[Math] Integral of time with respect to Brownian motion

stochastic-calculusstochastic-integrals

I am trying to compute $\int_0^T t\ dB_t$ where $B$ is the standard Brownian motion.

To this end I define the sequence of simple predictable functions
$$ f_n = \sum_{i=0}^{2^nn-1}t_i^n1_{(t_i^n,t_{i+1}^n]}(t) \quad \text{where} \quad t_i^n = i2^{-n}$$

Then I show that $\lVert f_n-t\rVert_{\mathcal{L}^2(B)} \rightarrow 0$. The convergence in $\mathcal{L}^2(B)$ is equivalent to showing that
$$\lim_{n\rightarrow\infty} E \int_0^m (f_n-t)^2\ d[B]_t = 0 \quad \forall{m} \in \mathbb{N}$$
First, I fix $m$ and choose $n > m$. I also drop the expectation.
\begin{align}E \int_0^m (f_n-t)^2\ d[B]_t \leq& \int_0^n (f_n-t)^2\ dt\\
= & \frac{1}{3}n2^{-2n} \end{align}
Now I let $n \rightarrow \infty$. Since $m$ was arbitrary, I have convergence in $\mathcal{L}^2(B)$.

Following the definition of stochastic integration of simple predictable processes I write
$$\int_0^Tf_n\ dB_t = \sum_{i=0}^{2^nn-1}t_i^n(B_{T\wedge t_{i+1}^n}-B_{T\wedge t_i^n})$$

I start having problems at this point. I need to get rid of the wedge sign and I do that intuitively. For a given $T$, I choose $n > T$ so that for some $m \in \mathbb{N}$
$$\sum_{i=0}^{2^nn-1}t_i^n(B_{T\wedge t_{i+1}^n}-B_{T\wedge t_i^n}) = \sum_{i=0}^{m-1}t_i^n(B_{t_{i+1}^n}-B_{t_i^n}) + t_{m}(B_T-B_{t_m^n})$$

Now I let $n$ go to infinity. But I don't see in full rigour how this is the same thing as
setting up a mesh $\pi: 0 = t_0 < t_1 < \ldots < t_r = T$ where $t_i = \frac{iT}{r}$ and letting $r$ go to infinity in
$$\sum_{i=0}^{r-1}t_i(B_{t_{i+1}}-B_{t_i})$$
Because eventually I need to show that the latter sum converges to something in $L^2$ as $r\rightarrow \infty$.

So this was my first question. The second one is how do I show that the sum
$$\sum_{i=0}^{r-1}t_i(B_{t_{i+1}}-B_{t_i})$$ converges to something (I don't know what that is yet) in $L^2$.
We are given a hint to use summation by parts. So I use the hint to get
$$\sum_{i=0}^{r-1}t_i(B_{t_{i+1}}-B_{t_i}) = TB_T – \frac{1}{r}\sum_{i=0}^{r-1}B_{t_{i+1}}$$

I have no clue what to do with the second part of this sum.

My apologies if there is any nonsense, gibberish in what I wrote as I myself am confused in this arduous process of learning stochastic calculus.

Best Answer

  1. Because of continuity of Brownian motion, it doesn't matter how we choose the partition as long as the mesh size converges to $0$. The continuity of Brownian motion implies that the term $$t_m (B_T-B_{t_m})$$ converges (almost surely and in $L^2$) to $0$ if we let $n \to \infty$. Since we know that the result does not depend on the chosen partition, the best choice (regarding notation) is $t_i := \frac{i}{n} T$, $i=0,\ldots,n$,
  2. Hint: $$\frac{1}{r} \sum_{i=0}^{r-1} B_{t_{i+1}} = \sum_{i=0}^{r-1} B_{t_{i+1}} (t_{i+1}-t_i).$$ Do you recognize the expression at the right-hand side? Think of $$\sum_{i=0}^{r-1} f(t_{i+1}) (t_{i+1}-t_i)$$
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