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Let $X$ be the solution to the one dimensional SDE

$dX_t = \mu(t, X_t)dt + \sigma(t, X_t) dW_t$, for $t \in [0, T]$.

with $X_0= x_0$ a.s. for some $x_0 \in \mathbb R$.

Here $W_t$ denotes a standard Brownian motion, and we assume $\mu$ and $\sigma$ are Lipschitz continuous and uniformly bounded.

For every $\varepsilon > 0$, denote by $\mathcal S_{\varepsilon}$ the event $\sup_{t \in [0, T]} |W_t| \leq \varepsilon$.

Question: Considering $X$ as a $C[0, T]$-valued random variable, is it true that the conditioned random variables $X| \mathcal S_\varepsilon$ converge in law to the deterministic solution $Y_t$ of

$dY_t = \mu(t, Y_t) dt$, with $Y_0 = x_0$ a.s.?

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2 Answers 2

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The answer is yes, provided that you write your equation in Stratonovich form, rather than Itô form (and assuming that $\mu$ and $\sigma$ are sufficiently smooth in their arguments). The reason is that in one dimension the solution to the Stratonovich equation is a continuous map of $W$ in the sup-norm topology, as observed by Doss in 1977.

This breaks in higher dimensions, but the answer to your question remains the same although I don't know if anyone wrote it up in precisely this way. (Various proofs of the Stroock-Varadhan support theorem use closely related variants of this statement. Note that it is again the Stratonovich formulation which is relevant.)

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  • $\begingroup$ Thank you! What is your gut feeling as to the whether the Ito form is true/false? I feel like because of the quadratic term in the Ito lemma (for Ito integrals) something might break due to roughness of the paths of $W_t$ - in the sense that controlling the sup norm of $W_t$ will not be sufficient. I don’t know how to push this idea further rigorously though… $\endgroup$
    – Nate River
    Sep 20, 2021 at 10:07
  • $\begingroup$ The two statements are mutually exclusive... $\endgroup$ Sep 20, 2021 at 10:08
  • $\begingroup$ Oh, I will try to work that out. Thanks! $\endgroup$
    – Nate River
    Sep 20, 2021 at 10:11
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As an approximation for a counterexample, consider $X_t=\sin(W_t+1)$. It has the stochastic differential $$dX_t=(-(1/2)\sin(W_t+1))dt+\cos(W_t+1)dW_t$$ with initial condition $X_0=\sin(1)$. Then $X|{\mathcal{S}_\varepsilon}$ converges to a constant function, which is not the solution of the diffusion-less equation $dX_t=(-(1/2)\sin(1))dt$.

This equation is not exactly in your form, but it is if you allow $2$-dimensional equations and consider $W$ as the first coordinate.

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  • $\begingroup$ That’s a really nice idea. I’m also convinced its no in the one dimensional case but have yet to find a specific counterexample. $\endgroup$
    – Nate River
    Sep 20, 2021 at 8:27
  • $\begingroup$ Hmm, though in your example $W$ and $X$ end up being very “correlated” whereas in the original problem $X$ is only linked to $W$ via the defining SDE. Maybe this changes things in a key way… $\endgroup$
    – Nate River
    Sep 20, 2021 at 8:29

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