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64 lines
2.0 KiB
Markdown
64 lines
2.0 KiB
Markdown
1 week ago
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# Linear regression
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The linear regression is a training procedure based on a linear model. The model makes a prediction by simply computing a weighted sum of the input features, plus a constant term called the bias term (also called the intercept term):
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$$ \hat{y}=\theta_0 + \theta_1 x_1 + \theta_2 x_2 + \cdots + \theta_n x_n$$
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This can be writen more easy by using vector notation form for $m$ values. Therefore, the model will become:
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$$
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\begin{bmatrix}
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\hat{y}^0 \\
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\hat{y}^1\\
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\hat{y}^2\\
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\vdots \\
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\hat{y}^m
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & x_1^0 & x_2^0 & \cdots &x_n^0\\
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1 & x_1^1 & x_2^1 & \cdots & x_n^1\\
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\vdots & \vdots &\vdots & \cdots & \vdots\\
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1 & x_1^m & x_2^m & \cdots & x_n^m
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\end{bmatrix}
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\begin{bmatrix}
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\theta_0 \\
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\theta_1 \\
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\theta_2 \\
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\vdots \\
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\theta_n
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\end{bmatrix}
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$$
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Resulting:
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$$\hat{y}= h_\theta(x) = x \theta $$
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**Now that we have our mode, how do we train it?**
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Please, consider that training the model means adjusting the parameters to reduce the error or minimizing the cost function. The most common performance measure of a regression model is the Mean Square Error (MSE). Therefore, to train a Linear Regression model, you need to find the value of θ that minimizes the MSE:
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$$ MSE(X,h_\theta) = \frac{1}{m} \sum_{i=1}^{m} \left(\hat{y}^{(i)}-y^{(i)} \right)^2$$
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$$ MSE(X,h_\theta) = \frac{1}{m} \sum_{i=1}^{m} \left( x^{(i)}\theta-y^{(i)} \right)^2$$
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$$ MSE(X,h_\theta) = \frac{1}{m} \left( x\theta-y \right)^T \left( x\theta-y \right)$$
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# The normal equation
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To find the value of $\theta$ that minimizes the cost function, there is a closed-form solution that gives the result directly. This is called the **Normal Equation**; and can be find it by derivating the *MSE* equation as a function of $\theta$ and making it equals to zero:
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$$\hat{\theta} = (X^T X)^{-1} X^{T} y $$
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$$ Temp = \theta_0 + \theta_1 * t $$
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```python
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import pandas as pd
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df = pd.read_csv('data.csv')
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df
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```
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