# Upper bound $I_R := \int_{B_R^c} |x| (P_t \ell_\nu) (x) \, \mathrm d x$ in terms of $R, \nu, t$?

Let $$(p_t)_{t >0}$$ be the Gaussian heat kernel on $$\mathbb R^d$$ and $$(P_t)_{t >0}$$ its induced semi-group, i.e., \begin{align} p_t (x) &:= (4\pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{4t}}, \quad t>0, x \in \mathbb R^d, \\ P_t f (x) &:= \int_{\mathbb R^d} p_t ( x- y) f(y) \, \mathrm d y. \end{align}

Let $$\nu$$ be a Borel probability measure on $$\mathbb R^d$$ with finite first moment. We denote by $$\ell_\nu$$ the probability density function of $$\nu$$. For $$R>0$$, let $$B_R$$ be the open ball centerted at $$0$$ with radius $$R$$. Let $$B_R^c := \mathbb R^d \setminus B_R$$ and $$I_R := \int_{B_R^c} |x| (P_t \ell_\nu) (x) \, \mathrm d x, \quad R >0.$$

Is there an upper bound of $$I_R$$ in terms of $$R, \nu, t$$?

Thank you so much for your elaboration!

Let $$Z$$ be a standard normal random vector in $$\mathbb R^d$$. Let $$Y$$ be a random vector in $$\mathbb R^d$$, independent from $$Z$$, with distribution $$\nu$$. Then, for each $$s\in(0,1)$$, $$I_R=E|\sqrt{2t}\,Z+Y|\,1(|\sqrt{2t}\,Z+Y|\ge R) \\ \le E(\sqrt{2t}\,|Z|+|Y|) \,[1(\sqrt{2t}\,|Z|\ge sR)+1(|Y|\ge(1-s)R)] \\ =\sqrt{2t}\,E|Z|1(|Z|\ge sR/\sqrt{2t}) \\ +\sqrt{2t}\,E|Z|\,P(|Y|\ge(1-s)R) \\ +E|Y|\,P(|Z|\ge sR/\sqrt{2t}) \\ +E|Y|\,1(|Y|\ge(1-s)R).$$
In turn, you can upper-bound the latter four summands using the following applications of Markov's inequality: $$E|Z|1(|Z|\ge sR/\sqrt{2t})\le \frac{E|Z|^2}{sR/\sqrt{2t}}=\frac n{sR/\sqrt{2t}}, \\ E|Z|\,P(|Y|\ge(1-s)R)\le\sqrt{E|Z|^2}\,\frac{E|Y|}{(1-s)R} =\frac{\sqrt n\,E|Y|}{(1-s)R} \\ P(|Z|\ge sR/\sqrt{2t})\le\frac{E|Z|^2}{(sR/\sqrt{2t})^2} =\frac{n}{(sR/\sqrt{2t})^2} \\ E|Y|\,1(|Y|\ge(1-s)R)\le\frac{E|Y|^2}{(1-s)R}.$$