Some questions on the details of an integration of Brownian Motion Against itself

Question

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:$$\int_0^tB_sdB_s = \frac{1}{2} B_t - \frac{1}{2} t$$

1. 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.

2. 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?

Answer 1

1. You can't apply Itô's isometry here (there are no stochastic integrals $\int \dots \, dB_s$ in this part of the proof!). Let $\Pi = \{0=t_0 < \ldots < t_n=t\}$ be a partition of the interval of $[0,t]$ and define $$f_n(s) = \sum_{j=0}^{n-1} B_{t_j} 1_{[t_j,t_{j+1})}(s).$$ By the linearity of the expectation, we have $$\mathbb{E} \left[ \int_0^t (B_s-f_n(s))^2 \, ds \right] = \sum_{j=0}^{n-1} \mathbb{E} \left( \int_{t_{j}}^{t_{j+1}} (B_s-f_n(s))^2 \, ds \right].$$ By the very definition of $f_n$, it holds that $f_n(s) = B_{t_j}$ for any $s \in [t_j,t_{j+1})$. Thus, $$\mathbb{E} \left[ \int_0^t (B_s-f_n(s))^2 \, ds \right] = \sum_{j=0}^{n-1} \mathbb{E} \left( \int_{t_j}^{t_{j+1}} (B_s-B_{t_j})^2 \, ds \right].$$ Now we can apply Fubini's theorem and use that $B_u-B_v \sim N(0,u-v)$ to conclude that $$\mathbb{E} \left[ \int_0^t (B_s-f_n(s))^2 \, ds \right] = \frac{1}{2} \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2.$$
2. Note that $t_n$ is the right end point of the partition of the interval $[0,t]$, and therefore $t_n=t$. In particular, $B_{t_n}^2 = B_t^2$.

Answer 2

While your question has already been answered above, it may be useful to see an alternative approach: Recall Ito's lemma, which says that: If $Y_t = f(B_t)$, then

$$ dY_t = f'(B_t)dB_t + \frac12 f''(B_t)dt.$$

Choose $f(x)=\frac12 x^2$. Then $f'(x)=x$ and $f''(x)=1$. Hence

$$ dY_t = B_tdB_t + \frac12 dt.$$

So $Y_t-Y_0= \int_0^t dY_s = \int_0^t B_sdB_s + \frac12 \int_0^tds$. Since $f(B_0)=f(0)=0$ we have that $$ \int_0^t B_sdB_s = \frac12 B_t^2 - \frac12 t.$$