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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.

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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
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.

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