使用 EM 算法，我想在给定数据集上训练具有四个分量的高斯混合模型。该集合是三维的，包含 300 个样本。
问题是在大约 6 轮 EM 算法之后，协方差矩阵 sigma 根据 matlab 变得接近奇异（rank(sigma) = 2而不是 3）。这反过来会导致不希望的结果，例如评估高斯分布的复值gm(k,i)。
此外，我使用高斯的日志来解决下溢问题 - 请参阅 E-step。我不确定这是否正确，是否必须将责任 p(w_k | x^(i), theta) 的 exp 带到其他地方？
你能告诉我到目前为止我的 EM 算法的实现是否正确吗？以及如何解释接近奇异协方差 sigma 的问题？
这是我对 EM 算法的实现：
首先，我使用 kmeans初始化了分量的均值和协方差：
load('data1.mat');
X = Data'; % 300x3 data set
D = size(X,2); % dimension
N = size(X,1); % number of samples
K = 4; % number of Gaussian Mixture components
% Initialization
p = [0.2, 0.3, 0.2, 0.3]; % arbitrary pi
[idx,mu] = kmeans(X,K); % initial means of the components
% compute the covariance of the components
sigma = zeros(D,D,K);
for k = 1:K
sigma(:,:,k) = cov(X(idx==k,:));
end
对于E-step，我使用以下公式来计算责任。
w_k 是 k 个高斯分量。
x^(i) 是单个数据点（样本）
theta 代表高斯混合模型的参数：mu, Sigma, pi。
下面是对应的代码：
% variables for convergence
converged = 0;
prevLoglikelihood = Inf;
prevMu = mu;
prevSigma = sigma;
prevPi = p;
round = 0;
while (converged ~= 1)
round = round +1
gm = zeros(K,N); % gaussian component in the nominator
sumGM = zeros(N,1); % denominator of responsibilities
% E-step: Evaluate the responsibilities using the current parameters
% compute the nominator and denominator of the responsibilities
for k = 1:K
for i = 1:N
Xmu = X-mu;
% I am using log to prevent underflow of the gaussian distribution (exp("small value"))
logPdf = log(1/sqrt(det(sigma(:,:,k))*(2*pi)^D)) + (-0.5*Xmu*(sigma(:,:,k)\Xmu'));
gm(k,i) = log(p(k)) * logPdf;
sumGM(i) = sumGM(i) + gm(k,i);
end
end
% calculate responsibilities
res = zeros(K,N); % responsibilities
Nk = zeros(4,1);
for k = 1:K
for i = 1:N
% I tried to use the exp(gm(k,i)/sumGM(i)) to compute res but this leads to sum(pi) > 1.
res(k,i) = gm(k,i)/sumGM(i);
end
Nk(k) = sum(res(k,:));
end
Nk(k)是使用 M 步中给出的公式计算的，并在 M 步中用于计算新的概率p(k)。
M步
% M-step: Re-estimate the parameters using the current responsibilities
for k = 1:K
for i = 1:N
mu(k,:) = mu(k,:) + res(k,i).*X(k,:);
sigma(:,:,k) = sigma(:,:,k) + res(k,i).*(X(k,:)-mu(k,:))*(X(k,:)-mu(k,:))';
end
mu(k,:) = mu(k,:)./Nk(k);
sigma(:,:,k) = sigma(:,:,k)./Nk(k);
p(k) = Nk(k)/N;
end
现在为了检查收敛性，使用以下公式计算对数似然：
% Evaluate the log-likelihood and check for convergence of either
% the parameters or the log-likelihood. If not converged, go to E-step.
