作者jangwei (呆呆)
看板Statistics
标题Re: [问题]推论统计(完全不懂)
时间Sat Feb 23 00:15:21 2008
※ 引述《imgodya (许我一个PhD)》之铭言：
: Find E{R^2/(1-R^2)} under the null hypothesis, where R^2 is the sample squared
: multiple correlation coefficient. Hint: E{[χ^2]^k} = 2^kΓ(v/2 + k)/Γ(v/2),
: whereχ^2 is aχ^2-distributed random variable with v degrees of freedom.
: 翻了很多书 不太懂这个问题的定理
: 想请问要如何推导??
Under the null hypothesis,
R^2 R^2 *SST SSR SSR/ σ^2
------- = -------------- = ------- = ------------
1-R^2 (1-R^2) *SST SSE SSE/ σ^2
(k-1 is the number of explanatory variables)
Since SSR/ σ^2 and SSE/ σ^2 are inpependent χ^2-distributed
random variable with degrees of freedom k-1 and n-k respectively
under null hypothesis.
R^2 SSR/ σ^2 1
E(-------) = E(-----------) = E( SSR/ σ^2 * ------------)
1-R^2 SSE/ σ^2 SSE/ σ^2
1
=E( SSR/ σ^2 )* E ( ------------)
SSE/ σ^2
Compute these expectations with the hint
E{[χ^2]^k} = 2^kΓ(v/2 + k)/Γ(v/2)
SSR/ σ^2 ~ χ^2(k-1) E(SSR/ σ^2)= k-1
SSE/ σ^2 ~ χ^2(n-k)
1
E (
--------- ) = 2^(
-1)Γ( (n-k)/2
-1 )/Γ( (n-k)/2 )
SSE/ σ^2
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 203.73.183.95
※ 编辑: jangwei 来自: 203.73.183.95 (02/23 01:04)
1F：推 goshfju:还蛮详细的 02/23 03:01
2F：推 imgodya:感谢J大 实在是"就甘心"的 02/23 11:10