Question

Practice on Ito integrals.Let(Bt,t > 0) be a Brownian motion defined on

(Q,5,P).We define for t ≥0 the process

Xt:=student submitted image, transcription available belowsgn(B_s)dB_s

where sgn(x)=-1 if x<0 and sgn(x)=＋1 if x ≥0.

The integral is well-defined even though s to the sgn(B_s)is not continuous. (a) Compute the mean and the covariance of the process(X_t,t ≥0).

(b)Show that X,and B,are uncorrelated for all t ≥0.

(c) Show that X, and B, are not independent.(Use B^2_t=2student submitted image, transcription available belowB_s dB_s +t.)It turns out that(Xt,t ≥0)is a standard Brownian motion.See Theorem 7.26.

Answer 1

The mean of the process (Xt, t ≥ 0) is zero and the variance is t. X and B are uncorrelated for all t ≥ 0, but they are not independent. This can be shown using Ito's isometry and the fact that B^2t = 2t.

Step-by-step explanation:

To compute the mean and covariance of the process (Xt, t ≥ 0), we need to find the expected value and variance of Xt, for t ≥ 0. Since Xt = ∫ subscript s=0 to t sgn(Bs) dBs, we can use Ito's isometry to compute the variance. According to Ito's isometry, the variance of a stochastic integral is the integral of the squared coefficient process: Var(Xt) = ∫ subscript s=0 to t (sgn(Bs))^2 ds = ∫ subscript s=0 to t ds = t.

To show that X and B are uncorrelated for all t ≥ 0, we need to show that Cov(Xt, Bt) = E[XtBt] - E[Xt]E[Bt] = 0. We can compute the covariance using Ito's isometry: Cov(Xt, Bt) = E[∫ subscript s=0 to t sgn(Bs) dBs * Bt] - E[Xt]E[Bt] = E[∫ subscript s=0 to t sgn(Bs) Bt dBs] - E[Xt]E[Bt] = 0, since the integrand is an odd function and the expectation of an odd function is zero.

To show that X and B are not independent, we can use the fact that B^2t = 2t. Since X is defined in terms of B, we can see that X is also a function of B^2t. Therefore, X and B are not independent.