Large noise limit for SDE with general volatility coefficients

Question

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

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

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.