Naive Bayes classifier

The discussion so far has derived the independent feature model, that is, the naive Bayes probability model. The naive Bayes classifier combines this model with a decision rule. One common rule is to pick the hypothesis that is most probable so as to minimize the probability of misclassification; this is known as the maximum a posteriori or MAP decision rule. The corresponding classifier, a Bayes classifier, is the function that assigns a class label ${\displaystyle {\hat {y}}=C_{k}}$ for some k as follows:

${\displaystyle {\hat {y}}={\underset {k\in \{1,\ldots ,K\}}{\operatorname {argmax} }}\ p(C_{k})\displaystyle \prod _{i=1}^{n}p(x_{i}\mid C_{k}).}$

Likelihood functions ${\displaystyle p(\mathbf {x} \mid Y)}$, Confusion matrix and ROC curve. For the naive Bayes classifier and given that the a priori probabilities ${\displaystyle p(Y)}$ are the same for all classes, then the decision boundary (green line) would be placed on the point where the two probability densities intersect, due to ${\displaystyle p(Y\mid \mathbf {x} )={\frac {p(Y)\ p(\mathbf {x} \mid Y)}{p(\mathbf {x} )}}\propto p(\mathbf {x} \mid Y)}$.

When dealing with continuous data, a typical assumption is that the continuous values associated with each class are distributed according to a normal (or Gaussian) distribution. For example, suppose the training data contains a continuous attribute, ${\displaystyle x}$. The data is first segmented by the class, and then the mean and variance of ${\displaystyle x}$ is computed in each class. Let ${\displaystyle \mu _{k}}$ be the mean of the values in ${\displaystyle x}$ associated with class C_k, and let ${\displaystyle \sigma _{k}^{2}}$ be the Bessel corrected variance of the values in ${\displaystyle x}$ associated with class C_k. Suppose one has collected some observation value ${\displaystyle v}$. Then, the probability density of ${\displaystyle v}$ given a class ${\displaystyle C_{k}}$, ${\displaystyle p(x=v\mid C_{k})}$, can be computed by plugging ${\displaystyle v}$ into the equation for a normal distribution parameterized by ${\displaystyle \mu _{k}}$ and ${\displaystyle \sigma _{k}^{2}}$. That is,

${\displaystyle p(x=v\mid C_{k})={\frac {1}{\sqrt {2\pi \sigma _{k}^{2}}}}\,e^{-{\frac {(v-\mu _{k})^{2}}{2\sigma _{k}^{2}}}}}$

Another common technique for handling continuous values is to use binning to discretize the feature values, to obtain a new set of Bernoulli-distributed features; some literature in fact suggests that this is necessary to apply naive Bayes, but it is not, and the discretization may throw away discriminative information.[4]

Sometimes the distribution of class-conditional marginal densities is far from normal. In these cases, kernel density estimation can be used for a more realistic estimate of the marginal densities of each class. This method, which was introduced by John and Langley,[8] can boost the accuracy of the classifier considerably.[11][2]

With a multinomial event model, samples (feature vectors) represent the frequencies with which certain events have been generated by a multinomial ${\displaystyle (p_{1},\dots ,p_{n})}$ where ${\displaystyle p_{i}}$ is the probability that event i occurs (or K such multinomials in the multiclass case). A feature vector ${\displaystyle \mathbf {x} =(x_{1},\dots ,x_{n})}$ is then a histogram, with ${\displaystyle x_{i}}$ counting the number of times event i was observed in a particular instance. This is the event model typically used for document classification, with events representing the occurrence of a word in a single document (see bag of words assumption). The likelihood of observing a histogram x is given by

${\displaystyle p(\mathbf {x} \mid C_{k})={\frac {(\sum _{i=1}^{n}x_{i})!}{\prod _{i=1}^{n}x_{i}!}}\prod _{i=1}^{n}{p_{ki}}^{x_{i}}}$

The multinomial naive Bayes classifier becomes a linear classifier when expressed in log-space:[12]

${\displaystyle {\begin{aligned}\log p(C_{k}\mid \mathbf {x} )&\varpropto \log \left(p(C_{k})\prod _{i=1}^{n}{p_{ki}}^{x_{i}}\right)\\&=\log p(C_{k})+\sum _{i=1}^{n}x_{i}\cdot \log p_{ki}\\&=b+\mathbf {w} _{k}^{\top }\mathbf {x} \end{aligned}}}$

where ${\displaystyle b=\log p(C_{k})}$ and ${\displaystyle w_{ki}=\log p_{ki}}$.

