Hi, I'm looking at page 8 of chapter 5 for (dBt)^2 = dt. It shows that: E[(dBt)^2] = ....= E[e^2] dt = 1*dt = dt, given e - N(0,1) so we know e is normal distribution with mean of zero, so E[e^2] should be zero too, but here it equals 1. Many thanks.
Hi. we are looking at the SECOND moment. E[e^2]!!! we know, generally: var(x) = E(x^2) + E(x)^2 we have E(x^2) = var(x) - E(x)^2 E(x)^2 = 0 (as you point out) so we know that the expression is the variance of standard normal, which is 1.
