设随机变量X与Y相互独立,均服从正态分布N(μ,σ∧2),记随机变量U=max(X,Y),V=m
简单计算一下即可,详情如图所示
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9a504fc2d56285350bf760b082ef76c6a7ef6308?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/77c6a7efce1b9d1608d028f0e1deb48f8c546408?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
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第1个回答 2019-10-07
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b03533fa828ba61e2867df2b4e34970a304e592c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
如图。
相似回答
求正态分布的数学期望 已知X~N(μ,σ^2),求E(X | X答:X,Y)的期望,答案为:1/根号下\Pi;在此基础上可以有一个简单做法解楼主的问题: 由X,Y相互独立且均服从标准正态分布,可以推出:—X,—Y相互独立且也是均服从标准正态分布,而 min(X,Y)= —max(—X, —Y),所以 Emin(X,Y)= —Emax(—X, —Y)=—1/根号下\Pi.