# Large noise limit for SDE with general volatility coefficients

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

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

• @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… Jul 25, 2022 at 10:49
• Thanks, that makes sense. Jul 25, 2022 at 10:57
• 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. Jul 27, 2022 at 21:51
• @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 $1$.) Jul 28, 2022 at 14:39
• @NateRiver Plus, you've got stochastic fluctuations of the same order on top of that. Jul 28, 2022 at 22:10

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.