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


$$\lim\_{M \to \infty} \, \mathbb E\_{\mathbb P^M} \big [\underset{0 \leq t \leq 1}{\sup} |X\_t - Y\_{Mt}| \, \big] = 0?$$



Large noise limit for SDE with general volatility coefficients
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.