loglikelihood = 0;
for i = 1:N
loglikelihood = loglikelihood + log(sum(gm(:,i)));
end
% Check for convergence of parameters
errorLoglikelihood = abs(loglikelihood-prevLoglikelihood);
if (errorLoglikelihood <= eps)
converged = 1;
end
errorMu = abs(mu(:)-prevMu(:));
errorSigma = abs(sigma(:)-prevSigma(:));
errorPi = abs(p(:)-prevPi(:));
if (all(errorMu <= eps) && all(errorSigma <= eps) && all(errorPi <= eps))
converged = 1;
end
prevLoglikelihood = loglikelihood;
prevMu = mu;
prevSigma = sigma;
prevPi = p;
end % while
我的高斯混合模型 EM 算法的 Matlab 实现有问题吗？
以前的烦恼：
问题是我无法使用对数似然检查收敛性，因为它是-Inf. 这是在评估责任公式中的高斯时四舍五入的零值导致的（参见 E-step）。
你能告诉我到目前为止我的 EM 算法的实现是否正确吗？以及如何解决舍入零值的问题？
这是我对 EM 算法的实现：
首先，我使用 kmeans初始化了分量的均值和协方差：
load('data1.mat');
X = Data'; % 300x3 data set
D = size(X,2); % dimension
N = size(X,1); % number of samples
K = 4; % number of Gaussian Mixture components
% Initialization
p = [0.2, 0.3, 0.2, 0.3]; % arbitrary pi
[idx,mu] = kmeans(X,K); % initial means of the components
% compute the covariance of the components
sigma = zeros(D,D,K);
for k = 1:K
sigma(:,:,k) = cov(X(idx==k,:));
end
对于E-step，我使用以下公式来计算责任
下面是对应的代码：
% variables for convergence
converged = 0;
prevLoglikelihood = Inf;
prevMu = mu;
prevSigma = sigma;
prevPi = p;
round = 0;
while (converged ~= 1)
round = round +1
gm = zeros(K,N); % gaussian component in the nominator -
% some values evaluate to zero
sumGM = zeros(N,1); % denominator of responsibilities
% E-step: Evaluate the responsibilities using the current parameters
% compute the nominator and denominator of the responsibilities
for k = 1:K
for i = 1:N
% HERE values evalute to zero e.g. exp(-746.6228) = -Inf
gm(k,i) = p(k)/sqrt(det(sigma(:,:,k))*(2*pi)^D)*exp(-0.5*(X(i,:)-mu(k,:))*inv(sigma(:,:,k))*(X(i,:)-mu(k,:))');
sumGM(i) = sumGM(i) + gm(k,i);
end
end
% calculate responsibilities
res = zeros(K,N); % responsibilities
Nk = zeros(4,1);
for k = 1:K
for i = 1:N
res(k,i) = gm(k,i)/sumGM(i);
end
Nk(k) = sum(res(k,:));
end
Nk(k)使用 M 步中给出的公式计算。
M步
% M-step: Re-estimate the parameters using the current responsibilities
mu = zeros(K,3);
for k = 1:K
for i = 1:N
mu(k,:) = mu(k,:) + res(k,i).*X(k,:);
sigma(:,:,k) = sigma(:,:,k) + res(k,i).*(X(k,:)-mu(k,:))*(X(k,:)-mu(k,:))';
end
mu(k,:) = mu(k,:)./Nk(k);
sigma(:,:,k) = sigma(:,:,k)./Nk(k);
p(k) = Nk(k)/N;
end
现在为了检查收敛性，使用以下公式计算对数似然：
% Evaluate the log-likelihood and check for convergence of either
% the parameters or the log-likelihood. If not converged, go to E-step.
loglikelihood = 0;
for i = 1:N
loglikelihood = loglikelihood + log(sum(gm(:,i)));
end
% Check for convergence of parameters
errorLoglikelihood = abs(loglikelihood-prevLoglikelihood);
if (errorLoglikelihood <= eps)
converged = 1;
end
errorMu = abs(mu(:)-prevMu(:));
errorSigma = abs(sigma(:)-prevSigma(:));
errorPi = abs(p(:)-prevPi(:));
if (all(errorMu <= eps) && all(errorSigma <= eps) && all(errorPi <= eps))
