设二维随机变量(X,Y)的分布函数为F(X,Y)=a(b+arctanx)(c+arctan2y),-∞<x<+∞,-∞<y<+∞(1)求常数a,b,c(2)(X,Y)的概率密度由分布函数性质知:F(+∞,+∞)=a(b+π/2)(c+π/2)=1F(x,-∞)=a(b+arctanx)(c-π/2)=0F(-∞,y
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 09:15:22
![设二维随机变量(X,Y)的分布函数为F(X,Y)=a(b+arctanx)(c+arctan2y),-∞<x<+∞,-∞<y<+∞(1)求常数a,b,c(2)(X,Y)的概率密度由分布函数性质知:F(+∞,+∞)=a(b+π/2)(c+π/2)=1F(x,-∞)=a(b+arctanx)(c-π/2)=0F(-∞,y](/uploads/image/z/2340893-29-3.jpg?t=%E8%AE%BE%E4%BA%8C%E7%BB%B4%E9%9A%8F%E6%9C%BA%E5%8F%98%E9%87%8F%EF%BC%88X%2CY%EF%BC%89%E7%9A%84%E5%88%86%E5%B8%83%E5%87%BD%E6%95%B0%E4%B8%BAF%28X%2CY%29%3Da%28b%2Barctanx%29%28c%2Barctan2y%29%2C-%E2%88%9E%EF%BC%9Cx%EF%BC%9C%2B%E2%88%9E%2C-%E2%88%9E%EF%BC%9Cy%EF%BC%9C%2B%E2%88%9E%281%29%E6%B1%82%E5%B8%B8%E6%95%B0a%2Cb%2Cc%282%29%28X%2CY%29%E7%9A%84%E6%A6%82%E7%8E%87%E5%AF%86%E5%BA%A6%E7%94%B1%E5%88%86%E5%B8%83%E5%87%BD%E6%95%B0%E6%80%A7%E8%B4%A8%E7%9F%A5%EF%BC%9AF%28%2B%E2%88%9E%2C%2B%E2%88%9E%29%3Da%28b%2B%CF%80%2F2%29%28c%2B%CF%80%2F2%29%3D1F%28x%2C-%E2%88%9E%29%3Da%28b%2Barctanx%29%28c-%CF%80%2F2%29%3D0F%28-%E2%88%9E%2Cy)
设二维随机变量(X,Y)的分布函数为F(X,Y)=a(b+arctanx)(c+arctan2y),-∞<x<+∞,-∞<y<+∞(1)求常数a,b,c(2)(X,Y)的概率密度由分布函数性质知:F(+∞,+∞)=a(b+π/2)(c+π/2)=1F(x,-∞)=a(b+arctanx)(c-π/2)=0F(-∞,y
设二维随机变量(X,Y)的分布函数为F(X,Y)=a(b+arctanx)(c+arctan2y),-∞<x<+∞,-∞<y<+∞
(1)求常数a,b,c
(2)(X,Y)的概率密度
由分布函数性质知:F(+∞,+∞)=a(b+π/2)(c+π/2)=1
F(x,-∞)=a(b+arctanx)(c-π/2)=0
F(-∞,y)=a(b-π/2)(c+arctan2y)=0
从上面第二式得c=π/2,从第三式得b=π/2,再得a=1/π^2..
我不知道题中的π/2哪来的?怎么知道的?算的思路是什么?
(2)F(x,y)=1/π^2(π/2+arctanx)(π/2+arctan2y)
从而概率密度为f(x,y)=2/π^2(1+x^2)(1+4y^2)
结果的算来是求导数吗?
设二维随机变量(X,Y)的分布函数为F(X,Y)=a(b+arctanx)(c+arctan2y),-∞<x<+∞,-∞<y<+∞(1)求常数a,b,c(2)(X,Y)的概率密度由分布函数性质知:F(+∞,+∞)=a(b+π/2)(c+π/2)=1F(x,-∞)=a(b+arctanx)(c-π/2)=0F(-∞,y
①首先,我们可以认为tan(π/2)=+∞,这是自然的,因此可以说arctan(+∞)=π/2
第一问的π/2就是这么来的,把x、y都带成+∞,然后分布函数的意思就是x
WF