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

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

differencebetween $X$ and $e^{Mt}$ will diverge. (Even worse, their ratio won't even converge to $1$.) $\endgroup$12more comments