1)^2} \max(u-v,0) f(u)dudv \\ = \int_{(0,1)^2} \max(u-v,0)1_{u-v>0} f(u) du dv + \underbrace{\int_{(0,1)^2} \max(u-v,0)1_{u-v \leq 0}f(u)dudv}_{=0} \\ \overset{(*)}=\int_0^1\int_0 ^u uf(u)dvdu-\int_0^1 \int_0^u vf(u)dvdu =\int_0^1 u^2f(u)du- \int_0^1 \frac{u^2}{2}f(u)du \\ = \int_0^1 \frac{u^2}{2}f(u)du = E((U^2/2)f(U)) $$ From which I conclude that $$ E((U-V)^+ \mid U)= U^2/2 $$ I did omit some details about the use of Fubini-Tonelli's Theorem in (*), but indeed with $f$ being bounded the use of this theorem is justified.


Question: Is my solution correct? If not, please highlight my mistake and try pointing me towards the right solution. Unfortunately I can't ask my tutors this week and I don't want to fall behind on this topic.

$\endgroup$
2
$\begingroup$

$$E((U-V)^+ \mid U) = \int_0^1 (U-x)^+dx= \int_0^U (U-x)dx =\frac{U^2}{2}$$

$\endgroup$
  • $\begingroup$ Thanks @Did. That looks much better. $\endgroup$ – Gordon Oct 2 '16 at 17:44
  • $\begingroup$ Nice answer, +1. $\endgroup$ – Did Oct 2 '16 at 17:45
  • $\begingroup$ I am sure this answer is correct, since I am very new to the conditional expectation I can only guess that some proposition was involved in calculating this. However, my question written in bold in the OP wasn't strictly answered. Maybe I should reformulate "Is my solution correct?" to "Is this the correct way to approach the solution when only working with the definition of the conditional expectation?" $\endgroup$ – Spaced Oct 3 '16 at 15:57
  • 1
    $\begingroup$ Your answer is correct and is also a good approach, though I prefer some simple approach as above. $\endgroup$ – Gordon Oct 3 '16 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.