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Var ⁣(0TWsds)\operatorname{Var}\!\left(\int_0^T W_s ds\right)

Evaluate the variance

专题
Probability / 概率
难度
L4

题目详情

Evaluate the variance of 0TWsds\int_0^T W_s ds

解析

I=0TWsdsI=\int_0^T W_s\,ds。由于 E[Ws]=0\mathbb{E}[W_s]=0,有 E[I]=0\mathbb{E}[I]=0

方差为

Var(I)=E[I2]=0T0TE[WsWu]dsdu.\operatorname{Var}(I)=\mathbb{E}[I^2]=\int_0^T\int_0^T \mathbb{E}[W_sW_u] \,ds\,du.

E[WsWu]=min(s,u)\mathbb{E}[W_sW_u]=\min(s,u),因此

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

Var ⁣(0TWsds)=T33.\boxed{\operatorname{Var}\!\left(\int_0^T W_s ds\right)=\frac{T^3}{3}}.