2
$\begingroup$

Let $(E,\mathcal E,\lambda)$ be a measure space, $p:E\to[0,\infty)$ be $\mathcal E$-measurable with $$c:=\int p\:{\rm d}\lambda$$ and $$\mu:=\underbrace{\frac1cp}_{=:\:\tilde p}\lambda$$ denote the measure with density $\tilde p$ with respect to $\lambda$.

Now let $q:E:\to[0,\infty)$ be $\mathcal E$-measurable with $$\int q\:{\rm d}\lambda=1$$ and $$\left\{q=0\right\}\subseteq\left\{p=0\right\}\tag1$$ and $\nu:=q\lambda$. Suppose we're actually interested in the distribution $\mu$. Is there any reason why we might prefer to run the Metorpolis-Hastings with target distributon $\nu$ instead of $\mu$?

Clearly, if $(X_n)_{n\in\mathbb N_0}$ and $(Y_n)_{n\in\mathbb N}$ denote the processes generated by the Metropolis-Hastings algorithm with target distribution $\mu$ and $\nu$, respectively, then $$\frac cn\sum_{i=b}^{b+n-1}f(X_i)\xrightarrow{n\to\infty}c\int f\:{\rm d}\mu=\int pf\:{\rm d}\lambda\tag2$$ and $$\frac1n\sum_{i=b}^{b+n-1}\frac{pf}q(Y_i)\xrightarrow{n\to\infty}\int\frac{pf}q\:{\rm d}\nu=\int pf\:{\rm d}\lambda$$ $\lambda$-almost everywhere for all $\mathcal E$-\measurable $f:E\to\mathbb R$ with $pf\in L^1(\lambda)$ and $b\in\mathbb N_0$. So, both processes can be used to approximate the integral on the right-hand side of $(2)$ and $(3)$.

I could imagine that the convergence to equilibrium can be sped up by choosing a suitable $q$. But this is just a vague speculation. Would be great if someone could elaborate on that.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.