超几何分布与二项分布的一个关系的证明!
收藏:
0
点赞数:
0
评论数:
0
1个回答

take Ln for 超几何分布C(M,m)*C(N-M,n-m)/C(N,n)

it changes to

ln C(M,m) + ln C(N-M,n-m) - ln C(N.n)

=ln M!- ln m!- ln(M-m)!+ ln(N-M)!- ln (n-m)!- ln (N-M-n+m)!

- ln N!+ ln n!+ ln (N-n)!

= ln (C(n,m)) + lnM!- ln(M-m)!+ ln (N-M)!- ln(N-M-n+m)!-ln N!+ln(N-n)!

use Stirling Formula :

for large number M; ln M!is approxiamtely equal to M(lnM-1)

plug this into each term in the equation above

like

ln M!= MlnM -M

ln(M-n)!=(M-n)ln(M-n)-(M-n)

and so on

taking m/M,n/N,nad (n-m)/(N-M) as small factors,you can get:

last equation

= ln(C(n,m))+mlnp +(n-m)ln(1-p)

=ln(C(n,m)*(p^m)*(1-p)^(n-m))

which means C(M,m)*C(N-M,n-m)/C(N,n)

will goes to C(n,m)*(p^m)*(1-p)^(n-m)

点赞数:
0
评论数:
0
关注公众号
一起学习,一起涨知识