Hello, thank you for reading this question. I have a question about using matlab’s (maybe undocumented) function predictExactWithCov.
I have found in this forum that if I have a RegressionGP gprmdl, set of test points xTest, then by using
[postMeanGPR,postCovGPR,confInt] = predictExactWithCov(gprmdl.Impl,xTest,0.05);
I can obtain an posterior mean of gprmdl at xTest (postMeanGPR), posterior covariance matrix at xTest (postCovGPR), and 95% confidence interval based on covariance matrix (confInt).
On the other hand, I can calculate posterior mean and variance using
(*)
where zero prior is assumed, is a kernel, and so on.
However, using the kernel information from gprmdl like
sig_f = gprmdl.KernelInformation.KernelParameters(12);
sig_l = gprmdl.KernelInformation.KernelParameters(1:11)’;
lambda = gprmdl.Sigma;
and calculating posterior mean and covariance using equation (*), I can obtain a same mean as postMeanGPR but posterior variance was quite different with postCovGPR’s diagonal elements.
Can anyone explain why does such a discrepency occur?
Thanks.
PS: the overall code is attached below
clear;
clc;

wineTable = readtable(‘winequality-red.csv’); % wine data from cortez et al 2009., file attached
% 1st-11th columns are inputs and 12th column is output
wineData = table2array(wineTable);
len_tot = length(wineData);

len_train = 1000;
len_test = len_tot – len_train;
DataTrain = zeros(len_train,12);
DataTest = zeros(len_test,12);
NN = 1;

rng(NN);
%% STEP 1. Load data and separate it into training and test sets
p = sort(randperm(len_tot,len_test));
jj = 1; kk = 1;
for ii = 1:len_tot
if ismember(ii,p’,’rows’)
DataTest(jj,:) = wineData(ii,:);
jj = jj + 1;
else
DataTrain(kk,:) = wineData(ii,:);
kk = kk + 1;
end
end

%% STEP 2. Calculate GP posterior mean and covariance using predictExactWithCov
xTrain = DataTrain(:,1:11);
yTrain = DataTrain(:,12);
xTest = DataTest(:,1:11);
yTest = DataTest(:,12);
gprmdl = fitrgp(xTrain,yTrain,’KernelFunction’,’ardsquaredexponential’,’BasisFunction’,’none’);
[postMeanGPR,postCovGPR,confInt] = predictExactWithCov(gprmdl.Impl,xTest,0.05);

%% STEP 3. Calculate GP posterior mean and variance using equation (*)
Gram = Kmat(xTrain,sig_f,sig_l);
sig_f = gprmdl.KernelInformation.KernelParameters(12);
sig_l = gprmdl.KernelInformation.KernelParameters(1:11)’;
lambda = gprmdl.Sigma;
KinvY = (Gram + lambda^2*eye(len_train))yTrain;
postMeanGPR2 = zeros(len_test,1);
postVarGPR2 = postMeanGPR2;
for ii = 1:len_test
postMeanGPR2(ii) = kCol(xTest(ii,:),xTrain,sig_f,sig_l)’*KinvY;
postVarGPR2(ii) = SE(xTest(ii,:),xTest(ii,:),sig_f,sig_l) – kCol(xTest(ii,:),xTrain,sig_f,sig_l)’*((Gram + lambda^2*eye(len_train))kCol(xTest(ii,:),xTrain,sig_f,sig_l));
end

function val = SE(x1, x2, sig_f, sig_l)

% x1, x2 in R^d, row vectors
% hyperparameter sig_l: length scale, length(sig_l) = d, sig_f > 0

Lambda = diag(sig_l);

val = sig_f^2 * exp(-0.5*( (x1 – x2) /Lambda^2 * (x1 – x2)’ ) );
end

function k = kCol(xTest, xTrain, sig_f, sig_l)

% xTest: Test point, row vector
% xTrain: Training points, each row corresponds to one training point

N = length(xTrain);
k = zeros(N,1);
for ii = 1:N
k(ii) = SE(xTest, xTrain(ii,:), sig_f, sig_l);
end
end

function K = Kmat(X, sig_f, sig_l)

