I'm going through a computation that calculates the integral of a Brownian motion with respect to another Brownian Motion. Apparently I can do this with martingales (after searching here), but I haven't learned these yet. The solution method I'm looking at uses $L^2$ convergence directly. I've indicated the places I get lost below:
Question:
Show:$$\int_0^tB_sdB_s = \frac{1}{2} B_t – \frac{1}{2} t$$
- Create a process that approximates $B_t$. Consider a simple process $$f_n(s)=\sum_{\text{over partitions labeled by }t_j}B_{t_j}\mathbb{1}_{[t_j,t_{j+1})}(s)$$
Then: $$\mathbf{E}\bigg[\int_0^t(B_s-f_n(s))^2ds\bigg] \\ = \mathbf{E}\bigg[\sum_j\int_{t_j}^{t_{j+1}}(B_s-B_{t_j})^2ds\bigg] \\ = \sum_j \frac{1}{2}(t_{j+1}-t_j)^2 \rightarrow 0$$
In the second line, I think ito isometry is used. When I plug in the definition of $f_n$ and work through it, I get lost.
- Using $f_n$ so constructed in part 1, we approximate the integral of interest then take limits. That is $$\int_0^tB_sdB_s = \lim_{n\rightarrow \infty} \int_0^t f_n dB_s = \lim_{n\rightarrow \infty} \sum_{j=1}^n B_{t_j}(B_{t_{j+1}}-B_{t_j})$$ Noting then that $B_{t_{j+1}}^2-B_{t_j}^2=(B_{t_{j+1}}-B_{t_j})^2 + 2B_{t_j}(B_{t_{j+1}}-B_{t_j})$ we obtain:
$$\sum_j B_{t_j}(B_{t_{j+1}}-B_{t_j}) = \frac{1}{2}B_t^2 – \sum_j \frac{1}{2}(B_{t_{j+1}}-B_{t_j})^2$$
The result follows by taking $n$ to the limit.
Where does the first term on the right hand side come from ($\frac{1}{2}B_t^2$)? I see a telescoping series, but after all the cancellation, I'm left with 2 terms. Does $\lim_{n \rightarrow \infty}B_{t_n}^2$ go to zero?
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