Let $W$ be a standard one dimensional Brownian motion, and let $X$ be the solution to the SDE


$$dX\_t = \sigma(X\_t) \, dW\_t \;, \quad X\_0 = x\_0\;.$$


where $\sigma:\mathbb R \to \mathbb R$ is a Lipschitz continuous function, and $x\_0 \in \mathbb R$ is a fixed constant.


For every $\varepsilon > 0$, let $A\_\varepsilon$ denote the event


$$\{\underset{0 \leq t \leq \varepsilon}{\text{max}} W\_t \geq 1\} \;, $$


and let $\mathbb P^\varepsilon$ be the probability measure given by


$$\mathbb P^\varepsilon (E) = \frac{\mathbb P(E \cap A\_\varepsilon)}{\mathbb P(A\_\varepsilon)} \;, $$


for all events $E$.


We denote by $\mathbb E\_{\mathbb P^\varepsilon}$ the expectation under $\mathbb P^\varepsilon$.


Consider the solution $Y$ to the deterministic ODE


$$dY\_t = \sigma(Y\_t) \, dt \; , \quad Y\_0 = x\_0.$$


**Question:** Is it true that


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



Short time limits for SDE
Let $\tau = \inf\{ t>0 : W\_t = 1 \}$. The conjecture is **true** and the essence of the proof outlined below appears to be the following peculiar property of the hitting time $\tau$: $$
\lim\_{\epsilon \searrow 0} P\_{\epsilon}[\tau > \epsilon-\epsilon^{3/2}] = 1 \;. $$
(This can be computed directly using the fact that distribution of $\tau$ is [inverse gamma](https://en.wikipedia.org/wiki/Inverse-gamma_distribution) with parameters $1/2$ and $1/2$.) In order to leverage this property, one must carefully split up $e\_t:= X\_t - Y\_{t/\epsilon}$ as outlined below.



> 
> **Theorem**. It holds that: $$\lim\_{\epsilon \searrow 0} E[ \sup\_{0 \le t \le \epsilon} |e\_t|^2 ] = 0 \;.$$
> 
> 
> 


**Proof**.
The proof shows that for any $T \in [0, \epsilon]$ $$
E\_{\epsilon} [ \sup\_{0 \le t \le T} |e\_t|^2 ] \le C\_1(\epsilon) + C\_2 (\epsilon) \int\_0^T E\_{\epsilon} [ \sup\_{0 \le r \le s} |e\_r|^2 ] ds 
$$ where $C\_1(\epsilon)$ and $C\_2 (\epsilon) $ are non-negative and $\lim\_{\epsilon \searrow 0} C\_1(\epsilon) =0$ and $\lim\_{\epsilon \searrow 0} C\_2(\epsilon) \epsilon = O(1)$. By Grönwall's inequality, $$
E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} |e\_t|^2 ] \le C\_1(\epsilon) \exp(C\_2 (\epsilon) \epsilon) \;,
$$ and then passing to the limit gives the required result. The remaining details follow.


By Itô's formula, \begin{align\*}
& |e\_t|^2 = \mathrm{I} + \mathrm{II} + \mathrm{III} \quad \text{where} \\
& \mathrm{I}:= \frac{2}{\epsilon} \int\_0^t e\_s (\sigma(X\_s) - \sigma(Y\_{s/\epsilon})) ds \;, \\
& \mathrm{II}:= \frac{2}{\epsilon} \int\_0^t e\_s \sigma(X\_s) (\epsilon dW\_s - ds) \;, \\
& \mathrm{III}:= \int\_0^t \sigma(X\_s)^2 ds \;. 
\end{align\*}




---


*Estimate for $\mathrm{I}$.*


This term exclusively contributes to $C\_2(\epsilon)$.
Since $\sigma$ is $L$-Lipschitz for some $L>0$ $$
I \le \frac{2 L}{\epsilon} \int\_0^t |e\_s|^2 ds 
$$ and thus $$
\sup\_{0 \le t \le \epsilon} I \le \frac{2 L}{\epsilon} \int\_0^{\epsilon} |e\_s|^2 ds \le \frac{2 L}{\epsilon} \int\_0^{\epsilon} \sup\_{0 \le r \le s} |e\_r|^2 ds \;.
$$
Thus, $C\_2(\epsilon) = 2 L / \epsilon$.




---


*Estimate for $\mathrm{II}$.*


This term contributes to $C\_1(\epsilon)$, and here is where we leverage the aforementioned peculiar property of $\tau$.  

\begin{align\*}
& \lim\_{\epsilon \searrow 0} E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \left| \mathrm{II} \right| ] = 
\lim\_{\epsilon \searrow 0} E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \left| \mathrm{II} \right| \mathbf{1}\_{ \{ \tau > \epsilon - \epsilon^{3/2} \} } ] \\
& \quad = 
\lim\_{\epsilon \searrow 0} E [ \sup\_{0 \le t \le \epsilon} \left| 2 \int\_0^t e\_s \sigma(X\_s) dW\_s \right| ] = 0
\end{align\*}
Here we took 3 steps that are explained in detail below.


