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Consider the following SDE system $$dx_t = b(y_t)dt + dw^1_t, \quad dy_t = dw^2_t.$$ Here the drift $b(\cdot)$ is a smooth function that may decay slowly. For example, $|b(x)| \le C/|x|^\sigma$ for some $\sigma > 0$ as $|x| \to \infty$. $w^1_t$ and $w^2_t$ are independent standard one-dimensional Brownian motion. We can solve $x_t$ by $$x_t = x_0 + w^1_t + \int_0^tb(y_0+w^2_s)ds.$$ Let $p(t, \mathbf{x}, \mathbf{y})$ denote the transition density of this process, that is, the probability density of this process starting at the point $\mathbf{x}$ at time $0$ and reaching the point $\mathbf{y}$ at time $t$.

Question: Can we get the decay estimate of the transition density $p(t, \mathbf{x}, \mathbf{y})$? For example, can we get Gaussian bound for the density similar to the situation when $b$ is identically zero? How does the decay rate change if we consider the drift function $b(\cdot)$ with different decay rates?

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  • $\begingroup$ What exactly does “decay rate” of $b$ mean? Does it possibly mean that it is strongly convex with “decay rate” being the strong convexity constant? What is the argument showing that the invariant measure of the SDE exists? It seems the $x$-component of the SDE is not ergodic; consider e.g. linear $b$. $\endgroup$ – Nawaf Bou-Rabee May 4 at 5:17
  • $\begingroup$ @NawafBou-Rabee Many thanks for pointing this out! I also realized that the question is not formulated in the right way. In general, one cannot expect the existence of an invariant measure. Instead, I would like to know the behavior of the transition density. The decay rate of $b$ means how fast $b(x)$ goes to zero when $|x| \to \infty$. Sorry for the ambiguity. I have edited the question to accommodate these changes. $\endgroup$ – Jacob Lu May 4 at 6:05
  • $\begingroup$ Since there is no statistical steady-state, what type of decay estimate for the transition density can we expect? Consider, e.g., the free case where $b=0$. $\endgroup$ – Nawaf Bou-Rabee May 4 at 6:15
  • $\begingroup$ I think if $b=0$, the transition density is computable and should be the two dimensional heat kernel. For some $b$, I think we may also expect this kind of result. But this may change if $b$ has different behavior at infinity. $\endgroup$ – Jacob Lu May 4 at 6:34
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    $\begingroup$ I thought Gaussian bound is also a kind of decay estimate since it decays very fast. I edited the problem to include this information. $\endgroup$ – Jacob Lu May 4 at 16:43

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