• $\kappa_d$ be a Markov kernel on $(\mathbb R^d,\mathcal B(\mathbb R^d))$ for $d\in\mathbb N$
  • $f_d:\mathbb R^d\times\mathbb R^d\to\mathbb R$ be Borel measurable for $d\in\mathbb N$
  • $B_d\in\mathcal B(\mathbb R^d)$ for $d\in\mathbb N$ with $$\sup_{x\in B_d}\int\kappa_d(x,{\rm d}y)|f(x,y)|\xrightarrow{d\to\infty}0\tag1$$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $X^{(d)}:\Omega\to\mathbb R^d$ be $(\mathcal A,\mathcal B(\mathbb R^d))$-measurable with $$\operatorname P\left[X^{(d)}\in B_d\right]\xrightarrow{d\to\infty}1\tag2$$ for $d\in\mathbb N$
  • $Y^{(d)}:\Omega\to\mathbb R^d$ be $(\mathcal A,\mathcal B(\mathbb R^d))$-measurable with$^1$ $$Y^{(d)}_\ast\operatorname P=\left(X^{(d)}_\ast\operatorname P\right)\kappa_d\tag3$$ for $d\in\mathbb N$

Are we able to conclude $$\left|f\left(X^{(d)},Y^{(d)}\right)\right|\xrightarrow{d\to\infty}0\tag4$$ in probability (or even almost surely)?

We may note that $$\operatorname E\left[\left|f\left(X^{(d)},Y^{(d)}\right)\right|\right]=\int\operatorname P\left[X^{(d)}\in{\rm d}x\right]\int\kappa_d(x,{\rm d}y)|f(x,y)|\tag5$$ for all $d\in\mathbb N$. We may split the outer integral and clearly $$\int_{B_d}\operatorname P\left[X^{(d)}\in{\rm d}x\right]\int\kappa_d(x,{\rm d}y)|f(x,y)|\le\operatorname P\left[X^{(d)}\in B_d\right]\sup_{x\in B_d}\int\kappa_d(x,{\rm d}y)|f(x,y)|\xrightarrow{d\to\infty}0\tag6.$$ So, a possible way would to show $$\int_{\mathbb R^d\setminus B_d}\operatorname P\left[X^{(d)}\in{\rm d}x\right]\int\kappa_d(x,{\rm d}y)|f(x,y)|\xrightarrow{d\to\infty}0\tag7.$$ If the inner integral wouldn't depend on $d$, this would be a simple application of the dominated convergence theorem. However, since it depends on $d$, this might be problematic.

$^1$ $\left(X^{(d)}_\ast\operatorname P\right)\kappa_d$ denotes the composition of the distribution $X^{(d)}_\ast\operatorname P$ of $X^{(d)}$ under $\operatorname P$ and $\kappa_d$.


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