If a given class and feature value never occur together in the training data, then the frequency-based probability estimate will be zero, because the probability estimate is directly proportional to the number of occurrences of a feature's value. This is problematic because it will wipe out all information in the other probabilities when they are multiplied. Therefore, it is often desirable to incorporate a small-sample correction, called pseudocount, in all probability estimates such that no probability is ever set to be exactly zero. This way of regularizing naive Bayes is called Laplace smoothing when the pseudocount is one, and Lidstone smoothing in the general case.

Rennie et al. discuss problems with the multinomial assumption in the context of document classification and possible ways to alleviate those problems, including the use of tf–idf weights instead of raw term frequencies and document length normalization, to produce a naive Bayes classifier that is competitive with support vector machines.[12]

In the multivariate Bernoulli event model, features are independent Booleans (binary variables) describing inputs. Like the multinomial model, this model is popular for document classification tasks,[9] where binary term occurrence features are used rather than term frequencies. If ${\displaystyle x_{i}}$ is a boolean expressing the occurrence or absence of the i'th term from the vocabulary, then the likelihood of a document given a class ${\displaystyle C_{k}}$ is given by[9]

${\displaystyle p(\mathbf {x} \mid C_{k})=\prod _{i=1}^{n}p_{ki}^{x_{i}}(1-p_{ki})^{(1-x_{i})}}$

where ${\displaystyle p_{ki}}$ is the probability of class ${\displaystyle C_{k}}$ generating the term ${\displaystyle x_{i}}$. This event model is especially popular for classifying short texts. It has the benefit of explicitly modelling the absence of terms. Note that a naive Bayes classifier with a Bernoulli event model is not the same as a multinomial NB classifier with frequency counts truncated to one.

Given a way to train a naive Bayes classifier from labeled data, it's possible to construct a semi-supervised training algorithm that can learn from a combination of labeled and unlabeled data by running the supervised learning algorithm in a loop:[13]

Given a collection ${\displaystyle D=L\uplus U}$ of labeled samples L and unlabeled samples U, start by training a naive Bayes classifier on L.

Until convergence, do:

Predict class probabilities ${\displaystyle P(C\mid x)}$ for all examples x in ${\displaystyle D}$.

Re-train the model based on the probabilities (not the labels) predicted in the previous step.

Convergence is determined based on improvement to the model likelihood ${\displaystyle P(D\mid \theta )}$, where ${\displaystyle \theta }$ denotes the parameters of the naive Bayes model.

This training algorithm is an instance of the more general expectation–maximization algorithm (EM): the prediction step inside the loop is the E-step of EM, while the re-training of naive Bayes is the M-step. The algorithm is formally justified by the assumption that the data are generated by a mixture model, and the components of this mixture model are exactly the classes of the classification problem.[13]

In the case of discrete inputs (indicator or frequency features for discrete events), naive Bayes classifiers form a generative-discriminative pair with (multinomial) logistic regression classifiers: each naive Bayes classifier can be considered a way of fitting a probability model that optimizes the joint likelihood ${\displaystyle p(C,\mathbf {x} )}$, while logistic regression fits the same probability model to optimize the conditional ${\displaystyle p(C\mid \mathbf {x} )}$.[16]

More formally, we have the following:

Proof

Consider a generic multiclass classification problem, with possible classes ${\displaystyle Y\in \{1,...,n\}}$, then the (non-naive) Bayes classifier gives, by Bayes theorem:

$${\displaystyle Pr(Y|X=x)=softmax(\{\ln Pr(Y=k)+\ln Pr(X=x|Y=k)\}_{k})}$$

The naive Bayes classifier gives

$${\displaystyle softmax\left(\left\{\ln Pr(Y=k)+{\frac {1}{2}}\sum _{i}(a_{i,k}^{+}-a_{i,k}^{-})x_{i}+(a_{i,k}^{+}+a_{i,k}^{-})\right\}_{k}\right)}$$

where

$${\displaystyle a_{i,s}^{+}=\ln Pr(X_{i}=+1|Y=s);\quad a_{i,s}^{-}=\ln Pr(X_{i}=-1|Y=s)}$$

This is exactly a logistic regression classifier.

