Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $U$ and $V$ are random variables. Then I want to show that if $$ E[g(U)\mid V]=g(V), $$ for all bounded non-negative Borel-measureable functions $g:\mathbb{R}\to\mathbb{R}_+$, then $U=V$ a.s.

I'm given a hint, which is to use the following result:

If $X\in \mathcal{L}^2(P)$, $\mathcal{B}$ is a sub-$\sigma$-field and we put $Y=E[X\mid \mathcal{B}]$ then $$ X\sim Y \Rightarrow X=Y \,\text{ a.s.} $$

My thoughts so far: If I can show that $g(U)\sim g(V)$ for all such $g$, then $g(U)=g(V)$ a.s., and I think this is enough to show that $U=V$ a.s. (i.e. I'm thinking of using $g_n(x)=-n\vee x\wedge n$ and let $n\uparrow \infty$). Please correct me if I'm wrong here. If this the way to proceed, I just need a hint or a tip on how to show $g(U)\sim g(V)$.

Thanks in advance.

share|improve this question
    
Thanks for your reply. Choosing $g$ like that implies that $U\sim V$, but how to I move on from here and show that $U=V$ a.s.? –  Stefan Hansen Mar 26 '12 at 13:31
    
@sos440: That is an answer, probably the answer if you are kind and add a few more words about the implication. –  Henry Mar 26 '12 at 13:31
    
@sos440: Thanks again, I think I follow what you say. The important thing, as I see it, is that $g$ is a bounded function defined on $\mathbb{R}$ which is injective. Any particular reason for adding $\pi/2$ in your $g$? –  Stefan Hansen Mar 26 '12 at 13:55
    
@sos440: Of course that is in order to make $g$ non-negative. I'd be happy to put this as the answer, if you would type it as an answer. –  Stefan Hansen Mar 26 '12 at 14:00
add comment

1 Answer

up vote 2 down vote accepted

We first claim that the given condition implies $U \sim V$. For each $a \in \mathbb{R}$, put

$$g(x) = \mathbf{1}_{\{x \leq a\}} = \begin{cases} 1 & x \leq a \\ 0 & x > a \end{cases}.$$

Then we have

$$ \mathbb{P}(V \leq a) = \mathbb{E}[g(V)] = \mathbb{E}[\mathbb{E}[g(U)|V]] = \mathbb{E}[g(U)] = \mathbb{P}(U \leq a), $$

hence the proof of the claim follows. In particular, for any Borel function $g : \mathbb{R} \to \mathbb{R}$ we have $g(U) \sim g(V)$.

Now let $g : \mathbb{R} \to \mathbb{R}$ be a bounded injective continuous function. For example, we may put

$$ g(x) = \frac{\pi}{2} + \arctan x.$$

Then both $g(U)$ and $g(V)$ are in $L^{\infty}(\mathbb{P}) \supset L^{2}(\mathbb{P})$ and satisfy both $g(V) = \mathbb{E}[g(U) | V]$ and $g(U) \sim g(V)$.

Then the hint yields $g(U) = g(V)$ a.s., hence $U = V$ a.s. as desired.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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