# “Continuity” of stochastic integral wrt Brownian motion

I'd like to prove a nice property of a stochastic integral with respect to Brownian motion.

Let $(H_t)_{t\geq0}$ be a progressive and bounded process that is continuous at $0$ and $B$ a standard Brownian motion. Then

$\frac{1}{B_{\varepsilon}}\int_{0}^{\varepsilon}H_s\mathbb{d}B_s\rightarrow H_0$ as $\varepsilon\rightarrow0$ in probability.

Anyone got some hints ? I'm really puzzled and I don't know where to start.

EDIT_2.0. Applying Ito-Isometry might be a bit tricky. This Brownian Motion in the denominator kinda troubles me :/

• @ Mr. Barrington : would please indicate a reference telling from where this question comes from ? Best regards – TheBridge Nov 19 '13 at 13:51
• write $H_s = (H_0) + (H_s - H_0) = (I) + (II)$. The contribution from I gives you what you want. Use the Ito isometry to show that the contribution from II is very small, by showing that it has a variance much smaller that $\epsilon$ – mike Nov 19 '13 at 14:05
• TheBridge: this a problem in (my first) course of stochastic calculus. @mike: thanks for your hints, but I cannot figure out how to efficiently estimate the variance. I'll add some calculations to show you where I'm stuck. – Mr. Barrrington Nov 19 '13 at 16:31
• @mike and MR. Barrington : The fact that the integrand $\frac{H_s-H_0}{B_{\varepsilon}}$ is not an adapted process ($B_{\varepsilon}$ is clearly looking into the future in this integral), is a problem to apply Itô isometry, isn't it ? – TheBridge Nov 19 '13 at 17:20
• @ Mr. Barrington : Hi, another approach that might work (not sure about it I haven't tried to get to end of the argumentation) would be to prove first that for simple processes, you have $L^2$ convergence (hence convergece in probability) and then use the density in $L^2$ of those processes to try to get to the conclusion. Best regards. – TheBridge Nov 19 '13 at 18:37

Hint Let $\varepsilon>0$, $\delta>0$. We have

\begin{align*} \mathbb{P} &\left( \left| \frac{1}{B_{\varepsilon}} \cdot \int_0^{\varepsilon} H_s \, dB_s - H_0 \right|>\delta \right) \\ &= \mathbb{P} \left( \left| \frac{1}{B_{\varepsilon}} \cdot \int_0^{\varepsilon} (H_s-H_0) \, dB_s \right|>\delta \right) \\ &\leq \mathbb{P} \left( \left| \frac{1}{B_{\varepsilon}} \cdot \int_0^{\varepsilon} (H_s-H_0) \, dB_s \right|>\delta, \left| \frac{\sqrt{\varepsilon}}{B_{\varepsilon}} \right| \leq K \right)+ \mathbb{P} \left( \left| \frac{\sqrt{\varepsilon}}{B_{\varepsilon}} \right| > K \right) \\ &\leq \mathbb{P} \left( \left| \frac{1}{\sqrt{\varepsilon}} \cdot \int_0^{\varepsilon} (H_s-H_0) \, dB_s \right|>\frac{\delta}{K} \right)+ \mathbb{P} \left( \frac{|B_{\varepsilon}|}{\sqrt{\varepsilon}} < \frac{1}{K} \right)\\ &=: I_1+I_2 \end{align*}

for any $K>0$. Since $\frac{B_{\varepsilon}}{\sqrt{\varepsilon}} \sim N(0,1)$, we can choose $K>0$ (independent of $\varepsilon$) such that

$$I_2 \leq \frac{\varepsilon}{4}$$

For the first term $I_1$ apply Markov's inequality and Itô's isometry to show that it converges to zero as $\varepsilon \to 0$, using the continuity of $H$ at $0$.

Remark A detailed proof can be found in Dean Isaacson, Stochastic Integrals and Derivatives (1969).

• +1: thank you saz for this clever hint/solution! I got it now :) And also thanks to TheBridge and mike for your tips! – Mr. Barrrington Nov 20 '13 at 9:06
• @Mr.Barrrington You are welcome. :) – saz Nov 20 '13 at 9:08