c******n
发帖数: 4 | 1 W(t) is a Brownian motion. Calculate E(exp(int[t,T]W(s)ds))。
也就是 expectation of exponential of 从t到T的W(t)的积分。
Thanks! | g********5
发帖数: 62 | 2 Let X = int(t,T)W(s)ds
X is random variable with Normal( mean = 0, var = ... )
mean(X) = 0
Var(X) = E[X^2] = t(T-t)^2 + [(T-t)^3]/3
E ( e^X ) = exp{ mean(X) + var(X)/2 } | c******n
发帖数: 4 | 3
~~~~~~~ Could you explain more about the E[X^2] = t(T-t)^2 + [(T-t)^3]/3?
Thanks!
【在 g********5 的大作中提到】
: Let X = int(t,T)W(s)ds
: X is random variable with Normal( mean = 0, var = ... )
: mean(X) = 0
: Var(X) = E[X^2] = t(T-t)^2 + [(T-t)^3]/3
: E ( e^X ) = exp{ mean(X) + var(X)/2 }
| h*****u
发帖数: 38 | 4 Let X = int(t,T)W(s)ds
E[x]=W(t)(T-t)
E[x^2]=E[int(t,T)W(s)ds*int(t,T)W(s)ds]=2*E[int(t,T)int(t,u)W(u)W(v)dvdu]
=2*int(t,T)int(t,u)E[(W(u)-W(t))(W(v)-W(t))]+W(t)^2 dv du
=1/3(T-t)^3+W(t)^2(T-t)^2
=>Var{X}=1/3(T-t)^3 |
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