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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.

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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 :/

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**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).

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