Using canonical correlation analysis with leave one out prediction

Question

I am trying to use canonical correlation to predict a set of held out x variables from a multivariable set of X and Y data. In this particular case I am only interested in X. In the real data X is a behavioural variable and Y is a biomarker.

My approach so far involves:

1. running the training CCA
2. apply the CCA weights to the held out Y data (V = Y * B)
3. using linear regression to estimate the unknown U value (assumes V and U are linearly related, as CCA should find)
4. applying the same CCA weights backwards to find X (X = U / A)

I've pasted some generalized code below using the fisheriris dataset in matlab. It produces high correlations between predicted and real values but I'm unsure if this is correct.

load fisheriris
X = meas(:,1:2);
Y = meas(:,3:4);

%% CCA LOO
N = length(X); %sample size

for subj = 1:N

    trainN = 1:N;
    trainN(subj) = []; % remove left-out subject

    Yheld = Y(subj,:); %held out data used to predict held out X

    %training data
    Xtrain = X(trainN,:); %held in data
    Ytrain = Y(trainN,:);

    %remove mean
    XTrainCentre = mean(Xtrain);
    YTrainCentre = mean(Ytrain);
    Xtrain = Xtrain - XTrainCentre;
    Ytrain = Ytrain - YTrainCentre;

    %This computes CCA on the training data
    [A,B,r,U,V,stats] = canoncorr(Xtrain,Ytrain);

    %Only interested in the first Mode
    Mode = 1;

    % build linear regression model using U and V
    y = U(:,Mode);                          % want to predict U
    x = [ones(length(V(:,1)),1),V(:,Mode)]; % from V
    beta = regress(y,x);                    % get regression coefficient

    % Calculate held-out V using the training data weights
    % see matlab help: V = (Y-repmat(mean(Y),N,1))*B
    Vpred = (Yheld - YTrainCentre) * B(:,Mode);

    % Calculate held-out U via linear regression equation from training set
    Upred = beta(1) + (Vpred * beta(2));

    % Apply weights, in the opposite direction, to get raw behaviour
    % U = X * A
    % X = U / A
    Xpred(subj,:,Mode) = Upred ./ A(:,Mode);
end

% prediction accuracy (by correlating real and predicted values).
for beh = 1:size(X,2)
        [r,p] = corr(X(:,beh),Xpred(:,beh));
        disp(['For variable: ',num2str(beh),', r = ',num2str(r),', p = ',num2str(p)])
end

```

Answer 1

The regression model is confusing. The intercept term is going to be zero as you demeaned the data prior to CCA (which is also a requirement). The regression model will give you back the CCA r-value between the first projected columns of U and V. Everything thereon out becomes circular. The assumption that U and V are linearly related is hazy; if anything the cross cov of U and V should produce an identity matrix. An alternate method for cross validation is splitting the data into 75%, 25% parts, and obtain cross-validated r values. That is, find A and B from the training data, apply A and B to held out data and see how correlated U_cv and V_cv are in relation to original model.