I would like to know why the following method leads to an incorrect solution: E[exp(-2Bt) | Fs] = exp(-2Bs) * E[exp(-2(Bt-Bs)| Fs] Now expanding the exponent, (I'm dropping | Fs since we are dealing with the increments and Fs is immaterial owing to the independence property) = exp(-2Bs) * E[1 + -2(Bt-Bs)/1! + (-2*(Bt-Bs))^2/2! + ...] Taking expectations inside the bracket, (Bt-Bs)^3 and higher orders is equal to 0. Also, E[Bt-Bs] = 0, so, it equates to: = exp(-2Bs)*(1 + 4*(t-s)/2) = exp(-2Bs)*(1 + 2*(t-s)) The solution gives the answer as : exp(-2Bs)*exp(2*(t-s)) Thanks!
I think there are a couple of reasons this doesn't work: First, the expansion you're using is valid for small exponents only, but Bt-Bs could be huge. Second, dBs^3 and higher orders is zero, but this applies to infinitesinal increments only. Bt-Bs is a finite increment and so this result doesn't apply. So the higher order terms can't just be ignored. Does that explain it?
assuming the t-s is a very small quantity, (1+ (t-s)) ~ exp(t-s), hence your answer is the same as given in the text book
