Algorithm 2.

Cross-resolution Move of Multiresolution Sampler

Let (θold, ${\mathit{Y}}_{\mathit{old}}^{\{k+1\}}$) be the current set of parameters and missing data. Draw (θtrial, ${\mathit{Y}}_{\mathit{trial}}^{\{k\}}$) from the empirical distribution of fk(θ, Y{k}|Y). Let ${\pi }_k^{\mathit{trial}}=f_k({\mathit{\theta }}_{\mathit{trial}},{\mathit{Y}}_{\mathit{trial}}^{\{k\}}\mid \mathit{Y})$. Draw ${\mathit{Z}}_{\mathit{trial}}^{\{k+1\}}$ from a distribution $T_{k+1}({\mathit{Z}}^{\{k+1\}}\mid {\mathit{\theta }}_{\mathit{trial}},{\mathit{Y}}_{\mathit{trial}}^{\{k\}},\mathit{Y})$. Let ${\tau }_{k+1}^{\mathit{trial}}=T_{k+1}({\mathit{Z}}_{\mathit{trial}}^{\{k+1\}}\mid {\mathit{\theta }}_{\mathit{trial}},{\mathit{Y}}_{\mathit{trial}}^{\{k\}},\mathit{Y})$. Recall that ${\mathit{Y}}_{\mathit{trial}}^{\{k+1\}}={\mathit{Y}}_{\mathit{trial}}^{\{k\}}\cup {\mathit{Z}}_{\mathit{trial}}^{\{k+1\}}$. Let ${\pi }_{k+1}^{\mathit{trial}}=f_{k+1}({\mathit{\theta }}_{\mathit{trial}},{\mathit{Y}}_{\mathit{trial}}^{\{k+1\}}\mid \mathit{Y})$. Similarly, let ${\pi }_k^{\mathit{old}}=f_k\left({\mathit{\theta }}_{\mathit{old}},{\mathit{Y}}_{\mathit{old}}^{\{k\}}\mid \mathit{Y}\right),{\tau }_{k+1}^{\mathit{old}}=T_{k+1}\left({\mathit{Z}}_{\mathit{old}}^{\{k+1\}}\mid {\mathit{\theta }}_{\mathit{old}},{\mathit{Y}}_{\mathit{old}}^{\{k\}},\mathit{Y}\right)$ and ${\pi }_{k+1}^{\mathit{old}}=f_{k+1}\left({\mathit{\theta }}_{\mathit{old}},{\mathit{Y}}_{\mathit{old}}^{\{k+1\}}\mid \mathit{Y}\right)$. Accept $\left({\mathit{\theta }}_{\mathit{trial}},{\mathit{Y}}_{\mathit{trial}}^{\{k+1\}}\right)$) as the next sample from fk+1(θ, Y{k+1} | Y) with probability $$r=min\left\{1,\frac{{\pi }_{k+1}^{\mathit{trial}}/({\pi }_k^{\mathit{trial}}{\tau }_{k+1}^{\mathit{trial}})}{{\pi }_{k+1}^{\mathit{old}}/({\pi }_k^{\mathit{old}}{\tau }_{k+1}^{\mathit{old}})}\right\}.$$ Otherwise, with probability 1 − r, keep $\left({\mathit{\theta }}_{\mathit{old}},{\mathit{Y}}_{\mathit{old}}^{\{k+1\}}\right)$ as the next sample.