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Consider two r.v.s $X,Y\geq1$ defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ which may not necessarily be integrable. Let $\mathbb{E}[X\mid \mathcal{G}]:=\lim_{n\to\infty}E[X\wedge n\mid\mathcal{G}]$ where $\mathcal{G}$ is any sub-$\sigma$-algebra of $\mathcal{F}$. I have $\mathbb{E}(X\mid Y)=Y$ and  $\mathbb{E}(Y\mid X)=X$. I need to prove the following: i) $\mathbb{E}\left(\frac{X}{Y}\right)=\mathbb{E}\left(\frac{Y}{X}\right)=1$; and ii) $X=Y$ a.s. by comparing $\mathbb{E}(1/X)$ and $1/\mathbb{E}(X)$. I also need to somehow use these to show that if instead $X,Y\geq-1$, $X=Y$ a.s.

I have read through the proofs outlined in here but the conditions for this case are not the same as the one here (since we cannot assume integrability anyway). What should I do in this case? I don’t really know to go about both cases. Thanks!

StubbornAtom
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user107224
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1 Answers1

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Suppose that $\mathbb E\left[\frac{X}{Y}\right]=\mathbb E\left[\frac{Y}{X}\right]=1$.

Since $U=\frac{X}{Y}> 0$ is integrable and $x\rightarrow \frac{1}{x}$ is strictly convex on $]0,+\infty[$, Jensen inequality says that \begin{align*} \mathbb E\left[\frac{1}{U}\right]&\geq\frac{1}{\mathbb E[U]} \end{align*} with equality if and only if $U$ is almost surely constant (and so $U=\mathbb E[U]~a.s.$). By the previous inequality, we know that $\mathbb E[U]=1$ and $\mathbb E\left[\frac{1}{U}\right]=\mathbb E\left[\frac{Y}{X}\right]=1$ so we have equality. This means that $\frac{X}{Y}=\mathbb E[U]$ almost surely and so we indeed get $X=Y$ almost surely.

We can compute \begin{align*} \mathbb E\left[\frac{X}{Y}\middle| Y\right] &= \lim_{n\to \infty} \mathbb E\left[\frac{X}{Y}\land n\middle| Y\right]\\ &= \frac{1}{Y}\lim_{n\to \infty} \mathbb E\left[X\land (Yn)\middle| Y\right]\\ &= \frac{1}{Y}\lim_{x\to \infty} \mathbb E\left[X\land x\middle| Y\right]\\ &= \frac{1}{Y}\mathbb E[X|Y]\\ &= 1 \end{align*} almost surely. We change the limit from $n$ to $x$ by multiplying by $Y\geq 1$, observe that $X\land (Yn)$ and $X\land x$ will converge to the same limit almost surely by conditional dominated convergence theorem (actuelly they converge to $X$). However there may be some integrability conditions I missed for the conditional dominated convergence theorem.

P. Quinton
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  • Hi, why can we automatically conclude U is integrable? Also what does the notation $]0,+\infty[$ mean, is it like $(0,\infty)$? – user107224 Nov 29 '20 at 11:48
  • Additionally for the second half of the proof, why can we bring $1/Y$ inside the expectation? I wouldn’t have thought it was “part of $\sigma(X)$”. – user107224 Nov 29 '20 at 11:52
  • Last question which is how do I extend this argument for $X,Y\geq-1$? Especially since we may encounter either $X$ or $Y$ to be zero? – user107224 Nov 29 '20 at 12:10
  • $]0,\infty[$ is the same as $(0,\infty)$. Observe that since $U=|U|$ then $\mathbb E[|U|]=1<\infty$. I think you are right, there is a problem with the second part of the proof (and for $X,Y\geq -1$, I don't know yet) – P. Quinton Nov 29 '20 at 12:37
  • I guess if $X,Y\geq -1$, then $X+2$ and $Y+2$ are $\geq 1$, so that $X+2=Y+2$ almost surely (by linearity of conditional expectation) and so $X=Y$ almost surely. – P. Quinton Nov 29 '20 at 12:41