Let $d\in\mathbb N$. I'm searching for a Markov kernel $\kappa$ on $\left([0,1)^d,{\mathcal B([0,1))}^{\otimes d}\right)$ suitable for the following application:

Given $x\in[0,1)^d$, I want to sample $y\in[0,1)^d$ with probability $\beta\in[0,1]$ from the uniform distribution $\mathcal U_{[0,\:1)^d}$ on $[0,1)^d$ and with probability $1-\beta$ from $\kappa(x,\;\cdot\;)$.

The idea is that with probability $\beta$ a "large step" transformation is performed and with probability $1-\beta$ a "small step" transformation is performed. This corresponds to sampling from the Markov kernel $$Q(x,B):=\beta\mathcal U_{[0,\:1)^d}(B)+(1-\beta)\kappa(x,B)\;\;\;\text{for }(x,B)\in[0,1)^d\times{\mathcal B([0,1))}^{\otimes d}.$$

Currently, I'm considering the wrapped normal distribution kernel $\mathcal W_{\sigma^2}$ with mean $0$ and variance $\sigma^2$, i.e. $$\mathcal W(x,\;\cdot\;):=\mathcal N(x,\;\cdot\;)\circ\iota^{-1}\;\;\;\text{for }x\in[0,1),$$ where $$\mathcal N(x,B):=\int_B\varphi_{\sigma^2}(y-x)\:{\rm d}y\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R),$$ $\varphi_{\sigma^2}$ denotes the density of the normal distribution with mean $0$ and variance $\sigma^2$ and $$\iota:\mathbb R\to[0,\infty)\;,\;\;\;x\mapsto x-\lfloor x\rfloor.$$ A typical choice is $\beta=0.3$ and $\sigma=0.01$. Now I choose $$\kappa(x,B):=\left(\bigotimes_{i=1}^d\mathcal W_{\sigma^2}(x_i,\;\cdot\;)\right)(B)\;\;\;\text{for }(x,B)\in[0,1)^d\times{\mathcal B([0,1))}^{\otimes d}.$$

This choice for $\kappa$ captures the idea of a "small step" (or a "local perturbation") of the current sample $x$ pretty well.

Assume that we have chosen a $\kappa$ such that $Q$ has a density $u$ with respect to the trace of the $d$-dimensional Lebesgue measure $\lambda^{\otimes d}$ on $[0,1)^d$. In my application, I have a finite set $I$ and a consider a varying number of dimensions $d_i$ and $\kappa_i$'s with density $u_i$, $i\in I$. Moreover, I have bounded measurable functions $f_i:[0,1)^{d_i}\to[0,\infty)$ and need to find the index $i\in I$ minimizing $$\int\lambda^{\otimes d_i}({\rm d}y)\frac{f_i(y)}{u_i\left(x^{(i)},y\right)}\tag1$$ for given fixed $x^{(i)}\in[0,1)^{d_i}$. (cf. my other question: Compute which of a finite number of integrals is minimal (not interested in the actual value of the integral)).

I need to solve this problem over and over again (inside a Metropolis-Hastings update) for different $x^{(i)}$'s. With the given choice of $\kappa$ this problem seems to be hard to solve and clearly depends on the $x^{(i)}$. However, if the $u_i$ would be such that the integral $\tag1$ does not depend on $x^{(i)}$, I could estimate $(1)$ and find the minimum in a precomputation step.

So, I'm searching for a different choice of $\kappa$ for which $(1)$ does not depend on $x^{(i)}$. $\kappa$ should still capture the idea of a "local, small step" and allow me to control the "size" of the step (as $\sigma$ does in the wrapped normal distribution).

Any suggestions?

  • $\begingroup$ @Xi'an I will take a look at it, but I guess I would need to consider a "wrapped" version of it, right? Note that the wrapped normal distribution actually does a great job in my application; the only reason why I'm searching for a replacement is the described optimization problem. Do you think it will be easier to solve with (a suitable version of) the beta distribution? $\endgroup$ – 0xbadf00d Mar 25 '20 at 15:45
  • $\begingroup$ @Xi'an To be honest, I'm not familiar with the Beta distribution and need to take a look to it first, but let me ask you again: Do you think it simplifies the optimization problem significantly? (BTW, your comment below my question on MSE seems to indicate that $(1)$ does not depend on $x^{(i)}$ if I choose $\kappa$ as the product of the wrapped normal distribution or am I missing something?) $\endgroup$ – 0xbadf00d Mar 25 '20 at 15:54
  • $\begingroup$ @Xi'an Okay, I've taken a look. So, your suggestion is to replace $\mathcal W_{\sigma^2}$ by $$B_\alpha(x,\;\cdot\;):=\mathcal{Be}(\alpha x,(1-\alpha)x\;\;\;\text{for }x\in(0,1)\tag2$$ for some $\alpha\in(0,1)$, right? (a) Wouldn't it make more sense to replace the right-hand side of $(2)$ by $\mathcal{Be}(\alpha x,\alpha(1-x)$ so that the mean is $x$ (and not $\alpha$)? (b) How does this solve my problem? Doesn't the corresponding $u_i$ still depend on the first argument? $\endgroup$ – 0xbadf00d Mar 26 '20 at 10:57
  • $\begingroup$ (a) Yes this was a typo. And (b) yes, the distribution depends on $x$. To be frank, I have serious trouble understanding the point of this question or of the dozen previous questions all related to the same topic. $\endgroup$ – Xi'an Mar 26 '20 at 12:37
  • $\begingroup$ @Xi'an (b) The point is that I want to solve this stats.stackexchange.com/q/454902/222528 problem when $r((i,x),y)=\int\lambda^{\otimes d_i}({\rm d}y)\frac{f_i(y)}{u_i\left(\varphi_i^{-1}(x),y\right)}$. If $u_i$ would not depend on the first argument, I could stay with my current solution (Monte Carlo estimation and then building the minimum), since I could solve this problem in a precomputation step. Otherwise I would need to solve it over and over again and that would be way too slow. $\endgroup$ – 0xbadf00d Mar 26 '20 at 13:53

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