Let $g\in C^1[0,1]$ be a given smooth curve. Consider a random path $X\in C[0,1]$ that solves the SDE $$dX_t = \mu(t) dt+\sigma dB_t,$$ i.e. $$X_t = X_0+\int_0^t \mu(s) ds +\sigma B_t.$$ We will be especially interested in the case when $\mu(t)=g'(t)$ and $X_0=g(0)$, or somewhat informally, when $\mu(t) \approx g'(t)$.

Question Are there any known estimates of the average supremum distance $$d:=\mathbb{E}(\|X-g\|_\infty)?$$ Here, $\|f\|_\infty = \max_{t\in [0,1]} |f(t)|$.

Some Context: The motivation vaguely comes from learning the dynamics of an SDE to fit an observed path $g(t)$. Work by Kidger et al on neural SDEs mention the issue of zero-diffusions being optimal. Specifically "Hodgkinson et al. (2020) introduce neural SDEs as a limit of random ODEs [...] however, they remark that in this setting the optimal diffusion is zero. This is a recurring problem: Innes et al. (2019) also train neural SDEs for which the optimal diffusion is zero". I am interested in bounds/estimates from a general perspective.

Some thoughts: Trivially $d=0$ when $\mu=g'$, $X_0=g(0)$ and $\sigma=0$. So over the class of paths $X$ with $\mu=g'$ starting from $g(0)$ and parameterized by $\sigma$, the minimum supremum distance is obtained when $\sigma=0$, i.e. a zero-diffusion trivially is the minimizer. How large or small is this distance when we must have a non-zero diffusion?

When $\sigma \neq0$, we have that the difference between $X_t$ and $g(t)$ is less than $x$, i.e. $$|X_t-g(t)|\leq x$$ if and only if $$\frac{g(t)-x-X_0-M(t)}{\sigma}\leq B_t \leq \frac{g(t)+x-X_0-M(t)}{\sigma},$$ where $$M(t):= \int_0^t \mu(s)ds.$$ Thus, it follows that the probability of the distance between $X$ and $g$ at $t$ being less than $x$ is $$\mathbb{P}(|X_t-g(t)|\leq x)= \Phi(h^+(t,x))-\Phi(h^-(t,x)),$$ where $$h^{\pm}(t,x)=\frac{g(t)\pm x-X_0-M(t)}{\sigma \sqrt{t}}.$$ Since the distance is non-negative, it follows that the expected distance between $X$ and $g$ at $t$ is given by \begin{align*} \mathbb{E}(|X_t-g(t)|) &= \int_0^\infty \mathbb{P}(|X_t-g(t)|\geq x)dx\\ &=\int_0^\infty 1-\Phi(h^+(t,x))+\Phi(h^-(t,x))dx. \end{align*} When we have the case $\mu=g'$ and $X_0=g(0)$, we obtain more simply

\begin{align*} \mathbb{E}(|X_t-g(t)|) &=\int_0^\infty 1-\Phi(x/\sigma \sqrt{t})+\Phi(-x/\sigma \sqrt{t})dx\\ & = 2\int_0^\infty 1-\Phi(x/\sigma \sqrt{t})dx. \end{align*}

Now if $X_t$ was differentiable (it is not!) we could perhaps proceed by differentiating under the integral and studying critical points. Proceeding formally, this would give us the expression $$\int_0^\infty \phi(x/\sigma \sqrt{t}) x/(\sigma t^{3/2}) dx,$$ if I am not mistaken and I do not see anything useful coming from this, even if we were allowed to proceed as such.



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