# Correlated brownian motions and Lévy's theorem

$W^{(1)}_t$ 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

• $\sqrt{1-\rho^2}$ ? Jul 22, 2016 at 21:54
• Presumably, $\rho\in(0,1)$? Jul 22, 2016 at 23:34
• With $\sqrt{1-\rho^2}$ instead, the variances add up to $t$, which is one of the requirements for being a (standard) Brownian motion.
– Ian
Jul 23, 2016 at 0:19
• I guess you mean $E[W_t^{(1)}W_t]=\rho dt$? Jun 16, 2019 at 2:53

If $W_t:=\rho W^{(1)}_t+\sqrt{1-\rho^2} W^{(2)}_t$ then we can show that, $W_t$ is a Brownian motion.

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.

For any $t>0$, we have $W_t^{(1)},W_t^{(2)}\stackrel{\mathrm{i.i.d.}}\sim \mathcal N(0,t)$, so $\rho W_t^{(1)}\sim\mathcal N\left(0,\rho^2 t\right)$ and $\sqrt{1-\rho^2}W_t^{(2)}\sim\mathcal N\left(0,(1-\rho^2)t \right)$, from which $$W_t \sim \mathcal N\left(0, t \right).$$ Therefore $\{W_t:t\in\mathbb R_+\}$ is a Gaussian process, and from independence of $W_t^{(1)}$ and $W_t^{(2)}$ we have $$\mathbb E\left[W_t^{(1)}W_t^{(2)}\right] =\mathbb E\left[W_t^{(1)}\right]\mathbb E\left[W_t^{(2)}\right]=0.$$ For $s,t>0$ we compute \begin{align} \mathsf{Cov}(W_s,W_t) &= \mathbb E[W_sW_t] - \mathbb E[W_s]\mathbb E[W_t]\\ &= \mathbb E\left[\left(\rho W_s^{(1)}+\sqrt{1-\rho^2}W_s^{(2)}\right)\left(\rho W_t^{(1)}+\sqrt{1-\rho^2}W_t^{(2)}\right)\right] - 0\\ &= \rho^2\mathbb E\left[W_s^{(1)}W_t^{(1)} \right]+(1-\rho^2) \mathbb E\left[W_s^{(2)}W_t^{(2)} \right] + \rho\sqrt{1-\rho^2}\mathbb E\left[W_s^{(1)}W_t^{(2)}+W_s^{(2)}W_t^{(1)} \right]\\ &= \rho^2 (s\wedge t) + \left(1-\rho^2\right)(s\wedge t) + \rho\sqrt{1-\rho^2}\cdot0\\ &= s\wedge t. \end{align} This of course is the covariance function of standard Brownian motion, and independence of increments readily follows. Sample path continuity follows from $W_t$ being a linear combination of Brownian motions, and therefore $W_t$ itself is a standard Brownian motion.

Edit: To prove this using Lévy’s theorem, i.e.

Let $X$ be a martingale with $X_0=0$. Then the following are equivalent.

1. $X$ is standard Brownian motion.
2. $X$ is path-continuous and $X_t^2-t$ is a martingale.
3. $X$ has quadratic variation $[X]_t=t$,

first note that $$W_0 = \rho W_0^{(1)} + \sqrt{1-\rho^2}W_0^{(2)} = 0\ \mathrm{w.\!p.\ } 1.$$ Since $W^{(1)}_t$ and $W^{(2)}_t$ are martingales, it follows that for $0\leqslant s<t$, \begin{align} \mathbb E[W_t\mid\mathcal F_s] &= \mathbb E\left[ \rho W_t^{(1)}+\sqrt{1-\rho^2} W_t^{(2)}\mid\mathcal F_s\right]\\ &= \rho\mathbb E\left[ \rho W_t^{(1)}\mid\mathcal F_s\right] + \sqrt{1-\rho^2}\mathbb E\left[W_t^{(2)}\mid\mathcal F_s\right]\\ &= \rho W_s^{(1)} + \sqrt{1-\rho^2} W_s^{(2)}\\ &= W_s, \end{align} so that $W_t$ is a martingale. Moreover, $\left(W_t^{(1)}\right)^2-t$ and $\left(W_t^{(1)}\right)^2-t$ are martingales, and hence \begin{align} \mathbb E\left[ W_t^2-t\mid\mathcal F_s\right] &= \mathbb E\left[\rho^2\left(W_t^{(1)}\right)^2 +2\rho\sqrt{1-\rho^2}W_t^{(1)}W_t^{(2)} + \left(1-\rho^2\right)\left(W_t^{(2)}\right)^2-t\mid\mathcal F_s\right]\\ &= \rho^2\mathbb E\left[\left(W_t^{(1)}\right)^2-t \mid\mathcal F_s\right]+2\rho\sqrt{1-\rho^2}\mathbb E\left[W_t^{(1)}W_t^{(2)} \mid\mathcal F_s\right]\\&\quad+\left(1-\rho^2\right)\mathbb E\left[\left(W_t^{(2)}\right)^2-t \mid\mathcal F_s\right]\\ &= \rho^2\left( \left(W_s^{(1)}\right)^2 - s\right)+2\rho\sqrt{1-\rho^2}W_s^{(1)}W_s^{(2)}+\left(1-\rho^2\right)\left( \left(W_s^{(2)}\right)^2 - s\right)\\ &= W_s^2 - s, \end{align} so that $W_t^2-t$ is a martingale. We conclude that $W_t$ is standard Brownian motion. (Path-continuity follows from $W_t$ being a linear combination of Brownian motions.)

• It is correct but Levi's theorem? OP? Jul 23, 2016 at 0:48
• @BehrouzMaleki By Lévy’s theorem do you mean the martingale characterization of Brownian motion? Jul 23, 2016 at 0:55
• Yes ,Indeed you should show that, $W_t^2-t$ is a martingale. Jul 23, 2016 at 0:58
• @BehrouzMaleki Very well, I shall add this to my answer. Jul 23, 2016 at 1:02
• Levy 1948, pages 77-78 of second edition. Jul 23, 2016 at 1:09