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Fix $\epsilon>0$ and let $(\Omega,F,F_t\mathbb{P})$ be a stochastic base. Is there a (Markov) diffusion process $X_t$ satisfying an SDE of the form: $$ d X_t = \mu(t,X_t)dt + \Sigma(t,X_t)dW_t, X_0^x $$ such that the (random) function $f_X:x\to X_1^x$ satisfies $$ \mathbb{P}\left( \int_{x \in \mathbb{R}^n} |f_X(x)| dx < \epsilon \right)=1? $$ If not, can we estimate the probability that this holds?

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$\newcommand\ep\epsilon$ $\newcommand\R{\mathbb R}$ $\newcommand\Si{\Sigma}$ Let $$X^x_t:=xe^{-ct|x|}$$ for some real $c>0$ and allreal $t\ge0$ and $x\in\R^n$. Then $X^x_0=x$ for all $x$ and your SDE holds with $\mu(t,x)=-c|x|xe^{-ct|x|}$ and $\Si(t,x)=0$. Moreover, $$\int_{\R^n}|X^x_1|\,dx=\int_{\R^n}|x|e^{-c|x|}\,dx<\ep,$$ as desired, if $c=c_\ep$ is large enough.


If you insist on $\Si(t,x)\ne0$, you can clearly make $P(\int_{\R^n}|X^x_1|\,dx<\ep)$ arbitrarily close to $1$, by approximation.

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  • $\begingroup$ But I'm a bit confused. So in general, for a typical function we would take $X_t^x = x(e^{-ct|x|} +f(x))$? $\endgroup$
    – BLBA
    Jun 5 '20 at 15:40
  • $\begingroup$ The current version of your question is completely different from the original one. I suggest you restore the original question and post any other questions you may have separately. $\endgroup$ Jun 5 '20 at 15:40
  • $\begingroup$ Fair enough, I have rolled-back the question and posted the modified one here: mathoverflow.net/questions/362274/… $\endgroup$
    – BLBA
    Jun 5 '20 at 15:46

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