*In the first step*, we used Cauchy-Schwarz to show that $$
\left( E\_{\epsilon} \sup\_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}\_{\{\tau < \epsilon - \epsilon^{3/2} \} } \right)^2 \le \underbrace{E[\sup\_{0 \le t \le \epsilon} |\mathrm{II}|^2 ]}\_{\to O(1)} \, \underbrace{P\_{\epsilon}[ \tau < \epsilon - \epsilon^{3/2}]}\_{\to 0} 
$$


*In the second step*, we used a natural splitting and the triangle inequality to write, \begin{align\*}
& E\_{\epsilon} \sup\_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}\_{\{\tau > \epsilon - \epsilon^{3/2} \} } \le \mathrm{II}\_a + \mathrm{II}\_b \quad \text{where} \\
& \mathrm{II}\_a := E\_{\epsilon} \sup\_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int\_0^{t \wedge \tau} e\_s \sigma(X\_s) (\epsilon dW\_s - ds) \right| \mathbf{1}\_{\{\tau > \epsilon - \epsilon^{3/2} \} }\\
& \mathrm{II}\_b := E\_{\epsilon} \sup\_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int\_{t \wedge \tau}^t e\_s \sigma(X\_s) (\epsilon dW\_s - ds) \right| \mathbf{1}\_{\{\tau > \epsilon - \epsilon^{3/2} \} }
\;.
\end{align\*}
To estimate these terms, there are two cases to consider:


* *Case 1: $t \le \tau$*. Then $\mathrm{II}\_b=0$ and $\mathrm{II}\_a$ can be written in terms of the piece of the Brownian bridge up to time $t$; and,
* *Case 2: $t > \tau$*. Then $t>\epsilon - \epsilon^{3/2}$ and hence $\mathrm{II}\_b=O(\epsilon^{1/2})$ and again $\mathrm{II}\_a$ can be written in terms of the piece of the Brownian bridge up to time $\tau$.


In other words, conditioned on the event $(\tau < \epsilon)$, the law of $\epsilon W\_s - s$ is equal to the law of a standard Brownian bridge.


*In the third and last step*, we used [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Further_inequalities), [Itô isometry](https://en.wikipedia.org/wiki/It%C3%B4_isometry), and (standard) a priori bounds on $X\_t$ and $Y\_{t/\epsilon}$ over $(0,\epsilon)$. Since the estimate of this term is almost identical to the estimate of $\mathrm{III}$ given below, the details are suppressed.




---


*Estimate for $\mathrm{III}$.*


This term also contributes to $C\_1(\epsilon)$. Noting that $\sigma$ is $L$-Lipschitz, \begin{align\*}
E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \mathrm{III} ] &= E\_{\epsilon}[ \int\_0^{\epsilon} [ \sigma(X\_s)^2 ds ] \\
&\le 2 \epsilon \sigma(0)^2 + 2 L^2 E\_{\epsilon}[ \int\_0^{\epsilon} |X\_s|^2 ds ] \\
&\le 2 \epsilon ( \sigma(0)^2 + L^2 |x\_0|^2 ) + 2 L^2 \epsilon \int\_0^{\epsilon} \sigma(X\_s)^2 ds \\ 
& \quad + 2 L^2 \epsilon \int\_0^{\epsilon} X\_s \sigma(X\_s) dW\_s
\end{align\*}
and as long as $2 L^2 \epsilon \le 1/2$, it follows that
$$
E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \mathrm{III} ] \le 4 \epsilon ( \sigma(0)^2 + L^2 |x\_0|^2 ) + 4 L^2 \epsilon \int\_0^{\epsilon} X\_s \sigma(X\_s) dW\_s
$$ The last term in this expression can be treated in a similar way as the last step in the estimate for $\mathrm{II}$, namely [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Further_inequalities), Cauchy-Schwarz, [Itô isometry](https://en.wikipedia.org/wiki/It%C3%B4_isometry), and (standard) a priori bounds on $X\_t$ over $(0,\epsilon)$. In particular, \begin{align\*}
\left( E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \left| \int\_0^t X\_s \sigma(X\_s) d W\_s \right| ] \right)^2 &\le
E \sup\_{0 \le t \le \epsilon} \left| \int\_0^t X\_s \sigma(X\_s) dW\_s \right|^2\\
&\le 4 E \left| \int\_0^{\epsilon} X\_s \sigma(X\_s) dW\_s \right|^2 \\
&\le 4 E \int\_0^{\epsilon} X\_s^2 \sigma(X\_s)^2 ds \\
&\le 4 \tilde{C}\_2 (1+ x\_0^4) e^{\tilde{C}\_1 \epsilon} \epsilon
\end{align\*}
where in turn we used Cauchy-Schwarz, Doob's martingale inequality with $p=2$, Itô's isometry, and then an a priori bound on the second/fourth moment of $X\_t$ over $(0, \epsilon)$.


$\Box$


