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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

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

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2 Answers 2

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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 $(0, \epsilon)$.

$\Box$

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  • $\begingroup$ Wow, this is amazing. I’ll read through this properly today. $\endgroup$
    – Nate River
    Commented Aug 18, 2022 at 3:18
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    $\begingroup$ 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. $\endgroup$
    – Nate River
    Commented Aug 18, 2022 at 5:37
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    $\begingroup$ 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 $W_\epsilon = 1$. $\endgroup$
    – Nate River
    Commented Aug 18, 2022 at 5:50
  • $\begingroup$ I would be really grateful if you could help me understand these steps. $\endgroup$
    – Nate River
    Commented Aug 18, 2022 at 5:51
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    $\begingroup$ Thank you for the added details! It makes sense to me now. Very nice solution. $\endgroup$
    – Nate River
    Commented Aug 18, 2022 at 16:47
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No, for example, take $dX_t = 2 dB_t$. By the reflection principle the conditioned process is symmetric around 1, but $Y_1 = 2$.

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  • $\begingroup$ Can you elaborate on the reflection principle part? $\endgroup$
    – Nate River
    Commented Aug 6, 2022 at 7:23
  • $\begingroup$ 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… $\endgroup$
    – Nate River
    Commented Aug 6, 2022 at 7:27
  • $\begingroup$ 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 $\epsilon$. $\endgroup$
    – mike
    Commented Aug 6, 2022 at 15:13
  • $\begingroup$ But we condition on $W$ hitting $1$, not $X$. $X$ would hit $2$ at the time $W$ hits $1$. $\endgroup$
    – Nate River
    Commented Aug 6, 2022 at 16:53
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    $\begingroup$ my mistake, i did not notice. $\endgroup$
    – mike
    Commented Aug 7, 2022 at 5:27

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