Var (∫0TWsds)\operatorname{Var}\!\left(\int_0^T W_s ds\right)Var(∫0TWsds) Evaluate the variance 专题 Probability / 概率 难度 L4 来源 QuantQuestion 题目详情 Evaluate the variance of ∫0TWsds\int_0^T W_s ds∫0TWsds 解析 令 I=∫0TWs dsI=\int_0^T W_s\,dsI=∫0TWsds。由于 E[Ws]=0\mathbb{E}[W_s]=0E[Ws]=0,有 E[I]=0\mathbb{E}[I]=0E[I]=0。 方差为 Var(I)=E[I2]=∫0T∫0TE[WsWu] ds du.\operatorname{Var}(I)=\mathbb{E}[I^2]=\int_0^T\int_0^T \mathbb{E}[W_sW_u] \,ds\,du.Var(I)=E[I2]=∫0T∫0TE[WsWu]dsdu. 而 E[WsWu]=min(s,u)\mathbb{E}[W_sW_u]=\min(s,u)E[WsWu]=min(s,u),因此 Var(I)=∫0T∫0Tmin(s,u) ds du=T33.\operatorname{Var}(I)=\int_0^T\int_0^T \min(s,u)\,ds\,du=\frac{T^3}{3}.Var(I)=∫0T∫0Tmin(s,u)dsdu=3T3. 即 Var (∫0TWsds)=T33.\boxed{\operatorname{Var}\!\left(\int_0^T W_s ds\right)=\frac{T^3}{3}}.Var(∫0TWsds)=3T3.