从区间[0 1]中随机地抽出n个点X1 X2 … Xn。试分别求出最大值 最小值的数学期望和方差

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从区间[0,1]中随机地抽出n个点X1,X2,…,Xn。试分别求出最大值、最小值的数学期望和方差

热搜:触发器的输出逻辑电平从1到0或从0到1的转换称为

答案解析

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  • 由题意知,X1,X2,…,Xn都是[0,1]。独立同分布的随机变量,Xi(i=1,2,…,n)的概率密度均为
       latex.action?latex=cCh4KT1cbGVmdFx7XGJlZ2lue2FycmF5fXtjfTEsMOKJpHjiiaQxXFwgMCzlhbblroMgIFxlbmR7%0D%0AYXJyYXl9
       而其分布函数均为
        latex.action?latex=Rih4KT1cbGVmdFx7XGJlZ2lue2FycmF5fXtjfTAs44CA44CAeO%2B5pDBcXCB4LOOAgOOAgDDiiaR4%0D%0A77mkMVxcIDEs44CA44CAeOKJpTEgIFxlbmR7YXJyYXl9
      令Y1=max{X1,X2,…,Xn},Y2=min{X1,X2,…,Xn}
       (1)FY1(z)=P(Y1≤z)=P(X1≤z,X2≤z,…,Xn≤z)
       =P(X1≤z)P(X2≤z)…P(Xn≤z)=Fn(z)
       latex.action?latex=PVxsZWZ0XHtcYmVnaW57YXJyYXl9e2N9MCzjgIDjgIB677mkMFxcIHpee259LOOAgOOAgDDiiaR6%0D%0A77mkMVxcIDEs44CA44CAeuKJpTEgIFxlbmR7YXJyYXl9
       从而Y1的概率密度为
         latex.action?latex=cF97WV97MX19KHopPVxsZWZ0XHtcYmVnaW57YXJyYXl9e2N9bnpee24tMX0s44CA44CAMOKJpHri%0D%0AiaQxXFwgMCzjgIDjgIDlhbblroMgIFxlbmR7YXJyYXl9
       所以latex.action?latex=RVlfezF9PVxpbnRfezB9XnsxfcK3bnpee24tMX1kej1cZnJhY3tufXtuKzF9
       latex.action?latex=RFlfezF9PUVZX3sxfV57Mn0tRV57Mn1ZX3sxfT1caW50X3swfV57MX16XnsyfW56XntuLTF9ZHot%0D%0AXGxlZnQoXGZyYWN7bn17bisxfVxyaWdodCleezJ9PVxmcmFje259eyhuKzEpXnsyfShuKzIpfQ%3D%3D
       2. FY2(z)=P(Y2≤z)=1-P(Y2﹥z)
        =1-P(X1﹥z,X2﹥z,...Xn﹥z)
       =1-P(X1﹥z)P(X2﹥z)...P(Xn﹥z)
       =1-[1-P(X1≤z)][1-P(X2≤z)]...[1-P(Xn≤z)]
       =1-[1-F(z)]n
       latex.action?latex=PVxsZWZ0XHtcYmVnaW57YXJyYXl9e2N9MCzjgIDjgIB677mkMFxcIDEtKDEteilee259LOOAgOOA%0D%0AgDDiiaR677mkMVxcIDEs44CA44CAeuKJpTEgIFxlbmR7YXJyYXl9
       从而Y2的概率密度
       latex.action?latex=cF97WV97Mn19KHopPVxsZWZ0XHtcYmVnaW57YXJyYXl9e2N9bigxLXopXntuLTF9LOOAgOOAgDDv%0D%0AuaR64omkMVxcIDAs44CA44CA5YW25a6DICBcZW5ke2FycmF5fQ%3D%3D
       所以latex.action?latex=RVlfezJ9PVxpbnRfezB9XnsxfXpuKDEteilee24tMn1kej1cZnJhY3sxfXtuKzF9
       latex.action?latex=RFlfezJ9PUVZX3syfV57Mn0tRV57Mn1ZX3syfT1caW50X3swfV57MX16XnsyfcK3bigxLXopXntu%0D%0ALTF9ZHotXGZyYWN7MX17KG4rMSleezJ9fQ%3D%3D
       latex.action?latex=PVxmcmFje259eyhuKzEpXnsyfShuKzIpfQ%3D%3D