设二维随机变量（X,Y）的概率密度为：f(x,y)=12y^2,0<=y<=x<=1;f(x,y)=0,其他，求E(X),E(Y),E(X^2+Y^2)。
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解决时间 2021-01-27 22:42
- 提问者网友：王者佥
- 2021-01-27 07:35
设二维随机变量（X,Y）的概率密度为：f(x,y)=12y^2,0<=y<=x<=1;f(x,y)=0,其他，求E(X),E(Y),E(X^2+Y^2)。
最佳答案
- 五星知识达人网友：一秋
- 2021-01-27 07:53
EX=∫∫[0<=y<=x<=1] xf(x,y)dxdy=∫[0->1]∫[0->x] 12xy²dydx=4/5
EY=∫∫[0<=y<=x<=1] yf(x,y)dxdy=∫[0->1]∫[0->x] 12y³dydx=3/5
E(X²+Y²)=∫∫[0<=y<=x<=1] (x²+y²)f(x,y)dxdy=∫[0->1]∫[0->x] 12x²y²+12y^4dydx=16/15
EY=∫∫[0<=y<=x<=1] yf(x,y)dxdy=∫[0->1]∫[0->x] 12y³dydx=3/5
E(X²+Y²)=∫∫[0<=y<=x<=1] (x²+y²)f(x,y)dxdy=∫[0->1]∫[0->x] 12x²y²+12y^4dydx=16/15
全部回答
- 1楼网友：持酒劝斜阳
- 2021-01-27 08:40
EX=∫∫[0<=y<=x<=1] xf(x,y)dxdy=∫[0->1]∫[0->x] 12xy²dydx=4/5
EY=∫∫[0<=y<=x<=1] yf(x,y)dxdy=∫[0->1]∫[0->x] 12y³dydx=3/5
E(X²+Y²)=∫∫[0<=y<=x<=1] (x²+y²)f(x,y)dxdy=∫[0->1]∫[0->x] 12x²y²+12y^4dydx=16/15
EY=∫∫[0<=y<=x<=1] yf(x,y)dxdy=∫[0->1]∫[0->x] 12y³dydx=3/5
E(X²+Y²)=∫∫[0<=y<=x<=1] (x²+y²)f(x,y)dxdy=∫[0->1]∫[0->x] 12x²y²+12y^4dydx=16/15
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