# I'm searching for a sampling kernel capturing the idea of a small local perturbation (like the normal distribution does)

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?

• @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? – 0xbadf00d Mar 25 '20 at 15:45
• @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?) – 0xbadf00d Mar 25 '20 at 15:54
• @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? – 0xbadf00d Mar 26 '20 at 10:57
• (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. – Xi'an Mar 26 '20 at 12:37
• @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. – 0xbadf00d Mar 26 '20 at 13:53