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Let $B(t)$ and $W(t)$ be two independent Brownian motions. Show that $\frac{B(t)+W(t)}{2}$ is also a Brownian motion. Find correlation between $B(t)$ and $X(t)$.

thanks for any help

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  • $\begingroup$ Really, everything works here, no? What have you tried? $\endgroup$ – Did Nov 29 '13 at 20:15
  • $\begingroup$ Do not, I repeat, DO NOT edit the question after answers have been posted, these modifications make the answers appear off-topic. $\endgroup$ – Did Dec 13 '13 at 16:26
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Actually,

$$X_t := \frac{B_t+W_t}{2}$$

is not a Brownian motion since

$$\mathbb{E}(X_t^2) = 2t$$

(see the calculation below) which is a contradiction to $X_t \stackrel{!}{\sim} N(0,t)$.


Instead we consider the process

$$X_t := \alpha \cdot B_t + \beta \cdot W_t$$

for $\alpha, \beta \in \mathbb{R}$. We want to find out for which $\alpha,\beta$ this process is a Brownian motion. In order to do this, we have to check the following properties.

  1. $X_0 = 0$ a.s.
  2. $X_t-X_s \sim N(0,t-s)$ for any $0 \leq s < t$
  3. $X_{t_1}-X_{t_0},\ldots,X_{t_n}-X_{t_{n-1}}$ are independent for $0=t_0<\ldots<t_n$
  4. $t \mapsto X_t(\omega)$ is continuous for almost all $\omega \in \Omega$

Proof:

  1. $X_0 = \alpha \cdot B_0+\beta \cdot W_0 = 0$ since $B_0=W_0=0$.
  2. Let $0 \leq s < t$. By assumption, we know that $X:= B_t-B_s \sim N(0,t-s)$ and $Y:=W_t-W_s \sim N(0,t-s)$ are independent. Since the sum of independent normally distributed random variables is again normally distributed, it suffices to calculate mean and variance of $X_t-X_s = \alpha X+ \beta Y$. Using the independence and $X,Y \sim N(0,t-s)$ we find \begin{align*} \mathbb{E}(X_t-X_s) &=\alpha \mathbb{E}X+\beta \mathbb{E}Y=0 \\ \mathbb{E}((X_t-X_s)^2) &= \alpha^2 \mathbb{E}(X^2) +2 \alpha \beta \mathbb{E}(X \cdot Y) + \beta^2 \mathbb{E}(Y^2) \\ &= \alpha^2 t + 2 \cdot 0 + \beta^2 \cdot t \end{align*} From $\mathbb{E}((X_t-X_s)^2) \stackrel{!}{=} t$ we see that $\alpha^2+\beta^2 = 1$ is a necessary condition.
  3. Let $0=t_0<\ldots<t_n$. Since $(B_t)_t$ and $(W_t)_t$ are independent Brownian motions, we have that $$B_{t_1}-B_{t_0},\ldots,B_{t_n}-B_{t_{n-1}},W_{t_1}-W_{t_0},\ldots,W_{t_n}-W_{t_{n-1}}$$ are independent random variables. This implies in particular the independence of $$\alpha \cdot (B_{t_1}-B_{t_0})+\beta \cdot (W_{t_1}-W_{t_0}),\ldots, \alpha \cdot (B_{t_n}-B_{t_{n-1}})+\beta \cdot (W_{t_n}-W_{t_{n-1}})$$ i.e. $(X_t)_{t \geq 0}$ has independent increments.
  4. The continuity of the sample paths follows obviously from the continuity of the sample paths of $(B_t)_t$ and $(W_t)_t$.

Finally, we conclude that $(X_t)_t$ is a Brownian motion if and only if $\alpha^2+\beta^2 = 1$. Note that this equality is not satisfied for $\alpha=\beta = \frac{1}{2}$, but for $\alpha = \beta = \frac{1}{\sqrt{2}}$.

The covariance of $X_t$ and $B_t$ equals by definition,

$$\mathbb{E}(B_t \cdot X_t) = \alpha \mathbb{E}(B_t^2)+ \beta \cdot \mathbb{E}(B_t \cdot W_t)$$

Since $(B_t)_t$ and $(W_t)_t$ are independent and $B_t \sim N(0,t)$, we obtain

$$\text{corr}(X_t,B_t) = \frac{1}{\sqrt{\text{var}(X_t) \cdot \text{var} B_t}} \cdot \mathbb{E}(B_t \cdot X_t) = \frac{\alpha \cdot t}{\sqrt{t \cdot t}} = \alpha$$

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  • $\begingroup$ @saz.i edit this question . if it possible for you help me $\endgroup$ – pual ambagher Dec 13 '13 at 16:18
  • $\begingroup$ @pualambagher As I already wrote: For any $\alpha$, $\beta$ satisfying $\alpha^2+\beta^2 = 1$, $X_t$ is a Brownian motion. Hence, in particular for $\alpha= \beta = 1/\sqrt{2}$. (And I agree with Did: Please do not edit the question after answers have been posted.) $\endgroup$ – saz Dec 13 '13 at 16:29

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