public class LogisticGradient extends Gradient
In The Elements of Statistical Learning: Data Mining, Inference, and Prediction, 2nd Edition
by Trevor Hastie, Robert Tibshirani, and Jerome Friedman, which can be downloaded from
http://statweb.stanford.edu/~tibs/ElemStatLearn/ , Eq. (4.17) on page 119 gives the formula of
multinomial logistic regression model. A simple calculation shows that
$$ P(y=0|x, w) = 1 / (1 + \sum_i^{K-1} \exp(x w_i))\\ P(y=1|x, w) = exp(x w_1) / (1 + \sum_i^{K-1} \exp(x w_i))\\ ...\\ P(y=K-1|x, w) = exp(x w_{K-1}) / (1 + \sum_i^{K-1} \exp(x w_i))\\ $$
for K classes multiclass classification problem.
The model weights $w = (w_1, w_2, ..., w_{K-1})^T$ becomes a matrix which has dimension of (K-1) * (N+1) if the intercepts are added. If the intercepts are not added, the dimension will be (K-1) * N.
As a result, the loss of objective function for a single instance of data can be written as
$$ \begin{align} l(w, x) &= -log P(y|x, w) = -\alpha(y) log P(y=0|x, w) - (1-\alpha(y)) log P(y|x, w) \\ &= log(1 + \sum_i^{K-1}\exp(x w_i)) - (1-\alpha(y)) x w_{y-1} \\ &= log(1 + \sum_i^{K-1}\exp(margins_i)) - (1-\alpha(y)) margins_{y-1} \end{align} $$
where $\alpha(i) = 1$ if $i \ne 0$, and $\alpha(i) = 0$ if $i == 0$, $margins_i = x w_i$.
For optimization, we have to calculate the first derivative of the loss function, and a simple calculation shows that
$$ \begin{align} \frac{\partial l(w, x)}{\partial w_{ij}} &= (\exp(x w_i) / (1 + \sum_k^{K-1} \exp(x w_k)) - (1-\alpha(y)\delta_{y, i+1})) * x_j \\ &= multiplier_i * x_j \end{align} $$
where $\delta_{i, j} = 1$ if $i == j$, $\delta_{i, j} = 0$ if $i != j$, and multiplier = $\exp(margins_i) / (1 + \sum_k^{K-1} \exp(margins_i)) - (1-\alpha(y)\delta_{y, i+1})$
If any of margins is larger than 709.78, the numerical computation of multiplier and loss function will be suffered from arithmetic overflow. This issue occurs when there are outliers in data which are far away from hyperplane, and this will cause the failing of training once infinity / infinity is introduced. Note that this is only a concern when max(margins) > 0.
Fortunately, when max(margins) = maxMargin > 0, the loss function and the multiplier can be easily rewritten into the following equivalent numerically stable formula.
$$ \begin{align} l(w, x) &= log(1 + \sum_i^{K-1}\exp(margins_i)) - (1-\alpha(y)) margins_{y-1} \\ &= log(\exp(-maxMargin) + \sum_i^{K-1}\exp(margins_i - maxMargin)) + maxMargin - (1-\alpha(y)) margins_{y-1} \\ &= log(1 + sum) + maxMargin - (1-\alpha(y)) margins_{y-1} \end{align} $$
where sum = $\exp(-maxMargin) + \sum_i^{K-1}\exp(margins_i - maxMargin) - 1$.
Note that each term, $(margins_i - maxMargin)$ in $\exp$ is smaller than zero; as a result, overflow will not happen with this formula.
For multiplier, similar trick can be applied as the following,
$$ \begin{align} multiplier &= \exp(margins_i) / (1 + \sum_k^{K-1} \exp(margins_i)) - (1-\alpha(y)\delta_{y, i+1}) \\ &= \exp(margins_i - maxMargin) / (1 + sum) - (1-\alpha(y)\delta_{y, i+1}) \end{align} $$
where each term in $\exp$ is also smaller than zero, so overflow is not a concern.
For the detailed mathematical derivation, see the reference at http://www.slideshare.net/dbtsai/2014-0620-mlor-36132297
param: numClasses the number of possible outcomes for k classes classification problem in Multinomial Logistic Regression. By default, it is binary logistic regression so numClasses will be set to 2.
| Constructor and Description |
|---|
LogisticGradient() |
LogisticGradient(int numClasses) |
| Modifier and Type | Method and Description |
|---|---|
double |
compute(Vector data,
double label,
Vector weights,
Vector cumGradient)
Compute the gradient and loss given the features of a single data point,
add the gradient to a provided vector to avoid creating new objects, and return loss.
|
public LogisticGradient(int numClasses)
public LogisticGradient()