converged = 1;
end
prevLoglikelihood = loglikelihood;
prevMu = mu;
prevSigma = sigma;
prevPi = p;
end % while
第一轮之后loglikelihood大约是 700。在第二轮中，这是-Inf因为gm(k,i)E-step 中的某些值为零。因此，对数显然是负无穷大。
零值也导致sumGM等于零，因此导致mu和sigma矩阵内的所有 NaN 条目。
我怎么解决这个问题？你能告诉我我的实现是否有问题吗？是否可以通过某种方式提高 Matlab 的精度来解决？
编辑：
我为 gm(k,i) 中的 exp() 项添加了缩放比例。不幸的是，这并没有多大帮助。经过几轮后，我仍然遇到下溢问题。
scale = zeros(N,D);
for i = 1:N
max = 0;
for k = 1:K
Xmu = X(i,:)-mu(k,:);
if (norm(scale(i,:) - Xmu) > max)
max = norm(scale(i,:) - Xmu);
scale(i,:) = Xmu;
end
end
end
for k = 1:K
for i = 1:N
Xmu = X(i,:)-mu(k,:);
% scale gm to prevent underflow
Xmu = Xmu - scale(i,:);
gm(k,i) = p(k)/sqrt(det(sigma(:,:,k))*(2*pi)^D)*exp(-0.5*Xmu*inv(sigma(:,:,k))*Xmu');
sumGM(i) = sumGM(i) + gm(k,i);
end
end
此外，我注意到，与在 M 步中计算均值的后续轮次相比，kmeans 初始化的均值完全不同。
意思是：
mu = 13.500000000000000 0.026602138870044 0.062415945993735
88.500000000000000 -0.009869960132085 -0.075177888210981
39.000000000000000 -0.042569305020309 0.043402772876513
64.000000000000000 -0.024519281362918 -0.012586980924762
M步后：
round = 2
mu = 1.000000000000000 0.077230046948357 0.024498886414254
2.000000000000000 0.074260118474053 0.026484346404660
3.000000000000002 0.070944016105476 0.029043085983168
4.000000000000000 0.067613431480832 0.031641849205021
在接下来的回合mu中根本没有改变。它与第 2 轮相同。
我猜这是因为 gm(k,i) 中的下溢引起的？要么我的缩放实现不正确，要么算法的整个实现在某处出错:(
编辑 2
经过四轮，我得到了NaN价值，并更详细地研究了 gm。仅查看一个样本（并且没有 0.5 因子），gm在所有组件中都变为零。放入matlab gm(:,1) = [0 0 0 0]。这反过来导致 sumGM 等于零 -> NaN，因为 I 除以零。我已经给出了更多的细节
round = 1
mu = 62.0000 -0.0298 -0.0078
37.0000 -0.0396 0.0481
87.5000 -0.0083 -0.0728
12.5000 0.0303 0.0614
gm(:,1) = [11.7488, 0.0000, 0.0000, 0.0000]
round = 2
mu = 1.0000 0.0772 0.0245
2.0000 0.0743 0.0265
3.0000 0.0709 0.0290
4.0000 0.0676 0.0316
gm(:,1) = [0.0000, 0.0000, 0.0000, 0.3128]
round = 3
mu = 1.0000 0.0772 0.0245
2.0000 0.0743 0.0265
3.0000 0.0709 0.0290
4.0000 0.0676 0.0316
gm(:,1) = [0, 0, 0.0000, 0.2867]
round = 4
mu = 1.0000 0.0772 0.0245
NaN NaN NaN
3.0000 0.0709 0.0290
4.0000 0.0676 0.0316
gm(:,1) = 1.0e-105 * [0, NaN, 0, 0.5375]
首先，与kmeans的初始化相比，手段似乎没有改变并且完全不同。
并且每个样本（不仅仅是像这里的第一个样本）根据 的输出仅对应一个高斯分量gm(:,1)。样本不应该在每个高斯分量中“部分分布”吗？
编辑3：
所以我猜 mu 没有改变的问题是 M-step: 中的第一行mu = zeros(K,3);。
为了解决下溢问题，我目前正在尝试使用高斯日志：
function logPdf = logmvnpdf(X, mu, sigma, D)
Xmu = X-mu;
logPdf = log(1/sqrt(det(sigma)*(2*pi)^D)) + (-0.5*Xmu*inv(sigma)*Xmu');
end
新问题是协方差矩阵 sigma。Matlab 声称：警告：矩阵接近奇异或严重缩放。结果可能不准确。
6 轮后，我得到 gm（高斯分布）的虚值。
更新后的 E-Step 现在看起来像这样：
gm = zeros(K,N); % gaussian component in the nominator
sumGM = zeros(N,1); % denominator of responsibilities
for k = 1:K
for i = 1:N
%gm(k,i) = p(k)/sqrt(det(sigma(:,:,k))*(2*pi)^D)*exp(-0.5*Xmu*inv(sigma(:,:,k))*Xmu');
%gm(k,i) = p(k)*mvnpdf(X(i,:),mu(k,:),sigma(:,:,k));
gm(k,i) = log(p(k)) + logmvnpdf(X(i,:), mu(k,:), sigma(:,:,k), D);
sumGM(i) = sumGM(i) + gm(k,i);
end
end