# Short time limits for SDE

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

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

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 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, Itô 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, Cauchy-Schwarz, Itô 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 $$(0, \epsilon)$$.

$$\Box$$

• Wow, this is amazing. I’ll read through this properly today. Aug 18, 2022 at 3:18
• Sorry for the trouble, but I have several questions about the estimate for $II$. Firstly, how does one derive $\lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| ] = \lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| \mathbf{1}_{ \{ \tau > \epsilon - \epsilon^{3/2} \} } ]$ in the first step? I understand that we use the property $\lim_{\epsilon \searrow 0} P_{\epsilon}[\tau > \epsilon-\epsilon^{3/2}] = 1$, but I am not sure how to conclude that the expectation matches in the limit. Aug 18, 2022 at 5:37
• My second question concerns the Brownian bridge - I understand that $W_t - \frac{t}{\epsilon} W_\epsilon$ is a (scaled) Brownian bridge on $[0, \epsilon]$, independent of $W_\epsilon$, and hence conditional on any event involving only $W_\epsilon$, it is still a Brownian bridge. Thus after scaling by $\epsilon$ we get that $\epsilon W_s - s$ is a standard Brownian bridge on $[0, \epsilon]$ conditional on $W_\epsilon = 1$. However, we condition on $\tau < \epsilon$ which involves the path of $W$ before time $\epsilon$, further it is not certain that $W_\epsilon = 1$. Aug 18, 2022 at 5:50
• I would be really grateful if you could help me understand these steps. Aug 18, 2022 at 5:51
• Thank you for the added details! It makes sense to me now. Very nice solution. Aug 18, 2022 at 16:47

No, for example, take $$dX_t = 2 dB_t$$. By the reflection principle the conditioned process is symmetric around 1, but $$Y_1 = 2$$.

• Can you elaborate on the reflection principle part? Aug 6, 2022 at 7:23
• I assume you want to reflect after $W$ hits $1$, but wouldn’t that make $X$ symmetric around $2$? I don’t see any contradiction… Aug 6, 2022 at 7:27
• You are conditioning on whether the process has hit the level 1 before time $\epsilon$, but according to the strong markov property, after it hits 1, it is equally likely to go up or down, making the distribution conditioned on having hit 1 symmetric around 1. This application of the SMP is the reflection principal. BTW, I would expect conditioned as you have, virtually any process would satisfy $X_{\epsilon} \approx 1$ because I would expect that generally speaking is crosses preponderantly near time $\epsilon$.
– mike
Aug 6, 2022 at 15:13
• But we condition on $W$ hitting $1$, not $X$. $X$ would hit $2$ at the time $W$ hits $1$. Aug 6, 2022 at 16:53
• my mistake, i did not notice.
– mike
Aug 7, 2022 at 5:27