# Dividing the weights obtained on an already standardized data set by the standard deviation of the features? (Ridge regression)

I'm trying to understand a code snippet from my lecture on Machine Learning (see the code below).

It extracts the mean and standard deviation of the features and uses them to 'normalize' (standardize) the features. The data X contains all-ones in the first column. The variable t is the output value on which we train the weights. First the mean of the output is obtained so that ridge regression can be applied without a bias term (intercept), then the all-ones column is removed and the data is normalized. The weights are obtained by the formula: What I don't understand is that the weights obtained by using the normalized data are once again divided by the standard deviation of the features W[1:, 0] = Wn[:, 0] * Sinv. Can someone help me out? Thanks!

PS: I know normalizing is something else and I actually mean standardizing, but my professor uses normalizing instead for some reason..

Here's the code:

def NormalisedRidgeRegression(X, t, p_lambda=0.0):
K = X.shape - 1

M = np.mean(X[:, 1:], axis=0)
S = np.std(X[:, 1:], axis=0)
Sinv = 1 / S

targetmean = np.mean(t)
tn = t - targetmean

Xn = np.dot(X[:, 1:] - M, np.diag(Sinv))

Wn = np.dot(np.dot(np.linalg.pinv(np.dot(Xn.T, Xn) + p_lambda * np.eye(K)), Xn.T), tn)
W = np.zeros((K + 1, 1))
W[1:, 0] = Wn[:, 0] * Sinv
W[0, 0] = targetmean
y_hat = np.dot(X, W)
return W, y_hat


Say for a given data point before normalising we have a*x + b*z = y Normalising makes this a * x/std(x) + b * z/std(z) = y This means that for standard deviations greater than 1 the features are shrunk and therefore the weights are blown up. In the code the y_hat values are obtained using the original data, which means the coefficients have to be shrunk to reflect the relationship between the output and the original feature values.