Acted notes page 39 – q 10.3 The answer states: The dBt’s are random quantities with mean 0 and variance dt. So the expected value of each element is 0 and the variance is t^2*dt. Given the above info, Variance (dBt) = dt? and Variance (Bt) = t^2*dt? How is the variance t^2*dt calculated/formed? I was also trying to understand the below: Variance (I) = Variance (integral 1 to 0: t*dBt) = integral 1 to 0: variance(t*dBt) Since the increments are independent. Variance (I) = integral 1 to 0: t^2 dt = integral 1 to 0: 1/3 * t^3 = 1/3 What impact/effect does the independent increments have on the above? many thanks bobby
Hi Bobby, Sorry for the late reply on this one. We can think of the integral as a summation of independent increments of standard Brownian motion. In other words the covariance between non-overlapping dBts is 0. In a simple case, if we have two variables X and Y then VAR(X+Y)=VAR(X)+VAR(Y)+2COV(X,Y). So the independence allows us to drop the covariance term and just sum the individual variances. Therefore the variance of the sum of DBts is equal to the sum of the variance of DBt. This takes us from Var[integral(from 0 to 1 DBt)] to Integral from 0 to 1 [tDBt]. Var[tDBt]=t^2 Var[DBt] = t^2 dt. (i.e. take the constant outside and square it) So we can integrate t^2 with respect to t and evaluate between 0 and 1 and this will give us the variance of 1/3. Does this help? Joe
