world:coupling
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world:coupling [2024/06/15 09:12] rdouc |
world:coupling [2024/06/20 15:02] (current) rdouc ↷ Page moved from yazid:coupling to world:coupling |
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| * Draw \(\tilde Y \sim \tilde \pi\). | * Draw \(\tilde Y \sim \tilde \pi\). | ||
| - | * Draw \(X \sim \pi\) and set $Y=X$ 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 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 $(\tilde \pi,\pi)$. | ||
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| <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, can we 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 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? | * 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> | ||
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| - | 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 \\ | ||
world/coupling.1718435545.txt.gz · Last modified: 2024/06/15 09:12 by rdouc
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