∫0TWsds\int_0^T W_s ds∫0TWsds 的均值与方差 Mean and variance 专题 Finance / 金融 难度 L4 来源 QuantQuestion 题目详情 Let WsW_sWs be a Brownian motion. What are the mean and variance of I(T)=∫0TWsdsI(T) = \int_0^T W_s dsI(T)=∫0TWsds ? 解析 设 I(T)=∫0TWs dsI(T)=\int_0^T W_s\,dsI(T)=∫0TWsds。 均值:E[I(T)]=∫0TE[Ws]ds=0\mathbb{E}[I(T)]=\int_0^T\mathbb{E}[W_s]ds=0E[I(T)]=∫0TE[Ws]ds=0。 方差: Var(I(T))=∫0T∫0Tmin(s,u) ds du=T33.\operatorname{Var}(I(T))=\int_0^T\int_0^T\min(s,u)\,ds\,du=\frac{T^3}{3}.Var(I(T))=∫0T∫0Tmin(s,u)dsdu=3T3. 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}}.E[I(T)]=0,Var(I(T))=3T3.