# Variance reduction of an estimator arising from the marginal destribution of a Metropolis-Hastings chain

Let

• $$(E,\mathcal E,\lambda)$$ and $$(E',\mathcal E',\lambda')$$ be measure spaces
• $$f\in L^2(\lambda)$$
• $$I$$ be a finite nonempty set
• $$\varphi_i:E'\to E$$ be bijective $$(\mathcal E',\mathcal E)$$-measurable with $$\lambda'\circ\varphi_i^{-1}=q_i\lambda\tag1$$ for $$i\in I$$
• $$p,q_i:E\to[0,\infty)$$ be $$\mathcal E$$-measurable with $$\int p\:{\rm d}\lambda=\int q_i\:{\rm d}\lambda=1$$ for $$i\in I$$
• $$\mu:=p\lambda$$
• $$w_i:E\to[0,1]$$ be $$\mathcal E$$-measurable
• $$\tau':(I\times E')^2\to[0,\infty)$$ be $$(2^I\otimes\mathcal E')^{\otimes2}$$-measurable with $$\sum_{j\in I}\int\lambda'({\rm d}y')\tau'((i,x'),(j,y'))=1\tag2$$ for all $$(i,x')\in I\times E'$$
• $$\tau_{i,\:j}(x,y):=w_i(x)q_j(y)\tau'\left(\left(i,\varphi_i^{-1}(x)\right),\left(j,\varphi_j^{-1}(y)\right)\right)$$ for $$x,y\in E$$ and $$i,j\in I$$
• $$\alpha_{i,\:j}(x,y):=\begin{cases}\displaystyle\min\left(1,\frac{p(y)w_j(y)q_i(x)}{p(x)w_i(x)q_j(y)}\right)&\text{, if }p(x)w_i(x)q_j(y)>0\\1&\text{, otherwise}\end{cases}$$ for $$x,y\in E$$ and $$i,j\in I$$
• $$k(x,y):=\sum_{i,\:j\:\in\:I}\alpha_{i,\:j}(x,y)\tau_{i,\:j}(x,y)$$ for $$x,y\in E$$

Assume $$\{q_i=0\}\subseteq\{w_ip=0\}\;\;\;\text{for all }i\in I,\tag3$$ $$\{p=0\}\subseteq\{f=0\}\tag4$$ and $$\{pf\ne0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}.\tag5$$

I want to determine which choice of $$(w_i)_{i\in I}$$ is maximizing the quantity $$\int\mu({\rm d}x)\int\lambda({\rm d}y)k(x,y)|f(x)-f(y)|^2\tag6$$ subject to the constraints $$(3)$$ and $$(5)$$. How can we do that?

Note that, by definition, $$p(x)k(x,y)=\sum_{i,\:j\:\in\:I}\min(p(x)w_i(x)q_j(y),p(y)w_j(y)q_i(x))\tau'\left(\left(i,\varphi_i^{-1}(x)\right),\left(j,\varphi_j^{-1}(y)\right)\right)\tag7$$ for all $$x,y\in E$$.

EDIT: Maybe we can rewrite the quantity using that $$2\min(a,b)=a+b-|a-b|$$ for all $$a,b\in\mathbb R$$.