# Is this process of Brownian motion?

Background Information:

The process $W = (W_t:t\geq 0)$ is a $\mathbb{P}$-Brownian motion if and only if

i) $W_t$ is continuous, and $W_0 = 0$

ii) the value of $W_t$ is distributed, under $\mathbb{P}$, as a normal random variable $N(0,t)$,

iii) the increment $W_{s+t} - W_{s}$ is distributed as a normal $N(0,t)$, under $\mathbb{P}$, and is independent of $\mathcal{F}_s$, the history of what the process did up to time $s$.

Question:

If $W_t$ and $\tilde{W}_t$ are two independent Brownian motions and $\rho$ is a constant between $-1$ and $1$, then the process $X_t = \rho W_t - \sqrt{1-\rho^2}\tilde{W}_t$ is continuous and has marginal distributions $N(0,t)$. Is this $X$ a Brownian motion?

We have the first two conditions met. Now we must check if $X_{s+t} - X_s~N(0,t)$, and if $X_{s+t} - X_s$ is independent of $X_s$. One problem I having here is what is $X_{s+t}$, would it be $X_{s+t}= \rho W_{s+t} + \sqrt{1-\rho^2}\tilde{W}_{s+t}$, and if so is $W_{s+t}$ and $\tilde{W}_{s+t}$ two independent Brownian motions?

Updated- We have \begin{align*} X_{s+t} - X_{s} &= \rho W_{s+t} + \sqrt{1-\rho^2}\tilde{W}_{s+t} - \rho W_s + \sqrt{1-\rho^2}\tilde{W}_s\\ &= \rho(W_{s+t} - W_s) - \sqrt{1-\rho^2}(\tilde{W}_{s+t} - \tilde{W}_s)\\ &\sim N(0,t) \end{align*} Since $W_{s+t} - W{s}\sim N(0,t)$ and $\tilde{W}_{s+t}-\tilde{W}_s\sim N(0,t)$. Now we need to check if $X_{s+t} - X{s}$ is independent of $X_{s}$. To do so we can check if the expectation of the product is equal to the product of the expectations.

We can show $$E\left[(X_{s+t}-X_{s})X_{s}\right]=E[X_{s+t}X_s]-E[X_s^2]=\text{Cov}(X_{t+s},X_s)-\text{Var}(X_s)=s-s=0$$ Hence $X$ is a Brownian motion.

• Those are not difficult questions. I prefer to leave this for you to try. Nov 22, 2016 at 16:29
• Yes, its just my first exposure to these types of questions so I am a bit slow at getting it Nov 22, 2016 at 16:31
• @Gordon is $X_{s+t} = \rho W_{s+t} + \sqrt{1-\rho^2}\tilde{W}_{s+t}$? And is $W_{s+t}$ and $\tilde{W}_{s+t}$ two independent Brownian motions? Nov 22, 2016 at 16:33
• Yes. $X_{s+t} = \rho W_{s+t} +\sqrt{1-\rho^2}\tilde{W}_{s+t}$. Moreover, $W_{s+t}$ and $\tilde{W}_{s+t}$ are two independent random variables. Nov 22, 2016 at 16:38
• @Gordon made an attempt, could you let me know if I am on the right track? Nov 22, 2016 at 17:07

Lévy’s theorem: Let $X_t$ be a martingale with $X_0=0$. Then the following are equivalent.

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

Proposition: If $W^{(1)}_t$ and $W^{(2)}_t$ be two independent standard Brownian motions then $W_t:=\rho W^{(1)}_t-\sqrt{1-\rho^2} W^{(2)}_t$ 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 $W_0=0$. Note

$$\mathbb{E}[W_t|\mathcal{F_s}]=\rho\,\mathbb{E}[W^{(1)}_t|\mathcal{F_s}]-\sqrt{1-\rho^2}\,\mathbb{E}[W^{(2)}_t|\mathcal{F_s}]=\rho W^{(1)}_s-\sqrt{1-\rho^2} W^{(2)}_s=W_s$$ 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 $$d(W_t^2-t)=2W_tdW_t$$ Therefore $W_t^2-t$, is a martingale (because it's SDE has a null drift ) and $W_t$ is a standard Brownian motion.

Another way

$W_t^{(1)},W_t^{(2)}\stackrel{\mathrm{i.i.d.}}\sim \mathcal N(0,t)$, so $$\rho W_t^{(1)}\sim\mathcal N\left(0,\rho^2 t\right)$$ and $$\sqrt{1-\rho^2}W_t^{(2)}\sim\mathcal N\left(0,(1-\rho^2)t \right)$$ thus $$W_t \sim \mathcal N\left(0, t \right).$$ Therefore $\{W_t:t\in\mathbb R_+\}$ is a Gaussian process, and from independence of $W_t^{(1)}$ and $W_t^{(2)}$ we have $$\mathbb E\left[W_t^{(1)}W_t^{(2)}\right] =\mathbb E\left[W_t^{(1)}\right]\mathbb E\left[W_t^{(2)}\right]=0.$$ Let $0<s<t$ we have \begin{align} \text{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] - 2\rho\sqrt{1-\rho^2}\mathbb E\left[W_s^{(1)}W_t^{(2)} \right]\\ &= \rho^2 s + \left(1-\rho^2\right)s - \rho\sqrt{1-\rho^2}\times 0\\ &= s. \end{align} on the other hand $$\mathbb{E}[(W_t-W_s)W_s]=\mathbb{E}[W_t\,W_s]-\mathbb{E}[W_s^2]=\text{Cov}(W_t,W_s)-\text{Var}(W_s)=s-s=0$$

• +1. This is good. But, it appears that @Wolfy is still at a very early stage to understand this. Can you please also give him the rudimentary approach to show that $(W_t: t\ge 0)$ is a Brownian motion by definition? Nov 22, 2016 at 18:56
• So thanks. Yes I will show it as soon as possible.
– user16651
Nov 22, 2016 at 18:58
• I don't see where I went wrong in my solution seems that I was able to show that $X_{s+t} - X_{s}\sim N(0,t)$ and $X_{s+t} - X_{s}$ is independent of $X_s$ Nov 22, 2016 at 19:37
• I don't understand your solution really, but thank you Nov 22, 2016 at 19:42
• @Wolfy Please Note $X_t = \rho W_t - \sqrt{1-\rho^2}\tilde{W}_t$ thus $X_{s+t}= \rho W_{s+t} - \sqrt{1-\rho^2}\tilde{W}_{s+t}$
– user16651
Nov 22, 2016 at 19:43