這是布朗運動的過程嗎？
背景資料：
過程 $ W = (W_t:t\geq 0) $ 是一個 $ \mathbb{P} $ -布朗運動當且僅當
一世） $ W_t $ 是連續的，並且 $ W_0 = 0 $
ii) 的價值 $ W_t $ 分佈，下 $ \mathbb{P} $ , 作為一個正態隨機變數 $ N(0,t) $ ,
iii) 增量 $ W_{s+t} - W_{s} $ 正態分佈 $ N(0,t) $ ， 在下面 $ \mathbb{P} $ , 並且獨立於 $ \mathcal{F}_s $ ，該過程迄今為止所做的歷史 $ s $ .
問題：
如果 $ W_t $ 和 $ \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.
Lévy’s theorem: Let $ X_t $ be a martingale with $ X_0=0 $ . Then the following are equivalent.
- $ X_t $ is a standard Brownian motion.
- $ X_t $ has continuous sample paths and $ X_t^2-t $ is a martingale.
- $ X_t $ has quadratic variation $ _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 $$ \mathbb{E}[(W_t-W_s)W_s]=\mathbb{E}[W_t,W_s]-\mathbb{E}[W_s^2]=\text{Cov}(W_t,W_s)-\text{Var}(W_s)=s-s=0 $$