# Integral of time with respect to Brownian motion

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.

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 $$\sigma^2 = \lim_{r\rightarrow\infty} \frac{T^3}{6}\left(1-\frac{1}{r}\right)\left(2-\frac{1}{r}\right) = \frac{T^3}{3}.$$

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. $$E\left[\left(\int_0^Tt\ dB_t\right)^2\right] = E\left[\int_0^Tt^2\ d[B]_t\right] = \frac{T^3}{3}$$

• This is really unnecessary if one wants to compute the variance.
– Did
Mar 20, 2015 at 17:06
• @Did I understand but I am dealing with this kind of stuff for the first time and whatever answer I come up with, as long as it is correct, helps me build confidence in myself in this field. Obviously, I am trying to learn better ways of solving such problems in the mean time. So please feel free to give your feedback if you want to. Mar 20, 2015 at 17:16
• Itô isometry is exactly suited to this.
– Did
Mar 20, 2015 at 17:22
• @Did Thanks. I updated my answer. Mar 20, 2015 at 17:30
• Yep, nice Edit.
– Did
Mar 20, 2015 at 17:30
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 $$\sum_{i=0}^{r-1} f(t_{i+1}) (t_{i+1}-t_i)$$
• Thank you so much, once again. Your hint for the second question almost looks like the Riemann integral but then wouldn't I need $f(t_i)$ instead of $f(t_{i+1})$? Or it doesn't matter in this case because the integrator, $t$, is smooth? By the way I came up with another answer for the second question (thanks to one of our previous discussions under another question). I will post it in a bit. Mar 20, 2015 at 14:37
• @Calculon "Smooth" is overkill, "continuous" does the job. So: Yes, because of continuity, it doesn't matter whether we use $f(t_i)$ or $f(t_{i+1})$ in the Riemann sum.
– saz
Mar 20, 2015 at 14:44
• But if continuity were sufficient, then integration wrt Brownian motion would yield the same result regardless of which end point (or intermediate point) we pick, which we know is not true. Mar 20, 2015 at 14:49
• @Calculon Ah, sorry, I was talking about continuity of $f$ (i.e. the integrand; not the integrator!) You are right, if we consider $$\sum f(t_{i+1}) (B_{t_{i+1}}-B_{t_i})$$ then the limit does not equal the limit of the Riemann sum $$\sum f(t_i) (B_{t_{i+1}}-B_{t_i})$$ even if $f$ is nice/smooth.
– saz
Mar 20, 2015 at 14:51
• so then a bit more regularity than continuity is required on the integrator to allow freedom of choice in where we evaluate the integrand? Mar 20, 2015 at 14:54