x***x
发帖数: 3401 | 1 If U is a random variable from uniform (0, 1) and V is random variable
from uniform (-1, 1); also, U^2+V^2<=1. It is said X=V/U is a cauchy
distribution. I tried to prove this by let X=V/U and Y=U. Using bivariate
transformation, I get f(X, Y)=Y/2. Also, I get the constraint
0
which has the cauchy form, but not exactly a cauchy distribution. The Pie
is missing, but i can't figure out where i went wrong. Any help will be
apprec | s*****n
发帖数: 2174 | 2 你是f(x,y)算错了.
f(x,y) = f(u,v) * Jacob(u,v; x,y)
其中 u = y, v = x*y
f(u,v) = 2/pi 你很可能是这里错了, 搞成 1/2 了.
Jacob(u,v; x,y) = y
f(x,y) = 2y / pi 而不是 y / 2
【在 x***x 的大作中提到】
: If U is a random variable from uniform (0, 1) and V is random variable
: from uniform (-1, 1); also, U^2+V^2<=1. It is said X=V/U is a cauchy
: distribution. I tried to prove this by let X=V/U and Y=U. Using bivariate
: transformation, I get f(X, Y)=Y/2. Also, I get the constraint
: 0: which has the cauchy form, but not exactly a cauchy distribution. The Pie
: is missing, but i can't figure out where i went wrong. Any help will be
: apprec
| x***x
发帖数: 3401 | 3 谢谢 你是对的 我把f(u v)算错了, 想成U V是完全独立了.
谢谢了哟.
【在 s*****n 的大作中提到】
: 你是f(x,y)算错了.
: f(x,y) = f(u,v) * Jacob(u,v; x,y)
: 其中 u = y, v = x*y
: f(u,v) = 2/pi 你很可能是这里错了, 搞成 1/2 了.
: Jacob(u,v; x,y) = y
: f(x,y) = 2y / pi 而不是 y / 2
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