# Proving it is a Brownian motion

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 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.

• Use another characterization of a Brownian motion: it is a centered Gaussian process with the covariance function $\min(t,s)$. Commented Sep 5 at 8:11
• @zhoraster This is true; however, going down the rabbit hole of calculating the covariance leads to weird results as well (at least when I attempted this). Commented Sep 5 at 18:45
• @392781 I've got exactly $\min(s,t)$ (though I used sage for that). Commented Sep 7 at 11:21

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 $$0=s_{0}<... and $$s=t_{0}<... 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 $$\int_{s}^{t}$$.