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Let $B(t)$ and $W(t)$ be two independent Brownian motions. Show that $X(t)=\frac{B(t)+W(t)}{\sqrt2}$ is also a Brownian motion. Find the correlation between $B(t)$ and $X(t)$.

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The sum of two independent normal random variables is normal. Therefore, for a given fixed $t$, $$ \frac{(B(t)+W(t))-(B(0)+W(0))}{\sqrt{2}} \sim \mathcal N(0,\sqrt{t}) \>, $$ i.e., a normal distribution with mean zero and standard deviation $\sqrt{t}$; here the starting values are known.

Furthermore, independence of increments is inherited from the independence of increments of the underlying stochastic processes, which you should check by looking at the finite dimensional distributions and using properties of the multivariate normal distribution.

For correlation, we can get the covariance between $B(t)=B_t$ and $X(t)=X_t$ for a fixed t, assuming we are starting at the origin to keep things simple.

$COV(B_t,X_t)=E(X_tB_t)-E(X_t)E(B_t)=\frac{1}{\sqrt{2}}E((B_t)^2+B_tW_t)=\frac{1}{\sqrt{2}}E((B_t)^2)=\frac{t}{\sqrt{2}}$

The last two steps are from the independence of $B_t,W_t$ and the fact that $Var(B_t|B_0=0)=t$, respectively.

The correlation is just the covaraince divided by the product of the standard deviations of $X_t$ and $B_t$, which are $\sigma_{B_t}=\sqrt{t}=\sigma_{X_t}$ So the correlation is just $\frac{1}{\sqrt{2}} \approx 0.71$

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  • $\begingroup$ i edit this qustion. $\endgroup$ – pual ambagher Dec 13 '13 at 16:20
  • $\begingroup$ @pualambagher I revised my response to address your second question. $\endgroup$ – user31668 Dec 13 '13 at 16:50

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