$W^{(1)}_t$ and $W^{(2)}_t$ are two independent Brownian motions. How can I use Lévy's Theorem to show that
$$W_t:=\rho W^{(1)}_t+\sqrt{(1-\rho^2)} W^{(2)}_t,$$
is also a Brownian motion for a given constant $\rho\in(0,1)$.
Also, it is clear why $\rho$ in front of the first Brownian term is there, to get $E[W^{(1)}_t W^{(2)}_t] = \rho dt $
But, I don't understand why the term $\sqrt{(1-\rho)}$ needs to be there in that form
Best Answer
Proof
Let $(\Omega, \mathcal{F},\mathbb{P},\{\mathcal{F_t}\})$ be a probability space . Clearly, $W_t$ has continuous sample paths and $W_0=0$.
$$\mathbb{E}[W_t|\mathcal{F_s}]=\rho\,\mathbb{E}[W^{(1)}_t|\mathcal{F_s}]+\sqrt{1-\rho^2}\,\mathbb{E}[W^{(2)}_t|\mathcal{F_s}]=\rho W^{(1)}_s+\sqrt{1-\rho^2} W^{(2)}_s=W_s$$ So $W_t$ is a martingale. Now we should show $W_t^2-t$ is a martingale.
By application of Ito's lemma, we have $$dW_t^2=2W_tdW_t+d[W_t,W_t]$$ $$dW_t^2=2W_tdW_t+\rho^2 d[W_t^{(1)},W_t^{(1)}]+(1-\rho^2) d[W_t^{(2)},W_t^{(2)}]+2\rho\sqrt{1-\rho^2}d[W_t^{(1)},W^{(2)}_t]$$
Since $W^{(1)}_t$ and $W^{(2)}_t$ are two independent Brownian motions, thus $d[W_t^{(1)},W^{(2)}_t]=0$ , hence $$dW_t^2=2W_tdW_t+dt$$ consequently $$d(W_t^2-t)=2W_tdW_t+dt-dt=2W_tdW_t$$ so to speak $$d(W_t^2-t)=2W_tdW_t$$ Indeed $$d(W_t^2-t)=2\rho W_t dW^{(1)}_t+2\sqrt{1-\rho^2}\,W_t dW^{(2)}_t$$ Therefore $W_t^2-t$, is a martingale (because it's SDE has a null drift ) and $W_t$ is a standard Brownian motion.