Showing a Process, $3(B_{2+\frac{t}{9}} - B_2)$, is a Standard Brownian Motion

Question

I am currently working with a process, $$ B^{(1)}(t) = 3(B_{2+\frac{t}{9}} - B_2), \quad t \geq 0.$$

where $ B = (B_t)_{t \geq 0} $ is a standard brownian motion (SBM). I am to prove that $B^{(1)}$ is a standard brownian motion also.

I can see that the $\frac{1}{9}$ and the 3 could be used to implement brownian scaling, i.e. the fact the if B is a SBM then so is, $$\frac{B_{\rho t}}{\sqrt{\rho}}$$ and the markov property, i.e. $$ B_{T+t} - B_T$$ is SBM if B is SBM.

The main issue I have is that I am struggling with how to manipulate the process as I am not too familiar with the Brownian Motions rules and what not. If someone could talk me through this proof with the manipulations along the way it'll give me a starting point to work with further questions.

Thank you in advance.

Answer 1

If you are familiar with Levy's characterization of Brownian motion, the proof would become straightforward.

In fact, let $$ X_t=3\left(B_{2+t/9}-B_2\right), $$ and it suffices to show that

• $X_0=0$, which is obviously true, and that
• $X_t$ is a martingale, which is also kind of clear (because $X_t$ is no more than a Brownian motion after a translation, a time shift and rescaling, and a global scaling), and that
• The quadratic variance $\left<X\right>_t=t$.

Well, we have $$ \left<X\right>_t=\left<3\left(B_{2+t/9}-B_2\right)\right>=9\left<B_{2+t/9}-B_2\right>=9\left<B_{t/9}\right>=9\cdot\frac{t}{9}=t. $$

Thus Levy's characterization applies, and $X_t$ is a Brownian motion.

Answer 2

I have end up finding "a" solution and think I wasn't too clear with what I was asking, not 100% on this solution so any feedback would be appreciated. The solution is as follows.

We have $X_t = 3(B_{2+(t/9)} - B_2)$ where B is an SBM, we wish to show it is an SBM.

First start that beacause B is an SBM then $$Y_t = 3B_{\frac{t}{9}}$$ is an SBM by brownian scaling. We can then apply then use the Markov Property that says that if $Y_t$ is an SBM then for fixed $T$ then so is $$Z_t = Y_{T+t} - Y_t$$ which can be written, $$Z_t = 3B_{\frac{T+t}{9}} - 3B_{\frac{T}{9}}$$.

And finally by choosing $T = 18$, $X_t$ is shown to be an SBM.

Like I said I am not 100% on this so any clarification is much appreciated. Thanks.