• 2022-06-26
    设随机变量(x,y)服从二维正态分布,概率密度为f(x,y)=(1/2pi)*exp[-1/2*(x^2+y^2)],求E(x^2+y^2)
  • 易知随机变量X和Y相互独立且均服从N(0,1),所以E(X^2)=D(x)+[E(X)]^2=1+0^2=1同理E(Y^2)=1所以E(X^2+Y^2)=E(X^2)+E(Y^2)=2当然,本题也可以采用二重积分来做,相对比较麻烦.

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      【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )

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      ‌在环形区域[img=136x26]18030733be53638.png[/img]上, 绘制函数图形[img=129x27]18030733c6c9cd6.png[/img]​ A: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},Exclusions→Function[{x,y},0.5<x^2+y^2<2]] B: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},RegionFunction→Function[{x,y},0.5<x^2+y^2<2]] C: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},RegionFunction→Function[{x,y},2>x^2+y^2>0.5]] D: Plot3D[x^2+y^2,{y,-2,2},{x,-2,2},Exclusions→Function[{x,y},0.5<x^2+y^2<2]]

    • 2

      设随机变量\(X\)的概率密度函数\(\phi(x)=\frac{1}{\pi(1+x^2)}\),求随机变量\(Y=aX^2,(a < 0)\)的概率密度函数\(f(y)\)

    • 3

      已知x=2008,y=2006,求x-x÷{x^2-y^2/x^3+y^3&#91;(x-x^2+y^2/y)÷(1/x-1/y)&#93;}的值

    • 4

      ‏计算二重积分[img=159x48]18030731271aaff.png[/img], D 是单位圆盘[img=89x26]180307312f6708b.png[/img],应使用的语句是‍ A: Integrate[Sqrt[x^2+y^2 ], {x^2+y^2≤1}] B: Integrate[Sqrt[x^2+y^2 ]Boole[x^2+y^2≤1],{x,-1,1},{y,-1,1}] C: NIntegrate[Sqrt[x^2+y^2 ]Boole[x^2+y^2≤1],{x,-1,1},{y,-1,1}] D: Integrate[Sqrt[x^2+y^2 ],{x^2+y^2≤1,{x,-1,1},{y,-1,1}}]