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0TWsds\int_0^T W_s ds 的均值与方差

Mean and variance

专题
Finance / 金融
难度
L4

题目详情

Let WsW_s be a Brownian motion. What are the mean and variance of I(T)=0TWsdsI(T) = \int_0^T W_s ds ?

解析

I(T)=0TWsdsI(T)=\int_0^T W_s\,ds

均值:E[I(T)]=0TE[Ws]ds=0\mathbb{E}[I(T)]=\int_0^T\mathbb{E}[W_s]ds=0

方差:

Var(I(T))=0T0Tmin(s,u)dsdu=T33.\operatorname{Var}(I(T))=\int_0^T\int_0^T\min(s,u)\,ds\,du=\frac{T^3}{3}.

因此

E[I(T)]=0,Var(I(T))=T33.\boxed{\mathbb{E}[I(T)]=0},\qquad \boxed{\operatorname{Var}(I(T))=\frac{T^3}{3}}.