Omid’s Logistic Regression tutorial
The main focus of this Logistic Regression tutorial is the usage of Logistic Regression in the field of Machine Learning and Data Mining. We will introduce the mathematical theory behind Logistic Regression and show how it can be applied to the field of Machine Learning when we try to extract information from very large data sets.
Machine learning is a subfield of Artificial Intelligence and deals with the development of techniques which allows computers to “learn” from previously seen datasets. Machine learning overlaps extensively with mathematical statistics, but differs in that it deals with the computational complexities of the algorithms.
Machine learning itself can be divided into many subfields, whereas the field we will work with is the one of supervised learning where we will start with a data set with labeled data points. Each data point is a vector
and each data point has a label
Given a set of data points and the corresponding labels we want be able to train a computer program to classify new (so far unseen) data points by assigning a correct class label to each data point. The ratio of correctly classified data points is called the accuracy of the system.
Regression analysis is a field of mathematical statistic that is well explored and has been used for many years. Given a set of observations, one can use regression analysis to find a model that best fits the observation data.
The most common form of regression models is the ordinary linear regression which is able to fit a straight line through a set of data points. It is assumed that the data point’s values are coming from a normally distributed random variable with a mean that can be written as a linear function of the predictors and with a variance that is unknown but constant.
We can write this equation as
where a is a constant, sometimes also denoted as b0, is a vector of the same size as our input variable x and where the error term
Figure 2 shows a response with mean
which follows a normal distribution with constant variance 1.
Figure 2 Ordinary Linear Regression
Ordinary Linear Regression for y = -2.45 + 0.35 * x. The error term has mean 0 and a constant variance.
The general form of regression, called generalized linear regression, assumes that the data points are coming from a distribution that has a mean that comes from a monotonic nonlinear transformation of a linear function of the predictors. If we can call this transformation g, the equation can be written as
where a is a constant, sometimes also denoted as b0, is a vector of the same size as our input variable x and where the error term is .
The inverse of g is called the link function. With generalized linear regression we no longer require the data points to have a normal random distribution, but we can have any distribution.
Figure 3 shows a response with mean
which follows a Poisson distribution.
Figure 3 Generalized Linear Regression
Generalized Linear Regression for a signal coming from a Poisson distribution with mean y = exp(-2.45 + 0.35 * x).
linear regression model is simple and used frequently it’s not adequate for
some purposes. For example, imagine the response variable y to be a probability
that takes on values between 0 and
Another problem with the linear regression model is the assumption that the response y has a constant variance. This can not be the case if y follows for example a binomial distribution (y ~ Bin(p,n)). If y also is normalized so that it takes values between 0 and 1, hence y = Bin(p,n)/n, then the variance would then be Var(y) = p*(1-p), which takes on values between 0 and 0.25. To then make an assumption that y would have a constant variance is not feasible.
In situations like this, when our response variable follows a binomial distribution, we need to use general linear regression. A special case of general linear regression is logistic regression, which assumes that the response variable follows the logit-function shown in Figure 4.
Figure 4 The logit function
Note that it’s only defined for values between 0 and 1. The logit function goes from minus infinity to plus infinity. The logit function has the nice property that logit(p) = -logit(1-p) and its inverse is defined for values from minus infinity to plus infinity, and it only takes on values between 0 and 1.
However, to get a better understanding for what the logit-function is we will now introduce the notation of odds. The odds of an event that occurs with probability P is defined as
odds = P / (1-P).
Figure 5 shows how the odds-function looks like. As we can see, the odds for an event is not bounded and goes from 0 to infinity when the probability for that event goes from 0 to 1.
However, it’s not always very intuitive to think about odds. Even worse, odds are quite unappealing to work with due to its asymmetry. When we work with probability we have that if the probability for yes is p, then the probability for no is 1-p. However, for odds, there exists no such nice relationship.
To take an example: If a Boolean variable is true with probability 0.9 and false with probability 0.1, we have that the odds for the variable to be true is 0.9/0.1 = 9 while the odds for being false is 0.1/0.9 = 1/9 = 0.1111... . This is a quite unappealing relationship. However, if we take the logarithm of the odds, when we would have log(9) for true and log(1/9) = -log(9) for false.
Hence, we have a very nice symmetry for log(odds(p)). This function is called the logit-function.
logit(p) = log(odds(p)) = log(p/(1-p))
As we can see, it is true in general that logit(1-p)=-logit(p).
logit(1-p) = log((1-p)/p) = - log(p/(1-p)) = -logit(p)
Figure 5 The odds function
The odds function maps probabilities (between 0 and 1) to values between 0 and infinity.
