1
$\begingroup$

Let $\tilde\kappa$ denote the transition kernel of the Markov chain generated by the Metropolis-Hastings algorithm with proposal kernel $\tilde Q$ and target distribution $\tilde\mu$ (see definitions below).

I want to choose $(w_i)_{i\in I}$ such that there is a $\beta\in[0,1)$ as small as possible with $$\left\|\tilde\kappa(\tilde g-\tilde\mu\tilde g)\right\|_{L^2(\tilde\mu)}\le\beta\left\|\tilde g-\tilde\mu\tilde g\right\|_{L^2(\tilde\mu)}\tag1$$ for all nonnegative $g\in\mathcal L^2(\mu)$ and $$\tilde g(i,x'):=(g\circ\varphi_i)(x')\;\;\;\text{for }(i,x')\in\tilde E.$$

Fix any such $\tilde g$. Note that $\int\tilde g\:{\rm d}\tilde\mu=\int g\:{\rm d}\mu$ and $$(\tilde\kappa\tilde g)(\tilde x)-\int\tilde g\:{\rm d}\tilde\mu=\int\tilde\lambda({\rm d}\tilde y)\left(\tilde k(\tilde x,\tilde y)-\tilde p(\tilde y)\right)\left(\tilde g(\tilde y)-\tilde g(\tilde x)\right)\tag2.$$

I guess we first need to show that there exists (at least one) choice of $(w_i)_{i\in I}$ such that there is a $\beta$ satisfying $(1)$ and then try to minimize $\beta$ over all possible choices of $(w_i)_{i\in I}$. How can we possibly do that?

Substituting the definitions, we may note that \begin{equation}\begin{split}&\left\|\tilde\kappa(\tilde g-\tilde\mu\tilde g)\right\|_{L^2(\tilde\mu)}^2\\&\;\;\;\;=\sum_{i\in I}\int_{\{\:w_ip\:>\:0\:\}}\mu({\rm d}x)w_i(x)\left|\sum_{j\in I}\int\lambda({\rm d}y)\left(q_j\wedge\frac{w_j(y)p(y)}{w_i(x)p(x)}q_i(x)\right)\sigma_{ij}(x,y)(g(y)-g(x))+g(x)-\mu g\right|^2.\end{split}\tag3\end{equation}

Definitions: Let

  • $(E,\mathcal E,\lambda)$, $(E',\mathcal E',\lambda')$ be measurable spaces;
  • $I$ be a finite nonempty set;
  • $p,q_i$ be probability densities on $(E,\mathcal E,\lambda)$ for $i\in I$;
  • $w_i:E\to[0,1]$ be $\mathcal E$-measurable with $\sum_{i\in I}w_i=1$;
  • $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$-measurable with $\lambda'\circ\varphi_i^{-1}=q_i\lambda$ for $i\in I$;
  • $w'_i:=w_i\circ\varphi_i$ and $p'_i:=\frac p{q_i}\circ\varphi_i$ for $i\in I$;
  • $\zeta$ denote the counting measure on $(I,2^I)$;
  • $(\tilde E,\tilde{\mathcal E},\tilde\lambda):=(I\times E',2^I\otimes\mathcal E',\zeta\otimes\lambda')$
  • $\tilde p:=w'p'$ and $\tilde\mu:=\tilde p\tilde\lambda$;
  • $\tilde q:\tilde E^2\to[0,\infty)$ be symmetric and $\tilde{\mathcal E}^{\otimes2}$-measurable with $\int\tilde\lambda({\rm d}\tilde y)\tilde q(\tilde x,\tilde y)=1$ for all $\tilde x\in\tilde E$ and $\tilde Q(\tilde x,\;\cdot\;):=\tilde q(\tilde x,\;\cdot\;)\tilde\lambda$;
  • $\tilde\alpha(\tilde x,\tilde y):=1\wedge\frac{\tilde p(\tilde y)}{\tilde p(\tilde x)}$ for $\tilde x,\tilde y\in\tilde E$;
  • $\tilde k:=\tilde\alpha\tilde q$;
  • $\sigma_{ij}(x,y):=\tilde q((i,\varphi_i^{-1}(x)),(j,\varphi_j^{-1}(y)))$ for $(i,x),(j,y)\in I\times E$;
  • $\tilde\kappa(\tilde x,\tilde B):=\int_{\tilde B}\tilde Q(\tilde x,{\rm d}\tilde y)\tilde\alpha(\tilde x,\tilde y)+\left(1-\int\tilde Q(\tilde x,{\rm d}\tilde y)\tilde\alpha(\tilde x,\tilde y)\right)1_{\tilde B}(\tilde x)$ for $(\tilde x,\tilde B)\in\tilde E\times\tilde{\mathcal E}$.
$\endgroup$

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.