2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $\left\{(x,y)\in E^2:p(x)q(x,y)>0\right\}$ and $\left\{(y,x)\in E^2:p(x)q(x,y)>0\right\}$ are usually the same set $\endgroup$ – Taylor Aug 26 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 at 15:53
  • 1
    $\begingroup$ 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 at 12:19
  • 1
    $\begingroup$ 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 at 12:22
  • 1
    $\begingroup$ @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 at 12:23
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Can we fixed the claim somehow such that it remains true in the general case? $\endgroup$ – 0xbadf00d Aug 27 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 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 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 at 19:13
  • 1
    $\begingroup$ 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 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.