The logit-function has all the properties we wanted but did not have when we previously tried to use linear regression for a problem where the response variable followed a binomial distribution. If we instead use the logit-function we will have p bounded to values between 0 and 1 and we will still have a linear expression for our input variable x
logit(p) = a + * x.
If we would like to rewrite this expression to get a function for the probability p it would look like
In practice, one usually simplifies notation somewhat by only having one parameter instead of both and .
If our original problem is formulated such as
We rewrite this as
If we now call ’ =  T and x’ = [1 x] then we can formulate the exact same problem but with only “one” model parameter ’
Note that this is nothing but a change of notation. We still have two parameters to determine, but we have simplified our notation so that we now only need to estimate ’.
From now on, we will denote ’ as and x’ as x and our problem statement will hence be to obtain the model parameter when
If we have made n observations with responses yi and predictors xi we can define
The system we want to solve to find the parameter is then written as
The minimum square error solution to this system is found as follows
We just need to evaluate the expression (XT * X)-1 * X T * Y and we have found the that minimizes the sum of squares residuals. However, in practice there might be computational difficulties with evaluating this expression, as we will see further on.
As we have seen we can obtain by simply evaluating
However, if some prediction variables are (almost) linearly dependent, then XT * X is (almost) singular and hence he variance of is very large. So to avoid having XT * X singular we add a small constant value to the diagonal of the matrix
where I = unity matrix, and λ = small constant.
By doing this we avoid the numerical problems we will get when trying to invert an (almost) singular matrix. But we are paying a price for doing this. By doing this we have biased the prediction and hence we are solving the solution to a slightly different problem. As long as the error due to the bias is smaller than the error we would have got from having a (nearly) singular XT * X we will end up getting a smaller mean square error and hence ridge regression is desirable.
We can also see ridge regression as a minimization problem where we try to find a according to
Which we (through Lagrange multiplier) can rewrite to an unconstraint minimization problem
where λ is inversely proportional to s.
This can be compared to the classic regression where we are minimizing
Now the problem is just to find a good λ (or s) so that the variance gets small, but at the same time we should make sure the bias error doesn’t get to big either. To find a good λ (or s) we can use heuristics, graphics or cross validation. However, this can be computationally expensive, so in practice one might prefer to just choose a small constant λ and then normalize the input data so that
Or in other words, we make sure x is centered and normalized.
Ridge regression has the advantage of preferring smaller coefficient values for and hence we end up with a less complex model. This is desirable, due too Occam’s razor which says that it is preferable to pick the simpler model out of two models that are equally good but where one is simpler than the other, since the simpler model is more likely to be correct and also hold for new unseen data.
Another way to get an intuition for why we prefer small coefficient values is in the case when we have correlated attributes. Imagine two attributes that are strongly correlated and when either one of them takes the value 1, the other one does the same with high likelihood and vice verse. It would now be possible that the coefficients for these two attributes to take identical extremely large values but with different signs since they both “cancel out” each other. This is of course undesirable in the situations when the attributes take different values and X * takes on ridiculously large values.
Ridge regression has proved itself to be superior to many alternative methods when it has been used to avoid numerical difficulties when solving linear equation systems for building logistic regression classifiers (, , ).
Ridge regression was first used in the context of least square regression in  and later on used in the context of logistic regression in .
As we have seen we need to evaluate this expression in classic logistic regression
This expression came from the linear equation system
Indirectly we assumed that all observations where equally important and hence had the same weight, since we tried to minimize the sum of squared residuals.
However, when we do weighted logistic regression we will weight the importance of our observations so that different observations have different weights associated to them. We will have a weight matrix W that is a diagonal matrix with the weight of observation i at location Wii.
Now, instead of evaluating
we will evaluate
and μi is our estimate for p, which we previously saw could be written as
The weights Wii are nothing but the standard deviation of our own prediction. In general, if
and since we have a Bernoulli trial we have n = 1 so the variance becomes
The term yi - μi is our prediction error and the variance Wii “scales” it so that a low variance will have a larger impact on U than a high variance data point. Or in other words, the importance of correctly classifying data points with a low variance increases while the importance of correctly classifying data points with a high variance decreases.
We have now seen the theory behind the equation that we now need to solve. With notation as before we now want to solve
However, so far we have not discussed the computational difficulties with doing this. One of the major differences between classical statistics and machine learning is that the later one deals with the computational difficulties one is facing when one is trying to solve the equations obtained from the field of classical statistics.
When one needs to evaluate an expression such as the one we have for , it is very common to write down the problem as a linear equation system that needs to be solved, to avoid having to calculate the inverse of a large matrix.
Hence, the problem can be rewritten as
Let us now take a closer look at the problem we are facing.