The link between the two can be seen by observing that the decision function for naive Bayes (in the binary case) can be rewritten as "predict class ${\displaystyle C_{1}}$ if the odds of ${\displaystyle p(C_{1}\mid \mathbf {x} )}$ exceed those of ${\displaystyle p(C_{2}\mid \mathbf {x} )}$". Expressing this in log-space gives:

${\displaystyle \log {\frac {p(C_{1}\mid \mathbf {x} )}{p(C_{2}\mid \mathbf {x} )}}=\log p(C_{1}\mid \mathbf {x} )-\log p(C_{2}\mid \mathbf {x} )>0}$

The left-hand side of this equation is the log-odds, or logit, the quantity predicted by the linear model that underlies logistic regression. Since naive Bayes is also a linear model for the two "discrete" event models, it can be reparametrised as a linear function ${\displaystyle b+\mathbf {w} ^{\top }x>0}$. Obtaining the probabilities is then a matter of applying the logistic function to ${\displaystyle b+\mathbf {w} ^{\top }x}$, or in the multiclass case, the softmax function.

Discriminative classifiers have lower asymptotic error than generative ones; however, research by Ng and Jordan has shown that in some practical cases naive Bayes can outperform logistic regression because it reaches its asymptotic error faster.[16]

Problem: classify whether a given person is a male or a female based on the measured features. The features include height, weight, and foot size. Although with NB classifier we treat them as independent, they are not in reality.

Example training set below.

The classifier created from the training set using a Gaussian distribution assumption would be (given variances are unbiased sample variances):

The following example assumes equiprobable classes so that P(male)= P(female) = 0.5. This prior probability distribution might be based on prior knowledge of frequencies in the larger population or in the training set.

Below is a sample to be classified as male or female.

In order to classify the sample, one has to determine which posterior is greater, male or female. For the classification as male the posterior is given by

${\displaystyle {\text{posterior (male)}}={\frac {P({\text{male}})\,p({\text{height}}\mid {\text{male}})\,p({\text{weight}}\mid {\text{male}})\,p({\text{foot size}}\mid {\text{male}})}{evidence}}}$

For the classification as female the posterior is given by

${\displaystyle {\text{posterior (female)}}={\frac {P({\text{female}})\,p({\text{height}}\mid {\text{female}})\,p({\text{weight}}\mid {\text{female}})\,p({\text{foot size}}\mid {\text{female}})}{evidence}}}$

The evidence (also termed normalizing constant) may be calculated:

${\displaystyle {\begin{aligned}{\text{evidence}}=P({\text{male}})\,p({\text{height}}\mid {\text{male}})\,p({\text{weight}}\mid {\text{male}})\,p({\text{foot size}}\mid {\text{male}})\\+P({\text{female}})\,p({\text{height}}\mid {\text{female}})\,p({\text{weight}}\mid {\text{female}})\,p({\text{foot size}}\mid {\text{female}})\end{aligned}}}$

However, given the sample, the evidence is a constant and thus scales both posteriors equally. It therefore does not affect classification and can be ignored. The probability distribution for the sex of the sample can now be determined:

${\displaystyle P({\text{male}})=0.5}$

${\displaystyle p({\text{height}}\mid {\text{male}})={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\exp \left({\frac {-(6-\mu )^{2}}{2\sigma ^{2}}}\right)\approx 1.5789}$,

where ${\displaystyle \mu =5.855}$ and ${\displaystyle \sigma ^{2}=3.5033\cdot 10^{-2}}$ are the parameters of normal distribution which have been previously determined from the training set. Note that a value greater than 1 is OK here – it is a probability density rather than a probability, because height is a continuous variable.

