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world:coupling
This is an old revision of the document!
Table of Contents
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A new coupling technique
Let $\pi$, $\tilde \pi$ be two probability measures on the same measurable space $(\Xset,\Xsigma)$.
We draw jointly the couple of random variables $(\tilde Y,Y)$ according to the following procedure:
- Draw \(\tilde Y \sim \tilde \pi\)
- Draw a candidate \(X \sim \pi\) and accept the candidate $Y=X$ with probability \(1 -\alpha(X,\tilde Y)\) with \(\alpha(x,y)=\frac{\pi(y) \tilde \pi(x)}{\pi(x) \tilde \pi(y)} \wedge 1\). Otherwise reject the candidate and set \(Y=\tilde Y\).
Proposition. $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$.
1. What is nice is that we are able to couple these random variables whereas their densities are known only up to a multiplicative constant. I wonder if it is better to couple in that way: $\pi(y) \propto Q(x,\rmd y) \alpha_{MH}(x,y)$ and $\tilde \pi(y) \propto Q(x',\rmd y) \alpha_{MH}(x',y)$. These two densities are known only up to multiplicative constant. Up to some tricks, we can deduce a way of coupling two MH starting from different initial distributions? Can we compare it to the coupling of Pierre Jacob et al.?
1. Moreover, if we look at the the problem we have started today, if $(\tilde X_0,X_0)=\tilde \pi \otimes \pi$ then, $(\tilde X_1,X_1)$ is a coupling of $(\tilde \pi,\pi)$, no?
Proof
Obviously, $Y|_{\tilde Y} \sim K(\tilde Y,\cdot)$ where $$ K(\tilde y,\rmd y)=(1-\alpha(y,\tilde y)) \pi(\rmd y) + \lrb{\int \pi(\rmd x) \alpha(x,\tilde y)} \delta_{\tilde y} (\rmd y). $$
We now show that $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$. To do so, it is sufficient to check that for any bounded or non-negative function $f$, $\int \tilde \pi(\rmd \tilde y) K(\tilde y, \rmd y) f(y)=\pi(f)$.
Indeed, write: \begin{align*} \int \tilde \pi(\rmd \tilde y) K(\tilde y,\rmd y) f(y)&= \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int \tilde \pi(y) f(y)\rmd y \int \pi(x) \alpha(x,y) \rmd x \\ & = \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int f(y) \rmd y \lrcb{\int \underbrace{\tilde \pi(y) \pi(x) \alpha(x,y)}_{\tilde \pi(x) \pi(y) \alpha(y,x)} \rmd x} \\ & = \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int f(y) \rmd y \lrcb{\int \tilde \pi(x) \pi(y) \alpha(y,x) \rmd x}\\ & = \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int f(y) \pi(\rmd y) \lrcb{\int \alpha(y,x) \tilde \pi(\rmd x)} \\ & = \pi(f) \end{align*} which completes the proof.
Some comments
The probability of coupling is given by: $$ \PP(\tilde Y=Y)=\int \tilde \pi(\rmd \tilde y) \pi(\rmd y) \alpha (y,\tilde y)=\int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y $$
Question: we know that $\PP(\tilde Y=Y) \leq \int \pi(x) \wedge \tilde \pi(\tilde x) \rmd x$. But I can't see how to prove $$ \int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y \leq \int \pi(x) \wedge \tilde \pi(\tilde x) \rmd x $$
world/coupling.1718401029.txt.gz · Last modified: 2024/06/14 23:37 by rdouc
