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  • 【8】多元正态总体样本的极大似然估计

    【8】p-Normal Distribution's MLE

    考虑(p)元正态总体(Xsim N_p(mu,Sigma)),设(X_{(i)}=(x_{i1},dots,x_{ip})',)((i=1,dots,n))(p)元正态总体(X)的简单随机样本,则有观测数据库:

    [X= left( egin{array} {cccc} x_{11} & x_{12} & dots & x_{1p}\ x_{21} & x_{22} & dots & x_{2p}\ vdots & vdots & & vdots \ x_{n1} & x_{n2} & dots & x_{np}\ end{array} ight) ]

    是一个随机阵。

    引理(1)

    对于矩阵(A_{m imes n}),(A_{n imes m})有:(tr(AB)=tr(BA))

    [egin{align} tr(AB)=&sum_{i=1}^m(AB)_{ii}\ =&sum_{i=1}^m(sum_{j=1}^na_{ij}b_{ji})\ 同理:tr(BA)=&sum_{i=1}^n(sum_{j=1}^mb_{ij}a_{ji}) end{align} ]

    由于(Sigma)的可交换性,因此:

    [tr(AB)=sum_{i=1}^m(sum_{j=1}^na_{ij}b_{ji})=sum_{i=1}^n(sum_{j=1}^mb_{ij}a_{ji})=tr(BA) ]

    于是可以推广到多个矩阵相乘:

    [tr(prod_{i=1}^nA_i)=tr(A_nprod_{i=1}^{n-1}A_i) ]

    似然函数(L(mu,Sigma))

    对于样本(X_{(i)}=(x_{i1},dots,x_{ip})',)((i=1,dots,n)),其联合密度函数为:

    [f(x_{(i)})=frac1{(2pi)^{p/2}|Sigma|^{1/2}}expleft{-frac12(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu) ight} ]

    而由似然函数定义:

    [egin{align} L(mu,Sigma)=&prod_{i=1}^nf(x_{(i)})\ =&prod_{i=1}^nfrac1{(2pi)^{p/2}|Sigma|^{1/2}}expleft{-frac12(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu) ight}\ =&frac1{(2pi)^{np/2}|Sigma|^{n/2}}expleft{-frac12sum_{i=1}^n(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu) ight} end{align} ]

    由于:

    [(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu)=C_0(是一个数) ]

    所以:

    [(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu)=tr{(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu)}=tr(C_0) ]

    于是:

    [egin{align} L(mu,Sigma) =&frac1{(2pi)^{np/2}|Sigma|^{n/2}}expleft{-frac12sum_{i=1}^n(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu) ight}\ =&frac1{(2pi)^{np/2}|Sigma|^{n/2}}expleft{-frac12sum_{i=1}^ntr[(x_{(i)}-mu)'Sigma^{-1}(x_{(i)}-mu)] ight}\ (由引理)=&frac1{(2pi)^{np/2}|Sigma|^{n/2}}expleft{-frac12sum_{i=1}^ntr[Sigma^{-1}(x_{(i)}-mu)(x_{(i)}-mu)'] ight}\ =&frac1{(2pi)^{np/2}|Sigma|^{n/2}}expleft{tr(-frac12Sigma^{-1}sum_{i=1}^n(x_{(i)}-mu)(x_{(i)}-mu)') ight}\ end{align} ]

    其中:

    [egin{align} &sum_{i=1}^n(x_{(i)}-mu)(x_{(i)}-mu)'\ =&sum_{i=1}^n(x_{(i)}-overline{X}+overline{X}-mu)(x_{(i)}-overline{X}+overline{X}-mu)'\ =&sum_{i=1}^n(x_{(i)}-overline{X})(x_{(i)}-overline{X})'+n(overline{X}-mu)(overline{X}-mu)'\ =&A+n(overline{X}-mu)(overline{X}-mu)' end{align} ]

    带回似然函数可得:

    [egin{align} L(mu,Sigma)=&frac1{(2pi)^{np/2}|Sigma|^{n/2}}etrleft{-frac12Sigma^{-1}sum_{i=1}^n(x_{(i)}-mu)(x_{(i)}-mu)' ight}\ =&frac1{(2pi)^{np/2}|Sigma|^{n/2}}etrleft{-frac12Sigma^{-1}(A+n(overline{X}-mu)(overline{X}-mu)') ight}\ (两边求对数)ln{L(mu,Sigma)}=& -frac{np}2ln{(2pi)}-frac{n}2ln{|Sigma|}-frac12trleft[Sigma^{-1}(A+n(overline{X}-mu)(overline{X}-mu)') ight]\ =& -frac{np}2ln{(2pi)}-frac{n}2ln{|Sigma|}-frac12trleft[Sigma^{-1}A+Sigma^{-1}n(overline{X}-mu)(overline{X}-mu)' ight]\ =& -frac{np}2ln{(2pi)}-frac{n}2ln{|Sigma|}-frac12tr[Sigma^{-1}A]-frac12tr[Sigma^{-1}n(overline{X}-mu)(overline{X}-mu)']\ (仅当mu=overline{X}时取等号)leq&-frac{np}2ln{(2pi)}-frac{n}2ln{|Sigma|}-frac12tr[Sigma^{-1}A]\ end{align} ]

    求似然函数极大值:

    方法一
    • (引理二)

    (B)(p)阶正定矩阵,则:(tr(B)-ln{|B|}geq p).

    [ln{|B|}=sum_{i=1}^pln{lambda_i}=sum_{i=1}^pln{(1+lambda_i-1)}leqsum_{i=1}^p(lambda_i-1)=tr(B)-p ]

    则有:

    [tr{\,(B)}-ln{|B|}ge p ag{引理,证毕} ]

    于是观察(ln{L(mu,Sigma)})得:

    [egin{align} ln{L(mu,Sigma)}=& -frac{np}2ln{(2pi)}-frac{n}2ln{|Sigma|}-frac12tr[Sigma^{-1}A]\ =&-frac{np}2ln{(2pi)}-frac{n}2left[ln{|Sigma|}+tr[Sigma^{-1}frac{A}n] ight]\ =&-frac{np}2ln{(2pi)}-frac{n}2left[tr[Sigma^{-1}frac{A}n]-ln{|Sigma^{-1}frac An|}+ln{frac{A}{n}} ight]\ leq&-frac{np}2ln{(2pi)}-frac n2(p+ln{|frac An|}) end{align} ]

    以上不等式的等号当且仅当(Sigma^{-1}frac{A}n=I_p)时成立,于是(Sigma=frac{A}n),则有:

    [ln{L(overline{X},frac1nA)}=max_{overline{X},Sigma>0}ln{L(overline{X},Sigma)}=-frac{np}2(1+ln(2pi))-frac n2ln{|frac{A}n|} ]

    其中

    [overline{X}=frac1nsum_{i=1}^nX_{(i)}\ Sigma=frac1nsum_{i=1}^n(x_{(i)}-overline{X})(x_{(i)}-overline{X})' ]

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  • 原文地址:https://www.cnblogs.com/rrrrraulista/p/12369093.html
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