= Solution
Fix a deterministic $t>0$ and define $\widehat B_u=B_{t+u}-B_t$, $\widehat H_u=H_{t+u}$ and $\widehat{\mathcal F}_u=\mathcal F_{t+u}$. The increment process is a standard <Brownian motion> in the shifted filtration, and $\widehat H$ remains bounded, continuous and adapted. Thus part (a) applies even though the initial value $\widehat H_0=H_t$ is random. For $\varepsilon>0$,
$$
\frac{\int_t^{t+\varepsilon}H_s\,dB_s}{B_{t+\varepsilon}-B_t}-H_t
=\frac{\int_t^{t+\varepsilon}(H_s-H_t)\,dB_s}{B_{t+\varepsilon}-B_t}.
$$
Here the constant-in-time integrand $H_t$ is $\mathcal F_t$-measurable, so its <stochastic integral> equals $H_t(B_{t+\varepsilon}-B_t)$. The expectation of the absolute value of the right-hand side to the power $1/4$ tends to zero by part (a). For every $\eta>0$, <Markov inequality> consequently gives
$$
\mathbb P\left(\left|\frac{\int_t^{t+\varepsilon}H_s\,dB_s}{B_{t+\varepsilon}-B_t}-H_t\right|>\eta\right)
\leq\eta^{-1/4}\mathbb E\left|\frac{\int_t^{t+\varepsilon}(H_s-H_t)\,dB_s}{B_{t+\varepsilon}-B_t}\right|^{1/4}\longrightarrow0.
$$
Therefore
$$
\boxed{\frac1{B_{t+\varepsilon}-B_t}\int_t^{t+\varepsilon}H_s\,dB_s\xrightarrow[\varepsilon\downarrow0]{\mathbb P}H_t\quad\text{for every fixed }t>0.}
$$
The stated <convergence in probability> holds for each fixed deterministic time.
Back to article page
