Suppose you have $X$ and $Y$ on a common probability space Let $h,g$ be arbitrary bounded Borel-measurable functions. If you have:

$$ \mathbb{E}(h(X)g(Y)) = \mathbb{E}(h(X)g(X)) $$

Do you also have X=Y a.s.?

My intuition is yes, using $h=1,g(\cdot)=|\cdot-X|$ (is this allowed?), and then LHS=RHS=0 meaning $|X-Y|=0$ a.s.

  • $\begingroup$ To see if something is allowed, just track their related sets. Like, $X:\Omega\to\mathbb{R}$, $g:\mathbb{R}\to\mathbb{R}$. $\endgroup$
    – Atbey
    Apr 28, 2021 at 5:53

2 Answers 2


For $A,B\in \mathcal B(\mathbb R)$, using the hypothesis with $h=1_A$ and $g=1_B$, $$P\big((X\in A) \cap (Y\in B)\big) = P\big(X\in (A\cap B)\big).$$

Taking $A=\mathbb R$ yields $ \forall B\in \mathcal B(\mathbb R)$, $P(Y\in B) = P(X\in B)$, hence $X\stackrel{d}= Y$.

Consequently, for any bounded measurable $h,g$ we have $$E\big(h(X)g(Y) \big) = E\big(h(X)g(X) \big) = E\big(h(Y)g(Y) \big)$$ and exchanging $h$ and $g$ we have additionally $$E\big(g(X)h(Y) \big) = E\big(g(X)h(X) \big)$$ thus $$E\big(h(X)g(Y) \big) = E\big(g(X)h(Y) \big) = E\big(h(X)g(X) \big) = E\big(h(Y)g(Y) \big)$$

so that $$E\big([g(X)-g(Y)][h(X)-h(Y)] \big) = 0.$$

Taking $g=h=\arctan$ yields $$E\big([\arctan(X)-\arctan(Y)]^2 \big) = 0,$$ hence $\arctan(X)=\arctan(Y)$ a.s. and $X=Y$ a.s.


You cannot make that choice of $g$ but the result is true:

You get $P(X \in A, Y\in A)=P(X \in A)$ by taking $g=h=I_A$. But $(X \in A, Y\in A) \subseteq (X \in A)$. From this it follows that $X\in A$ implies $Y \in A$ a.s. [in the sense $P(X \in A, Y \notin A)=0$. Thus $\frac {i-1}n \leq X \leq \frac i n$ implies $\frac {i-1}n \leq Y \leq \frac i n$ so $|X-Y| \leq \frac 1 n$ a.s. Can you take it from here?

  • $\begingroup$ I can follow your first argument but in your last step, are you using $g$ to squeeze $X,Y$ to an interval? Wouldn't that fail if you let $Y=X (\%) \frac{i-1}{n}$ (the remainder)? $\endgroup$ Apr 28, 2021 at 5:43
  • $\begingroup$ $X$ always lies between $\frac {i-1} n$ and $\frac i n$ for some $i$. For that $i$ it follows that $Y$ also lies between these limits so $|X-Y| \leq \frac 1 n$. [The only techical point here is to find one null which works for each $n$ but that is certainly possible]. @Isomorphism $\endgroup$ Apr 28, 2021 at 5:47

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