E[X∣XY]\mathbb{E}[X\mid XY]E[X∣XY] X|XY 专题 General / 综合 难度 L4 来源 QuantQuestion 题目详情 Let XXX and YYY be independent standard Gaussians. Find E[X∣XY]E[X\mid XY]E[X∣XY] 解析 令 X^=−X,Y^=−Y\widehat X=-X,\widehat Y=-YX=−X,Y=−Y，则 (X^,Y^)(\widehat X,\widehat Y)(X,Y) 与 (X,Y)(X,Y)(X,Y) 同分布且 X^Y^=XY\widehat X\widehat Y=XYXY=XY。 因此 E[X∣XY]=E[−X^∣X^Y^]=−E[X∣XY],\mathbb{E}[X\mid XY]=\mathbb{E}[-\widehat X\mid \widehat X\widehat Y]= -\mathbb{E}[X\mid XY],E[X∣XY]=E[−X∣XY]=−E[X∣XY], 故 E[X∣XY]=0.\boxed{\mathbb{E}[X\mid XY]=0}.E[X∣XY]=0.