Let $W$ be a standard one dimensional Brownian motion, and let $X$ be the solution to the SDE

$$dX_t = \sigma(X_t) \, dW_t \;, \quad X_0 = 1 \;.$$

where $\sigma:\mathbb R \to \mathbb R$ is a Lipschitz continuous function.

For every $M > 0$, let $A_M$ denote the event

$$\{\underset{0 \leq t \leq 1}{\text{max}} W_t \geq M\} \;, $$

and let $\mathbb P^M$ be the probability measure given by

$$\mathbb P^M (E) = \frac{\mathbb P(E \cap A_M)}{\mathbb P(A_M)} \;, $$

for all events $E$.

We denote by $\mathbb E_{\mathbb P^M}$ the expectation under $\mathbb P^M$.

Consider the solution to the deterministic ODE

$$dY_t = \sigma(Y_t) \, dt \; , \quad Y_0 = 1.$$

Question: Is it true that

$$\lim_{M \to \infty} \, \mathbb E_{\mathbb P^M} \big [\underset{0 \leq t \leq 1}{\sup} |X_t - Y_{Mt}| \, \big] = 0?$$

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    $\begingroup$ @Nawaf Bou-Rabee Hm, heuristically what happens when $\sigma(y) = y$ is that the entire trajectory of $W_t$ (not $X_t$!) is close to the linear function $tW_1 \sim Mt$, more boldly one may write $dW_t \sim M dt$, so that $X$ converges to the solution of the deterministic $dX_t = M X_t \, dt$ which has solution $e^{Mt}$, so that $X_{1/M} \sim e$. In fact I think I have proven it rigorously in this particular case $\sigma(y) = y$, though I have not written it up yet… $\endgroup$
    – Nate River
    Jul 25, 2022 at 10:49
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    $\begingroup$ Thanks, that makes sense. $\endgroup$ Jul 25, 2022 at 10:57
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    $\begingroup$ Why would you expect this to be true? In the linear case, it seems to me that this would imply that $\mathbb{E}( \sup_{t > 1/2} |W_t - Mt - t/2|\,| \, A_M) \lesssim e^{-M/2}$ which is obviously not the case. $\endgroup$ Jul 27, 2022 at 21:51
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    $\begingroup$ @NateRiver Well, you will have $M^{-1} \log X_t \to t$, but the difference between $X$ and $e^{Mt}$ will diverge. (Even worse, their ratio won't even converge to $1$.) $\endgroup$ Jul 28, 2022 at 14:39
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    $\begingroup$ @NateRiver Plus, you've got stochastic fluctuations of the same order on top of that. $\endgroup$ Jul 28, 2022 at 22:10

1 Answer 1


The answer is no, as can be seen in the case $\sigma(u) = u$, so that $X_t = \exp(W_t - t/2)$. For the result to be true, Markov's inequality implies that the law of $W$, conditional on $A_M$, would need to give probability $1/2$ to the event $\sup_{t \le 1}|W_t - Mt + t/2| < K\exp(-M/2)$ for some fixed $K>0$. By large deviations, this event has probability smaller than $c\exp(-c\exp(M))$ for some $c>0$, while $A_M$ has probability larger than $c \exp(-cM^2)$ for some $c>0$, yielding a contradiction.

  • $\begingroup$ You're right, thank you! $\endgroup$
    – Nate River
    Aug 4, 2022 at 12:06

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