MATLAB中的高斯混合模型-经验方差协方差矩阵的计算 [英] Gaussian Mixture Model in MATLAB - Calculation of the Empirical Variance Covariance Matrix

问题描述

在协调有关高斯混合的一些基本理论结果以及Matlab中命令gmdistribution, random的输出时，我遇到了问题.

I am having issues in reconciling some basic theoretical results on Gaussian mixtures and the output of the commands gmdistribution, random in Matlab.

考虑权重为1/2,1/2的两个独立的3变量正态分布的混合.

Consider a mixture of two independent 3-variate normal distributions with weights 1/2,1/2.

第一分布A的特征在于均值和方差-协方差矩阵等于

The first distribution A is characterised by mean and variance-covariance matrix equal to

muA=[-1.4 3.2 -1.9]; %mean vector
rhoA=-0.5; %correlation among components in A
sigmaA=[1 rhoA rhoA; rhoA 1 rhoA; rhoA rhoA 1]; %variance-covariance matrix of A

第二个分布B的特征在于均值和方差-协方差矩阵等于

The second distribution B is characterised by mean and variance-covariance matrix equal to

muB=muB=[1.2 -1.6 1.5]; %mean vector
rhoB=0.3; %correlation among components in B
sigmaB=[1 rhoB rhoB; rhoB 1 rhoB; rhoB rhoB 1]; %variance-covariance matrix of B

让epsilon是作为混合物分布的3变量随机向量.我的计算表明，epsilon的期望值应为

Let epsilon be the 3-variate random vector distributed as the mixture. My calculations suggest that the expected value of epsilon should be

Mtheory=1/2*(muA+muB);

且方差-协方差矩阵应为

and the variance-covariance matrix should be

Vtheory=1/4*[2 rhoA+rhoB rhoA+rhoB; rhoA+rhoB 2 rhoA+rhoB; rhoA+rhoB rhoA+rhoB 2];

现在让我们来看一下Mtheory和Vtheory是否与通过从混合物中提取许多随机数而获得的经验矩一致.

Let's now try to see whether Mtheory and Vtheory coincide with the empirical moments that we get by drawing many random numbers from the mixture.

clear
rng default 

n=10^6; %number of draws 

w = ones(1,2)/2; %weights 

rhoA=-0.5; %correlation among components of A
rhoB=0.3; %correlation among components of B

muA=[-1.4 3.2 -1.9]; %mean vector of A
muB=[1.2 -1.6 1.5]; %mean vector of B
mu = [muA;muB];    
%Variance-covariance matrix for mixing
sigmaA=[1 rhoA rhoA; rhoA 1 rhoA; rhoA rhoA 1]; %variance-covariance matrix of A
sigmaB=[1 rhoB rhoB; rhoB 1 rhoB; rhoB rhoB 1]; %variance-covariance matrix of B 
sigma = cat(3,sigmaA,sigmaB);

obj = gmdistribution(mu, sigma,w);

%Draws
epsilon = random(obj, n); 

M=mean(epsilon);
V=cov(epsilon);
Mtheory=1/2*(muA+muB);
Vtheory=1/4*[2 rhoA+rhoB rhoA+rhoB; rhoA+rhoB 2 rhoA+rhoB; rhoA+rhoB rhoA+rhoB 2];

问题:M和Mtheory几乎重合. V和Vtheory完全不同.我究竟做错了什么?我应该做一些非常愚蠢的事情，但是我不知道在哪里.

Question: M and Mtheory almost coincide. V and Vtheory are completely different. What am I doing wrong? I should be doing something very silly but I don't see where.

推荐答案

计算协方差时，请注意数据不在中心.
此外，您的0.25系数是错误的.
这不是变量的缩放比例，而是选择.
计算应使用总差异法律/

When you calculate the Covariance pay attention that your data isn't centered.
Moreover, your 0.25 factor is wrong.
This is not a scaling of the variable but a selection.
The calculation should be done using the Law of Total Variance / Law of Total Covariance.
Where the "The Given Event" is the mixture index.

高斯混合物的协方差计算给出了计算示例.

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