Proving it is a Brownian motion

Question

I'm trying to solve this problem

Given a standard Brownian motion $B(t), t\geq 0$, let us define a stochastic process

$$W(t):= 3B(t)-\frac{12}{t}\int_0^tudB(u) + \frac{10}{t^2}\int_0^tu^2dB(u)\\=\int_0^t (3-12\frac{u}{t}+10\frac{u^2}{t^2}) dB(u)$$

with $W(0)=0$ on a probability space.

(a) Argue that $W(t), t\geq0$ is a standard Brownian motion.

(b) Show that $X:=\int_0^TudB(u)$ is independent of $W(t),0\leq t \leq T$ for every $T>0$.

---

My attempt:

a) To show that this a Brownian motion we have to show

1) $W(0)=0$ (because of the definition)

2) $W(t)$ has continuous paths. This is given by the definition of the Itô integral as the Brownian motion is pieced together.

3) We need to show that $W(t)-W(s) \sim N(0,t-s)$ We know by the definition of the stochastic integral that this is distributed as a normal. Also, if we let $0=t_0<t_1<...<t_j=s<t_{j+1}<...<t_k=t$ be a partition

$$ W_t-W_s= \int_0^t (3-12\frac{u}{t}+10\frac{u^2}{t^2}) dB(u)- \int_0^s (3-12\frac{u}{s}+10\frac{u^2}{s^2}) dB(u) \\= 3\sum_{i=0}^{k-1}(B_{t_{i+1}}-B_{t_i})-\frac{12}{t}\sum_{i=0}^{k-1}t_{i}(B_{t_{i+1}}-B_{t_i})+\frac{10}{t^2}\sum_{i=0}^{k-1}t^2_{i}(B_{t_{i+1}}-B_{t_i})- \Big(3\sum_{i=0}^{j-1}(B_{t_{i+1}}-B_{t_i})-\frac{12}{s}\sum_{i=0}^{j-1}t_{i}(B_{t_{i+1}}-B_{t_i})+\frac{10}{s^2}\sum_{i=0}^{j-1}t^2_{i}(B_{t_{i+1}}-B_{t_i})\Big) $$

As per, the Brownian motion we know that $B_{t_{i+1}}-B_{t_i} \sim N(0,t_{i+1}-t_{i})$

then

$$ \sum_{i=0}^{k-1}t_{i}(B_{t_{i+1}}-B_{t_i})\sim N(0,\sum_{i=0}^{k-1}t^2_it_{i+1}-t_{i}) $$

which make the variance term into the inferior Riemann sum of $\int_0^t x^2 dx$

You can do this analogously with the others, and in the end we will get

$$ W(t) \sim N(0, 9t-48t+20t) $$

which is not a right answer (because from that we get -19t, and variance can't be negative).

Then, it occurred to me that

$$ W(t)=\int_0^t (3-12\frac{u}{t}+10\frac{u^2}{t^2}) dB(u) \sim N(0, \int_0^t \Big(3-12\frac{u}{t}+10\frac{u^2}{t^2}\Big)^2 du ) = N(0,t) $$

The thing is: what am I doing wrong in the first approach?

4) To prove that $W(t)-W(s)$ is independent of $\mathscr{F}_s$. I decomposed the integral again into the sum, but I couldn't get to the conclusion that they are independent as we only have independency piecewise. For instance, $B_{t_{i+1}}-B_{t_i}$ in independent from $\mathscr{F}_{t_i}.$

b) For b, I think that once I work out how to do #4 of the last question, I'll get to it.

Answer 1

One of your mistakes in your first approach (3) forgetting the cross-terms i.e. each of the sums has a correlation with each other.

But regardless the question is about the limiting integral. So indeed by using Itô isometry you can easily compute that this process is Gaussian with variance

$$E[W^{2}]=\int_{0}^{t}(3-12\frac{u}{t}+10\frac{u^2}{t^2})^{2}du=t.$$

For (4), you again need to work with the limiting integrals

$$W_{t}-W_{s}=\int_{s}^{t}(g(u,t)-g(u,s))dB_{u}$$

and

$$W_{s}=\int_{0}^{s}g(u,s)dB_{u}.$$

These two integrals are independent of each other because they are each generated by limits of partitions $0=s_{0}<...<s_{n}=s$ and $s=t_{0}<...<t_{m}=t$ that are disjoint except at $s$.

For (b), you can simply use that when two Gaussians have zero covariance, they are independent. So here we have by Itô isometry that for $t>s$

$$E[W_{s}X_{t}]=\int_{0}^{s}u(3-12\frac{u}{t}+10\frac{u^2}{t^2})du=0,$$

where we also again used independence of increments to remove the part $\int_{s}^{t}$.