% each row of X corresponds to one data point of Gram matrix K

n = length(X);
K = zeros(n,n);

for ii = 1:n
for jj = 1:n
K(ii,jj) = SE(X(ii,:),X(jj,:), sig_f, sig_l);
end
end
endHello, thank you for reading this question. I have a question about using matlab’s (maybe undocumented) function predictExactWithCov.
I have found in this forum that if I have a RegressionGP gprmdl, set of test points xTest, then by using
[postMeanGPR,postCovGPR,confInt] = predictExactWithCov(gprmdl.Impl,xTest,0.05);
I can obtain an posterior mean of gprmdl at xTest (postMeanGPR), posterior covariance matrix at xTest (postCovGPR), and 95% confidence interval based on covariance matrix (confInt).
On the other hand, I can calculate posterior mean and variance using
(*)
where zero prior is assumed, is a kernel, and so on.
However, using the kernel information from gprmdl like
sig_f = gprmdl.KernelInformation.KernelParameters(12);
sig_l = gprmdl.KernelInformation.KernelParameters(1:11)’;
lambda = gprmdl.Sigma;
and calculating posterior mean and covariance using equation (*), I can obtain a same mean as postMeanGPR but posterior variance was quite different with postCovGPR’s diagonal elements.
Can anyone explain why does such a discrepency occur?
Thanks.
PS: the overall code is attached below
clear;
clc;

wineTable = readtable(‘winequality-red.csv’); % wine data from cortez et al 2009., file attached
% 1st-11th columns are inputs and 12th column is output
wineData = table2array(wineTable);
len_tot = length(wineData);

len_train = 1000;
len_test = len_tot – len_train;
DataTrain = zeros(len_train,12);
DataTest = zeros(len_test,12);
NN = 1;

rng(NN);
%% STEP 1. Load data and separate it into training and test sets
p = sort(randperm(len_tot,len_test));
jj = 1; kk = 1;
for ii = 1:len_tot
if ismember(ii,p’,’rows’)
DataTest(jj,:) = wineData(ii,:);
jj = jj + 1;
else
DataTrain(kk,:) = wineData(ii,:);
kk = kk + 1;
end
end

%% STEP 2. Calculate GP posterior mean and covariance using predictExactWithCov
xTrain = DataTrain(:,1:11);
yTrain = DataTrain(:,12);
xTest = DataTest(:,1:11);
yTest = DataTest(:,12);
gprmdl = fitrgp(xTrain,yTrain,’KernelFunction’,’ardsquaredexponential’,’BasisFunction’,’none’);
[postMeanGPR,postCovGPR,confInt] = predictExactWithCov(gprmdl.Impl,xTest,0.05);

%% STEP 3. Calculate GP posterior mean and variance using equation (*)
Gram = Kmat(xTrain,sig_f,sig_l);
sig_f = gprmdl.KernelInformation.KernelParameters(12);
sig_l = gprmdl.KernelInformation.KernelParameters(1:11)’;
lambda = gprmdl.Sigma;
KinvY = (Gram + lambda^2*eye(len_train))yTrain;
postMeanGPR2 = zeros(len_test,1);
postVarGPR2 = postMeanGPR2;
for ii = 1:len_test
postMeanGPR2(ii) = kCol(xTest(ii,:),xTrain,sig_f,sig_l)’*KinvY;
postVarGPR2(ii) = SE(xTest(ii,:),xTest(ii,:),sig_f,sig_l) – kCol(xTest(ii,:),xTrain,sig_f,sig_l)’*((Gram + lambda^2*eye(len_train))kCol(xTest(ii,:),xTrain,sig_f,sig_l));
end

function val = SE(x1, x2, sig_f, sig_l)

% x1, x2 in R^d, row vectors
% hyperparameter sig_l: length scale, length(sig_l) = d, sig_f > 0

Lambda = diag(sig_l);

val = sig_f^2 * exp(-0.5*( (x1 – x2) /Lambda^2 * (x1 – x2)’ ) );
end

function k = kCol(xTest, xTrain, sig_f, sig_l)

% xTest: Test point, row vector
% xTrain: Training points, each row corresponds to one training point

N = length(xTrain);
k = zeros(N,1);
for ii = 1:N
k(ii) = SE(xTest, xTrain(ii,:), sig_f, sig_l);
end
end

function K = Kmat(X, sig_f, sig_l)

