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world:coupling
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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 \(X \sim \pi\) and set $Y=X$ with probability \(1 -\alpha(X,\tilde Y)\) where \(\alpha(x,y)=\frac{\pi(y) \tilde \pi(x)}{\pi(x) \tilde \pi(y)} \wedge 1\). Otherwise set \(Y=\tilde Y\).
Proposition. $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$.
- 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 interesting to couple in that way: $\pi(y) \propto Q(x,\rmd y) \alpha_{MH}(x,y)$ and $\tilde \pi(y) \propto Q(x',\rmd y) \alpha_{MH}(x',y)$. Up to some tricks, can we deduce a way of coupling two MH starting from different initial distributions, maybe with delayed coupling? Can we compare it to the coupling of Pierre Jacob et al.?
- Moreover, if we look at the problem 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, using the detailed balance condition in the second line : \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{1-\int \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 coupling probability is given by: $$ \PP(\tilde Y=Y)=\int \tilde \pi(\rmd \tilde y) \pi(\rmd x) \alpha (x,\tilde y)=\int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y $$
Question: we know that $\PP(\tilde Y=Y) \leq \int \pi(x) \wedge \tilde \pi(x) \rmd x$. 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.txt · Last modified: 2024/06/20 15:02 by rdouc
