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Let

  • $(E,\mathcal E,\lambda)$ be a measure space
  • $p:E\to[0,\infty)$ be $\mathcal E$-measurable with $$c:=\int p\:{\rm d}\lambda\in(0,\infty)$$ and $$\mu:=\underbrace{\frac1cp}_{=:\:\tilde p}\lambda$$
  • $q:E^2\to[0,\infty)$ be ${\mathcal E}^{\otimes2}$-measurable and $$Q(x,\;\cdot\;)=q(x,\;\cdot\;)\lambda\;\;\;\text{for }x\in E$$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $X$ be an $(E,\mathcal E)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with $X\sim\mu$
  • $Y$ be an $(E,\mathcal E)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with$^1$ $(X,Y)\sim\mu\otimes Q$

Note that $$\xi:=\frac{p(Y)q(Y,X)}{p(X)q(X,Y)}$$ is almost surely well-defined. How can we show that $\operatorname E\left[\xi\right]=1$?

Let $$B:=\left\{(x,y)\in E^2:p(x)q(x,y)>0\right\}$$ and $N:=E^2\setminus B$. Note that $N$ is a $\mu\otimes Q$-null set and

\begin{equation} \begin{split} \operatorname E\left[\xi\right]&=\int\mu({\rm d}x)\int Q(x,{\rm d}y)1_B(x,y)\frac{p(y)q(y,x)}{p(x)q(x,y)}\\&=\int\lambda({\rm d}x)\lambda({\rm d}y)1_B(x,y)\tilde p(y)q(y,x)\\&=\int\mu({\rm d}y)\int Q(y,{\rm d}x)1_B(x,y). \end{split}\tag1 \end{equation}

However, $(1)$ is not equal to $1$ (unless there is some kind of "reversibility" allowing us to swap the arguments of $1_B$). So, is the claim wrong as stated?

The claim can be found inside the proof of Theorem 1 on page 13 here: https://arxiv.org/pdf/1810.07151.pdf. (The proof itself is too complicated. I guess the author missed the point that $2(a\wedge b)=a+b-|a-b|$ for all $a,b\in\mathbb R$.)

EDIT: If the claim is wrong, what I really want to show is the claim about the total variation distance in Theorem 1.

$^1$ see https://en.wikipedia.org/wiki/Transition_kernel#Product_of_kernels.

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    $\begingroup$ $\left\{(x,y)\in E^2:p(x)q(x,y)>0\right\}$ and $\left\{(y,x)\in E^2:p(x)q(x,y)>0\right\}$ are usually the same set $\endgroup$
    – Taylor
    Aug 26, 2019 at 15:35
  • $\begingroup$ @Taylor Sorry I cannot follow. How does this help unless $q$ is symmetric? And I don't see why your sets are equal? $\endgroup$
    – 0xbadf00d
    Aug 26, 2019 at 15:53
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    $\begingroup$ Isn't it a violation of your agreement as a reviewer to provide a link to a paper that is under "double blind review"? $\endgroup$
    – whuber ♦
    Aug 27, 2019 at 12:19
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    $\begingroup$ Thank you for clarifying the status: your post was being flagged based on that concern. But since the paper has been published, is there some reason you are linking to a review copy? $\endgroup$
    – whuber ♦
    Aug 27, 2019 at 12:22
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    $\begingroup$ @whuber Yes, I've seen that the paper has been published after I saw it first on openreview. But I can exchange the link. $\endgroup$
    – 0xbadf00d
    Aug 27, 2019 at 12:23

1 Answer 1

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When the support of $q(x,\cdot)$ differs from the support of $p(\cdot)$ then the expectation of the ratio is not necessarily one. As an illustration, take \begin{align} p(x) &= \frac{1}{3}\Bbb I_{(1,4)}(x)\\ q(x,y) &= \frac{1}{3}\Bbb I_{(x-1,x+2)}(y) \end{align} Then the expectation of $$\Bbb I_{(1,4)}(x) \Bbb I_{(x-1,x+2)}(y)$$ under the density $p(y)q(y,x)$ is 5/9.

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  • $\begingroup$ Thank you for your answer. Can we fixed the claim somehow such that it remains true in the general case? $\endgroup$
    – 0xbadf00d
    Aug 27, 2019 at 12:09
  • $\begingroup$ Does defining $$r(x,y):=\left.\begin{cases}\displaystyle\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)\ne0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x,y\in E$$ and $\xi:=r(X,Y)$ help? With this definition, I've observed that $$\operatorname E\left[|1-\xi|\right]=\frac1c\int\lambda({\rm d}x)\int\lambda({\rm d}y)|p(x)q(x,y)-p(y)q(y,x)|\tag2$$ which might help to solve my total variation question. $\endgroup$
    – 0xbadf00d
    Aug 27, 2019 at 12:40
  • $\begingroup$ Hi. Sorry, I can't follow. What exactly does not work? Actually, I think this is the way we should have defined $\xi$ in the first place. Before this definition was hidden behind the "almost surely well-defined" claim. But clearly, the set on which it is defined to be $1$ is a null set with respect to the distribution of $(X,Y)$. Please note that my main concern is to prove the total variation claim in Theorem 1 (which might hold even though $\operatorname E\left[\xi\right]\ne1$). $\endgroup$
    – 0xbadf00d
    Aug 27, 2019 at 14:03
  • $\begingroup$ I checked the paper and think the Total Variation identity suffers from the same difficulty. A sufficient condition for it to hold is that the support of $p(x)q(x,y)$ is the same as the support of $p(y)q(y,x)$. $\endgroup$
    – Xi'an
    Aug 27, 2019 at 19:13
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    $\begingroup$ I've accepted your answer, since I initially asked for showing that $\operatorname E\left[\xi\right]=1$. I've asked a separate question for the total variation thing I'm still interested in and included some observations I've made meanwhile: stats.stackexchange.com/q/423913/222528. Would be great if you could take look. $\endgroup$
    – 0xbadf00d
    Aug 27, 2019 at 19:27

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