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?
1 Answer
$\begingroup$
$\endgroup$
3
$\newcommand\ep\epsilon$ $\newcommand\R{\mathbb R}$ $\newcommand\Si{\Sigma}$ Let $$X^x_t:=xe^{ctx}$$ 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 $\mu(t,x)=cxxe^{ctx}$ and $\Si(t,x)=0$. Moreover, $$\int_{\R^n}X^x_1\,dx=\int_{\R^n}xe^{cx}\,dx<\ep,$$ 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.

$\begingroup$ But I'm a bit confused. So in general, for a typical function we would take $X_t^x = x(e^{ctx} +f(x))$? $\endgroup$– ABIMCommented Jun 5, 2020 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. $\endgroup$ Commented Jun 5, 2020 at 15:40

$\begingroup$ Fair enough, I have rolledback the question and posted the modified one here: mathoverflow.net/questions/362274/… $\endgroup$– ABIMCommented Jun 5, 2020 at 15:46