A model called logistic regression is introduced in Week 3 of Machine Learning course taught by Andrew Ng. This post tries to explain how to obtain logistic regression model from linear regression model which has been explained in Week 1 and Week 2 of the course. Therefore, I recommend all readers for studying Week 1 and Week 2 before reading this post.

Week 1 of the course shows that linear regression model has the capability to solve classification problem with prediction model (hypothesis) as follows:

$$\begin{equation} \bbox[5px,border:2px solid blue] {h_\theta(x) = \theta^T x} \tag{1}\label{eq:linear-regression} \end{equation}$$

with $\theta = \begin{bmatrix} \theta_0 & \theta_1 & \theta_2 & \cdots & \theta_n \end{bmatrix}^T$ is called parameter model and $x = \begin{bmatrix} 1 & x_1 & x_2 & \cdots & x_n \end{bmatrix}^T$ is a test example whose $y$ we would like to predict.

The question arises as follows:

How can we derive logistic regression model from linear regression model?

As we already know that $h_\theta(x)$ in logistic regression has a range of values between $0$ and $1$; on the other hand, $h_\theta(x)$ in linear regression has a range of values between $-\infty$ and $\infty$. Therefore, $h_\theta(x)$ which belongs to logistic regression needs to be converted into $-\infty < h_\theta(x) < \infty$. Moreover, after the conversion is done, the new $h_\theta(x)$ shall be substituted into Equation $\eqref{eq:linear-regression}$.

Let us start by converting $h_\theta(x)$ which belongs to logistic regression and then substitute the new $h_\theta(x)$ into Equation $\eqref{eq:linear-regression}$.

$$\begin{align} 0 < h_\theta(x) < 1 &\Longleftrightarrow 0 < \frac{h_\theta(x)}{1-h_\theta(x)} < \infty \tag{2}\label{eq:odds-ratio} \\ &\Longleftrightarrow -\infty < \underbrace{\log \left( \frac{h_\theta(x)}{1-h_\theta(x)} \right)}_{\text{the new }h_\theta(x)} < \infty \tag{3}\label{eq:logit}. \end{align}$$

$\frac{h_\theta(x)}{1-h_\theta(x)}$ in Equation $\eqref{eq:odds-ratio}$ is called odds ratio and $\log \left( \frac{h_\theta(x)}{1-h_\theta(x)} \right)$ in Equation $\eqref{eq:logit}$ is named as logit function (Raschka & Mirjalili, 2017).

Now that the new $h_\theta(x)$ has the range of values between $-\infty < h_\theta(x) < \infty$, this new $h_\theta(x)$ can be substituted into Equation $\eqref{eq:linear-regression}$ as follows:

$$\begin{align} \log \left( \frac{h_\theta(x)}{1-h_\theta(x)} \right) = \theta^T x &\Longleftrightarrow e^{\log \left( \frac{h_\theta(x)}{1-h_\theta(x)} \right)} = e^{\theta^T x} \\ &\Longleftrightarrow \frac{h_\theta(x)}{1-h_\theta(x)} = e^{\theta^T x} \\ &\Longleftrightarrow h_\theta(x) = e^{\theta^T x} - h_\theta(x) e^{\theta^T x} \\ &\Longleftrightarrow h_\theta(x) + h_\theta(x) e^{\theta^T x} = e^{\theta^T x} \\ &\Longleftrightarrow h_\theta(x) \left( 1 + e^{\theta^T x} \right) = e^{\theta^T x} \\ &\Longleftrightarrow h_\theta(x) = \frac{e^{\theta^T x}}{1 + e^{\theta^T x}} \\ &\Longleftrightarrow h_\theta(x) = \frac{e^{\theta^T x}}{1 + e^{\theta^T x}} \times \frac{e^{-\theta^T x}}{e^{-\theta^T x}} \\ &\Longleftrightarrow h_\theta(x) = \frac{1}{1+e^{-\theta^T x}}. \end{align}$$

Accordingly, logistic regression model that we have derived is

$$\begin{equation} \bbox[5px,border:2px solid blue] {h_\theta(x) = \frac{1}{1+e^{-\theta^T x}}} \tag{4}\label{eq:sigmoid} \end{equation}$$

Logistic regression model in Equation $\eqref{eq:sigmoid}$ is popularly also called sigmoid function. Finally, we have successfully derived logistic regression model (Equation $\eqref{eq:sigmoid}$) from linear regression model (Equation $\eqref{eq:linear-regression}$).

References

Raschka, S. and Mirjalili, V. (2017) Python Machine Learning Second Edition. Packt Publishing Ltd. Page 59-60.