B
bobby sanger
Member
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
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