% each row of X corresponds to one data point of Gram matrix K

n = length(X);
K = zeros(n,n);

for ii = 1:n
for jj = 1:n
K(ii,jj) = SE(X(ii,:),X(jj,:), sig_f, sig_l);
end
end
end Hello, thank you for reading this question. I have a question about using matlab’s (maybe undocumented) function predictExactWithCov.
I have found in this forum that if I have a RegressionGP gprmdl, set of test points xTest, then by using
[postMeanGPR,postCovGPR,confInt] = predictExactWithCov(gprmdl.Impl,xTest,0.05);
I can obtain an posterior mean of gprmdl at xTest (postMeanGPR), posterior covariance matrix at xTest (postCovGPR), and 95% confidence interval based on covariance matrix (confInt).
On the other hand, I can calculate posterior mean and variance using
(*)
where zero prior is assumed, is a kernel, and so on.
However, using the kernel information from gprmdl like
sig_f = gprmdl.KernelInformation.KernelParameters(12);
sig_l = gprmdl.KernelInformation.KernelParameters(1:11)’;
lambda = gprmdl.Sigma;
and calculating posterior mean and covariance using equation (*), I can obtain a same mean as postMeanGPR but posterior variance was quite different with postCovGPR’s diagonal elements.
Can anyone explain why does such a discrepency occur?
Thanks.
PS: the overall code is attached below
clear;
clc;

wineTable = readtable(‘winequality-red.csv’); % wine data from cortez et al 2009., file attached
% 1st-11th columns are inputs and 12th column is output
wineData = table2array(wineTable);
len_tot = length(wineData);

len_train = 1000;
len_test = len_tot – len_train;
DataTrain = zeros(len_train,12);
DataTest = zeros(len_test,12);
NN = 1;

rng(NN);
%% STEP 1. Load data and separate it into training and test sets
p = sort(randperm(len_tot,len_test));
jj = 1; kk = 1;
for ii = 1:len_tot
if ismember(ii,p’,’rows’)
DataTest(jj,:) = wineData(ii,:);
jj = jj + 1;
else
DataTrain(kk,:) = wineData(ii,:);
kk = kk + 1;
end
end

%% STEP 2. Calculate GP posterior mean and covariance using predictExactWithCov
xTrain = DataTrain(:,1:11);
yTrain = DataTrain(:,12);
xTest = DataTest(:,1:11);
yTest = DataTest(:,12);
gprmdl = fitrgp(xTrain,yTrain,’KernelFunction’,’ardsquaredexponential’,’BasisFunction’,’none’);
[postMeanGPR,postCovGPR,confInt] = predictExactWithCov(gprmdl.Impl,xTest,0.05);

%% STEP 3. Calculate GP posterior mean and variance using equation (*)
Gram = Kmat(xTrain,sig_f,sig_l);
sig_f = gprmdl.KernelInformation.KernelParameters(12);
sig_l = gprmdl.KernelInformation.KernelParameters(1:11)’;
lambda = gprmdl.Sigma;
KinvY = (Gram + lambda^2*eye(len_train))yTrain;
postMeanGPR2 = zeros(len_test,1);
postVarGPR2 = postMeanGPR2;
for ii = 1:len_test
postMeanGPR2(ii) = kCol(xTest(ii,:),xTrain,sig_f,sig_l)’*KinvY;
postVarGPR2(ii) = SE(xTest(ii,:),xTest(ii,:),sig_f,sig_l) – kCol(xTest(ii,:),xTrain,sig_f,sig_l)’*((Gram + lambda^2*eye(len_train))kCol(xTest(ii,:),xTrain,sig_f,sig_l));
end

function val = SE(x1, x2, sig_f, sig_l)

% x1, x2 in R^d, row vectors
% hyperparameter sig_l: length scale, length(sig_l) = d, sig_f > 0

Lambda = diag(sig_l);

val = sig_f^2 * exp(-0.5*( (x1 – x2) /Lambda^2 * (x1 – x2)’ ) );
end

function k = kCol(xTest, xTrain, sig_f, sig_l)

% xTest: Test point, row vector
% xTrain: Training points, each row corresponds to one training point

N = length(xTrain);
k = zeros(N,1);
for ii = 1:N
k(ii) = SE(xTest, xTrain(ii,:), sig_f, sig_l);
end
end

function K = Kmat(X, sig_f, sig_l)

% each row of X corresponds to one data point of Gram matrix K

n = length(X);
K = zeros(n,n);

for ii = 1:n
for jj = 1:n
K(ii,jj) = SE(X(ii,:),X(jj,:), sig_f, sig_l);
end
end
end matlab, gaussian process regression, posterior covariance matrix MATLAB Answers — New Questions

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