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I am trying to implement a simple multivariate linear regression model without using any inbuilt machine libraries. So far, I have been able to get a root mean squared error for training about $2.93$ and the model from the normal (closed-form) equation is able to produce a training RMSE of $~2.3$. I am looking for ways in which I can improve my implementation of the gradient descent algorithm. Below is my implementation:

My gradient descent method looks like this: $\theta = \theta - [(\alpha/2N) * X (X\theta - Y)]$ where $\theta$ is the model parameter, $N$ is the number of training elements, $X$ is the input and $Y$ are the target elements. $\alpha$ is the step size.

def gradientDescent(self):
    for i in range(self.iters):
        # T = T - (\alpha/2N) * X*(XT - Y)
        self.theta = self.theta - (self.alpha/len(self.X)) * np.sum(self.X * (self.X @ self.theta.T - self.Y), axis=0)
    return errors

I had set the $\alpha$ as $0.1$ and number of iterations as 1000. The gradient descent reaches convergence at around 700-800 iterations (checked).

My error function is like:

def error_function(self):
        # Error function: (1/2N) * (XT - Y)^2 where T is theta
        error_values = np.power(((self.X @ self.theta.T) - self.Y), 2)
        return np.sum(error_values)/(2 * len(self.X))

I was expecting the training error from the gradient descent and the normal equations would turn out to be similar, but they have a bit of a huge difference. So, I wanted to know whether I am doing anything wrong or not.

PS I have not normalized the data, yet. Normalizing leads to a much lower RMSE (~$0.22$)

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    $\begingroup$ Shouldn't that be X transposed? $๐‘‹^{T}(๐‘‹๐œƒโˆ’๐‘Œ)$. $\endgroup$ – Elmex80s Nov 18 '20 at 21:40
  • $\begingroup$ Yes, it would be $X^T(X\theta - Y)$. I have this habit of typing all vector equations in one go :P $\endgroup$ – MaJoR21 Nov 19 '20 at 12:09
  • $\begingroup$ self.theta = self.theta - (self.alpha/len(self.X)) * np.sum(self.X * (self.X @ self.theta.T - self.Y), axis=0). Lot of errors in this line of code. $\endgroup$ – Elmex80s Nov 19 '20 at 16:14
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That could be due to many different reasons. The most important one is that your cost function might be stuck in local minima. To solve this issue, you can use a different learning rate or change your initialization for the coefficients.

There might be a problem in your code for updating weights or calculating the gradient.

However, I used both methods for a simple linear regression and got the same results as follows:

import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from sklearn.datasets import make_regression


# generate regression dataset
X, y = make_regression(n_samples=100, n_features=1, noise=30)



def cost_MSE(y_true, y_pred):
    '''
    Cost function
    '''
    # Shape of the dataset
    n = y_true.shape[0]
    
    # Error 
    error = y_true - y_pred
    # Cost
    mse = np.dot(error, error) / n
    return mse


def cost_derivative(X, y_true, y_pred):
    '''
    Compute the derivative of the loss function
    '''
    # Shape of the dataset
    n = y_true.shape[0]
    
    # Error 
    error = y_true - y_pred
    
    # Derivative
    der = -2 / n * np.dot(X, error)

    return der


# Lets run an example

X_new = np.concatenate((np.ones(X.shape), X), axis = 1)
learning_rate = 0.1
X_new_T = X_new.T
n_iters = 100
mse = []

#initialize the weight vector
alpha = np.array([0, np.random.rand()])

for _ in range(n_iters):
    
    # Compute the predicted y
    y_pred = np.dot(X_new, alpha)
    
    # Compute the MSE
    mse.append(cost_MSE(y, y_pred))
    
    # Compute the derivative
    der = cost_derivative(X_new_T, y, y_pred)
    
    # Update the weight
    alpha  -= learning_rate * der
alpha

for the gradient descent the coefficients were:

array([-3.36575322, 28.06370831])

Here is the code for closed-form solution:

np.dot(np.linalg.inv(np.dot(X_new_T,X_new)), np.dot(X_new_T, y))

And the coefficients for the closed-form solution:

array([-3.36575322, 28.06370831])

As the coefficients are equal, the RMSE, MSE, R2 are equal.

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