What we want to achieve is to build a classifier that will classify very large data sets. Our input data is X and (indirectly) U. To get an idea of what size our matrices have, imagine our application having 100,000 classes and 100,000 attributes, also imagine us having in average 10 training data points per class. The size of the A matrix would then be 100,000 x 100,000 and the size of the b vector would be 100,000 x 1. The X matrix would be of size 1’000’000 x 100,000. Luckily for us, our data will be sparse, and hence only a small fraction of the elements will have non-zero values. Using this knowledge, we can choose an equation solver that is efficient given this assumption.
In general when one needs to solve a linear equation system such as
one needs to choose a solver method appropriate to the properties of the problem. This basically means that one needs to investigate what properties are satisfied for A and from that choose one of the many available solver methods that are available. If one needs an iterative solver, which does not give an exact solution but is computationally efficient and in many case the only practical alternative,  offers an extensive list of solvers that can be used.
For our application we are going to use the conjugate gradient method, which is a very efficient method for solving linear equation systems when A is a symmetric positive definite matrix, since we only need to store a limited number of vectors in memory.
When we solve a linear system with iterative CG we will use the fact that the solution to the problem A * = b, for symmetric positive definite matrices A, is identical to the solution for the minimization problem
The complete algorithm for solving using the CG algorithm is found in Figure 6.
Figure 6 The conjugate gradient method
The conjugate gradient method can
efficiently solve the equation system
A * = b, for symmetric positive definite sparse matrices A.
With perfect arithmetic, we will be able to find the correct solution x in the CG algorithm above in m steps, if A is of size m. However, since we will solve the system A * = b iteratively, and our A and b will change after each iteration, we don’t iterate the CG algorithm until we have an exact solution for . We stop when is “close enough” to the correct solution and then we recalculate A and b, using the recently calculated value, and once again run the CG algorithm to obtain an even better value. So although the CG algorithm requires m steps to find the exact solution, we will terminate the algorithm in advance and hence get a significant speed up.
How fast we get to a solution that is good depends on the eigenvalues of the matrix A. We earlier stated that m iterations are required to find the exact solution, but to be more precise the number of iterations required is also bounded by the number of distinct eigenvalues of matrix A. However, in most practical situations with large matrices we will just iterate until the residual is small enough. Usually we will get a solution that is reasonable good within 20-40 iterations.
Note that one might choose many different termination criteria for when we want to stop the CG algorithm. For example:
The deviance for our logistic regression system is
where as previously
For an extensive investigation of how different termination criteria are affecting the resulting classifier accuracy, see .
For the reader interested in more details about the conjugate gradient method and possible extensions to it, the authors would like to recommend ,  and .
Many papers have been investigating how one can build large scale logistic classifiers with different linear equations solvers (, , , ). We will be using the conjugate gradient method for this task. This has previously been reported to be a successful method for building large scale logistic classifiers in terms of nr of attributes and in nr of data points (). However, due to the fact that the high computational complexity of calculating it is infeasible to build very large logistic regression classifiers if we don’t have an algorithm for building the classifier in a distributed environment using the power of a large number of machines.
The main contribution of this research will be to develop an efficient algorithm for building a very large scale logistic regression classifier using a distributed system.
We have now gone through all theory we need to be able to build a large scale logistic regression classifier. To obtain we will now be using the iteratively reweighted least-squares method, also known as the IRLS method in Figure 10.
A complete algorithm for getting is shown in Figure 7.
Figure 7 Algorithm for Obtaining (IRLS)
The Iteratively Reweighted Least-Squares method algorithm.
Note that this will give us for one class only. We need to run this algorithm once for each class that we have so that we have one for each class. If we for example have 100,000 classes, the algorithm would need to run 100,000 times with different yi values each time. To run this code with a data set with around one million data points that have in the order of 100,000 attributes could take in the order of 1 minute to finish, so to be able to scale up our classifier we need an algorithm that can efficiently run this piece of code on distributed client machines.
The way we will do classification with our system when we have all values is to create a matrix that we call the “weight matrix” (denoted W).
Say we have a data point x and we want to know which of the n classes it should belong to.
We have previously seen that the probability that data point x belongs to the class corresponding to is
Hence, the larger value we have for * x, the stronger is our belief that the data point x belongs to the class corresponding to . So to do classification, we only need to see which gives the highest value and chose that class as our best guess.
To do ranking, we do basically the same thing as for classification
Hence, the score for class i will be i * x, and we rank the classes so that the class with the highest score is ranked highest.
LR Logistic Regression
NB Naïve Bayesian
CG Conjugate Gradient
IRLS Iteratively Reweighted Least-Squares
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