Logistics Regression

Posted by GwanSiu on June 17, 2018

1. Logistic Regression

Data: Inputs are continuous vectors of length $K$. Outputs are discrete.

$\begin{equation} \mathbb{D}={x^{(i)},y^{(i)}}_{i=1}^{N}, \text{ where } \mathbf{x}\in\mathbb{R} \end{equation}$

Model: Logistic function applied to dot product of parameters with input vector.

$\begin{equation} p_{\theta}(y=1\vert x) =\frac{1}{1+exp(-\theta^{T}\mathbf{x})} \end{equation}$

Learning: Finds the parameters that minimized some objective function.

$\begin{equation} \theta^{\ast}=\arg\min_{\theta}J(\theta) \end{equation}$

Usually, we minimize the negative log conditional likelihood:

$\begin{equation} J(\theta) = -\log\prod_{i=1}^{N}p_{\theta}(y^{(i)}\vert \mathbf{x}^{(i)}) \end{equation}$

Why can’t maximize likelihood(as in naive bayes)

Prediction: Output is the most probable class.

$\begin{equation} \tilde{y} = \arg\max p_{\theta}(y\vert x),y\in \{0,1\} \end{equation}$

2. Learning Methods

learning: Three approaches to solve $\theta^{\ast}=\arg\min_{\theta}J(\theta)$

• Approach 2: Stochastic Gradient Descent(SGD)(Take many small steps apposite the gradient)
• Approach 3: Newton’s Method(use second derivatives to better follow curvature)

3. Newton’s Method and Logistic Regression

The motivation behind Newton’s method is to use a quadreatic approximation of our function to make a good guess where we should step next.

The Taylor series expansion for an infinitely differentiable function $f(x),x\in\mathbb{R},$ about a pint $\nu\in\mathbb{R}$ is:

$\begin{equation} f(x) = f(\nu)+\frac{(x-\nu)f^{'}(x)}{1!}+\frac{(x-v)^{2}f^{''}(x)}{2!}+\frac{(x-v)^{3}f^{'''}(x)}{3!}+... \end{equation}$

The 2nd-order Taylor series approximation cuts off the expansion after the quadratic term:

$\begin{equation} f(x)\approx f(v)+\frac{(x-v)f^{'}(x)}{1!}+\frac{(x-v)^{2}f^{''}(x)}{2!} \end{equation}$

The vector version of Taylor series expansion for an infinitely differentiable function $f(x),\mathbf{x}\in \mathbb{R}^{K}$, about a point $\mathbf{v}\in\mathbb{R}^{K}$ is:

$\begin{equation} f(x) = f(\nu)+\frac{(x-\nu)^{T}\nabla f(x)}{1!}+\frac{(x-v)^{T}\nabla^{2} f(x)(x-v)}{2!}+... \end{equation}$

The 2nd-order Taylor series approximation cuts off the expansion after the quadratic term:

$\begin{equation} f(x) \approx f(\nu)+\frac{(x-\nu)^{T}\nabla f(x)}{1!}+\frac{(x-v)^{T}\nabla^{2} f(x)(x-v)}{2!} \end{equation}$

Taking the derivative of $\tilde{f}(v)$ and setting to 0 gives us the closed form minimizer of this(convex) quadratic function:

$\arg\min_{x}\tilde{f}(x)=x-(\nabla^{2}f(x))^{-1}\nabla f(x)$

The added term $\nabla x_{nt}=-(\nabla^{2}f(x))^{-1}\nabla f(x)$ is called Newton’s step.

The algorithm of Newton’s Method(Newton-Raphson method)

Goal: $x^{\ast}=\arg\min_{x}f(x)$

1. Approximate the function with the 2nd order Taylor series:
$f(x) \approx f(\nu)+\frac{(x-\nu)^{T}\nabla f(x)}{1!}+\frac{(x-v)^{T}\nabla^{2} f(x)(x-v)}{2!}$
1. Compute its minimizer
$\arg\min_{x}\tilde{f}(x)=x-(\nabla^{2}f(x))^{-1}\nabla f(x)$
1. Step to that minimizer:
$x = -(\nabla^{2}f(x))^{-1}\nabla f(x)$
1. Repeat

3. Iterative Weighted Least Square(Newton’s Method for Logistic Regression)

As we know, linear regression has closed-form solution, but there are no longer closed-form solution for logistic regression due to nonlinearity of the logistic sigmoid function. In fact, the lost function of logistic regression is convex, and it is guaranteed to find a global optimal solution. Furthermore, the error funciton can be solved by an efficient iterative optimization algorithm based on Newton-Rashon iterative optimization scheme. For Logistic Regression:

$\begin{equation} -H^{-1}g = -(X^{T}SX)^{-1}(X^{T}(\mu-y)) \end{equation}$

take $-H^{-1}g$ back to $\theta\leftarrow \theta - H^{-1}g$, we obtain:

\begin{align} \theta&\leftarrow\theta-H^{-1}g \\ &=\theta - (X^{T}SX)^{-1}(X^{T}(\mu-y)) \\ &=(X^{T}SX)^{-1}((X^{T}SX)\theta-(X^{T}(\mu-y))) \\ &= (X^{T}SX)^{-1}X^{T}(SX\theta-(\mu-y)) \\ &= (X^{T}SX)^{-1}X^{T}S(X\theta-S^{-1}(\mu-y) \\ &= (X^{T}SX)^{-1}X^{T}Sz \end{align}

where $z=(X\theta-S^{-1}(\mu-y)$

Compare with the closed-form in linear regression: $\theta^{\ast}=(X^{T}X)^{-1}X^{T}y$, the step $\theta\leftarrow \theta-H^{-1}g$ can be reformlized as $\theta=(X^{T}SX)^{-1}X^{T}Sz$, where $z=(X\theta-S^{-1}(\mu-y)$. It’s the same form as the closed-form of linear regression.

Therefore, we can see the above update yields the minimizer for the weighted least squares problem:

\begin{align} \theta^{\ast}\leftarrow \arg\min_{\theta}(z-X\theta)^{T}S(z-X\theta) \\ \arg\min_{\theta} \sum_{i=1}^{N}S_{ii}(z_{i}-\theta^{T}x^{(i)})^{2} \end{align}

where $S_{ii}$ is the weight of the $i\text{th}$ “training example” consisting of the pair $(x^{(i)},z_{i})$.

The explaination in wiki:

The method of iteratively reweighted least squares (IRLS) is used to solve certain optimization problems with objective functions of the form:

$\begin{equation} \arg\min_{\beta}sum_{i=1}^{n}(y_{i}-f_{i}(\beta))^{p} \end{equation}$

by an iterative method in which each step involves solving a weighted least squares problem of the form:

$\begin{equation} \beta^{t+1} = \arg\min_{\beta}\omega_{i}(\beta^{(t)})(y_{i}-f_{i}(\beta))^{2} \end{equation}$

IRLS is used to find the maximum likelihood estimates of a generalized linear model, and in robust regression to find an M-estimator, as a way of mitigating the influence of outliers in an otherwise normally-distributed data set. For example, by minimizing the least absolute error rather than the least square error. 