${\displaystyle p({\text{weight}}\mid {\text{male}})={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\exp \left({\frac {-(130-\mu )^{2}}{2\sigma ^{2}}}\right)=5.9881\cdot 10^{-6}}$

${\displaystyle p({\text{foot size}}\mid {\text{male}})={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\exp \left({\frac {-(8-\mu )^{2}}{2\sigma ^{2}}}\right)=1.3112\cdot 10^{-3}}$

${\displaystyle {\text{posterior numerator (male)}}={\text{their product}}=6.1984\cdot 10^{-9}}$

${\displaystyle P({\text{female}})=0.5}$

${\displaystyle p({\text{height}}\mid {\text{female}})=2.23\cdot 10^{-1}}$

${\displaystyle p({\text{weight}}\mid {\text{female}})=1.6789\cdot 10^{-2}}$

${\displaystyle p({\text{foot size}}\mid {\text{female}})=2.8669\cdot 10^{-1}}$

${\displaystyle {\text{posterior numerator (female)}}={\text{their product}}=5.3778\cdot 10^{-4}}$

Since posterior numerator is greater in the female case, the prediction is that the sample is female.

Here is a worked example of naive Bayesian classification to the document classification problem. Consider the problem of classifying documents by their content, for example into spam and non-spam e-mails. Imagine that documents are drawn from a number of classes of documents which can be modeled as sets of words where the (independent) probability that the i-th word of a given document occurs in a document from class C can be written as

${\displaystyle p(w_{i}\mid C)\,}$

(For this treatment, things are further simplified by assuming that words are randomly distributed in the document - that is, words are not dependent on the length of the document, position within the document with relation to other words, or other document-context.)

Then the probability that a given document D contains all of the words ${\displaystyle w_{i}}$, given a class C, is

${\displaystyle p(D\mid C)=\prod _{i}p(w_{i}\mid C)\,}$

The question that has to be answered is: "what is the probability that a given document D belongs to a given class C?" In other words, what is ${\displaystyle p(C\mid D)\,}$?

Now by definition

${\displaystyle p(D\mid C)={p(D\cap C) \over p(C)}}$

and

${\displaystyle p(C\mid D)={p(D\cap C) \over p(D)}}$

Bayes' theorem manipulates these into a statement of probability in terms of likelihood.

${\displaystyle p(C\mid D)={\frac {p(C)\,p(D\mid C)}{p(D)}}}$

Assume for the moment that there are only two mutually exclusive classes, S and ¬S (e.g. spam and not spam), such that every element (email) is in either one or the other;

${\displaystyle p(D\mid S)=\prod _{i}p(w_{i}\mid S)\,}$

and

${\displaystyle p(D\mid \neg S)=\prod _{i}p(w_{i}\mid \neg S)\,}$

Using the Bayesian result above, one can write:

${\displaystyle p(S\mid D)={p(S) \over p(D)}\,\prod _{i}p(w_{i}\mid S)}$

${\displaystyle p(\neg S\mid D)={p(\neg S) \over p(D)}\,\prod _{i}p(w_{i}\mid \neg S)}$

Dividing one by the other gives:

${\displaystyle {p(S\mid D) \over p(\neg S\mid D)}={p(S)\,\prod _{i}p(w_{i}\mid S) \over p(\neg S)\,\prod _{i}p(w_{i}\mid \neg S)}}$

Which can be re-factored as:

${\displaystyle {p(S\mid D) \over p(\neg S\mid D)}={p(S) \over p(\neg S)}\,\prod _{i}{p(w_{i}\mid S) \over p(w_{i}\mid \neg S)}}$

Thus, the probability ratio p(S | D) / p(¬S | D) can be expressed in terms of a series of likelihood ratios. The actual probability p(S | D) can be easily computed from log (p(S | D) / p(¬S | D)) based on the observation that p(S | D) + p(¬S | D) = 1.

Taking the logarithm of all these ratios, one obtains:

${\displaystyle \ln {p(S\mid D) \over p(\neg S\mid D)}=\ln {p(S) \over p(\neg S)}+\sum _{i}\ln {p(w_{i}\mid S) \over p(w_{i}\mid \neg S)}}$

(This technique of "log-likelihood ratios" is a common technique in statistics. In the case of two mutually exclusive alternatives (such as this example), the conversion of a log-likelihood ratio to a probability takes the form of a sigmoid curve: see logit for details.)

Finally, the document can be classified as follows. It is spam if ${\displaystyle p(S\mid D)>p(\neg S\mid D)}$ (i. e., ${\displaystyle \ln {p(S\mid D) \over p(\neg S\mid D)}>0}$), otherwise it is not spam.