Integral of time with respect to Brownian motion

Question

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.

Answer 1

For the second question I came up with this answer. We have that for each $r \geq 1$ $$\sum_{i=0}^{r-1}t_i(B_{t_{i+1}}-B_{t_i}) \sim \mathcal{N}\left(0,\frac{T^3}{6}\left(1-\frac{1}{r}\right)\left(2-\frac{1}{r}\right)\right)$$ Furthermore, $$\sum_{i=0}^{r-1}t_i(B_{t_{i+1}}-B_{t_i}) \rightarrow \int_0^Tt\ dB_t \quad \text{in} \quad L^2$$

We have the result that if a sequence of normal random variables $(X_r)_r$ converges to $X$ in distribution, then $X$ is normal as well. We have $L^2$ convergence so weak convergence is implied. So then $\int_0^Tt\ dB_t$ must have a normal distribution for which mean is $0$ and variance is given as $$\sigma^2 = \lim_{r\rightarrow\infty} \frac{T^3}{6}\left(1-\frac{1}{r}\right)\left(2-\frac{1}{r}\right) = \frac{T^3}{3}.$$

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Here is a much easier to way compute the variance.

First of all $\int_0^Tt\ dB_t$ is a martingale. It immediately follows that its mean is $0$. To compute the variance we make use of Ito isometry, i.e. $$E\left[\left(\int_0^Tt\ dB_t\right)^2\right] = E\left[\int_0^Tt^2\ d[B]_t\right] = \frac{T^3}{3}$$

Answer 2

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 $$\sum_{i=0}^{r-1} f(t_{i+1}) (t_{i+1}-t_i)$$