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I have this problem:

Let $(\Omega, \mathscr{F}, \mathbb{P}$ be a probability space, $\mathscr{G}$ be a sub-$\sigma$-algebra. Let $X,Y$ be two real random variables such that for every bounded Borel-measurable function $g: \mathbb{R} \rightarrow \mathbb{R}^+$ we have $$\mathbb{E}\left(g(X)\mid\mathscr{G}\right) = g(Y) \text{ a.s.}$$ The question is if that implies $X=Y \text{ a.s}$ .

I guess the answer is yes, so I tried to prove that $\{X<Y\}$ is a nullset. If this set is in $\mathscr{G}$ it is obviously a nullset. But otherwise i get stuck. Does anyone have an idea?

Michael Hardy
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abcxyzf
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1 Answers1

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First, integrating the equality given in the assumption gives that $\mathbb E\left[g(X)\right]= \mathbb E\left[g(Y)\right]$ for any bounded function $g$, hence considering $g$ as the indicator function of $(-\infty,t]$ where $t\in\mathbb R$, we derive that $X$ and $Y$ have the same distribution.

Therefore, for any bounded function $g$, $g(X)=g(Y)$ in distribution and by assumption, $\mathbb E\left[g(X)\mid\mathcal G\right]$ has the same distribution as $g(X)$, which entails that $\mathbb E\left[g(X)\mid\mathcal G\right]=g(X)$ a.s.. Going back to the initial assumption, we derive that for any bounded and non-negative function $g$, $g(X)=g(Y)$ a.s. From this, we can deduce equality of the positive parts, and by a similar reasoning that of the negative parts.

Community
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Davide Giraudo
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  • Btw i found a very simple solution: We want to show {X<Y} is a nullset, assume that is not, then we can find some rational q s.t. {X<q<Y} also has a positiv measure. Choose g as the indicator function on (-∞,q) and consider the integrals on {Y>q} to obtain a contradiction. – abcxyzf Dec 14 '16 at 23:08