# Is there a reason why we should run the Metorpolis-Hastings algorithm with a target density approximating the density we're actually after?

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.