world:coupling
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world:coupling [2024/06/14 23:37] rdouc |
world:coupling [2024/06/20 15:02] (current) rdouc ↷ Page moved from yazid:coupling to world:coupling |
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| We draw jointly the couple of random variables $(\tilde Y,Y)$ according to the following procedure: | We draw jointly the couple of random variables $(\tilde Y,Y)$ according to the following procedure: | ||
| - | * Draw \(\tilde Y \sim \tilde \pi\) | + | * 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 \(Y=\tilde Y\). | + | * 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 \(Y=\tilde Y\). |
| <WRAP center round tip 80%> **Proposition. ** $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$. | <WRAP center round tip 80%> **Proposition. ** $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$. | ||
| Line 14: | Line 14: | ||
| <WRAP center round important 80%> | <WRAP center round important 80%> | ||
| - | * 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.? | + | * 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 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? | + | * 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? |
| </WRAP> | </WRAP> | ||
| Line 27: | Line 27: | ||
| - | Indeed, write: | + | Indeed, write, using the detailed balance condition in the second line : |
| \begin{align*} | \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 \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 \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) \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)} \\ | + | & = \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) | & = \pi(f) | ||
| \end{align*} | \end{align*} | ||
| Line 40: | Line 40: | ||
| - | The probability of coupling is given by: | + | The coupling probability 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 | + | \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 |
| $$ | $$ | ||
| <WRAP center round tip 80%> | <WRAP center round tip 80%> | ||
| - | **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 | + | **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 | \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.1718401056.txt.gz · Last modified: 2024/06/14 23:37 by rdouc
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