stat841f11
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STAT 441/841 / CM 463/763 - Tuesday, 2011/09/20
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Classification
Definitions
classification: Predict a discrete random variable [math]Y[/math] (a label) by using another random variable [math]X[/math] (new data point) picked iid from a distribution
[math]X_i = (X_{i1}, X_{i2}, ... X_{id}) \in \mathcal{X} \subset \mathbb{R}^d[/math] ([math]d[/math]-dimensional vector) [math]Y_i[/math] in some finite set [math]\mathcal{Y}[/math]
classification rule:
[math]h : \mathcal{X} \rightarrow \mathcal{Y}[/math]
Take new observation [math]X[/math] and use a classification function [math]h(x)[/math] to generate a label [math]Y[/math]. In other words, If we fit the function [math]h(x)[/math] with a random variable [math]X[/math], it generates the label [math]Y[/math] which is the class to which we predict [math]X[/math] belongs.
Example: Let [math] \mathcal{X}[/math] be a set of 2D images and [math]\mathcal{Y}[/math] be a finite set of people. We want to learn a classification rule [math]h:\mathcal{X}\rightarrow\mathcal{Y}[/math] that with small true error predicts the person who appears in the image.
true error rate for classifier [math]h[/math] is the error with respect to the underlying distribution (that we do not know).
[math]L(h) = P(h(X) \neq Y )[/math]
empirical error rate (or training error rate) is the amount of error that our classification function [math]h(x)[/math] makes on the training data.
[math]\hat{L}_n(h) = (1/n) \sum_{i=1}^{n} \mathbf{I}(h(X_i) \neq Y_i)[/math]
where [math]\mathbf{I}()[/math] is an indicator function. Indicator function is defined by
[math]\mathbf{I}(x) = \begin{cases} 1 & \text{if } x \text{ is true} \\ 0 & \text{if } x \text{ is false} \end{cases}[/math]
So in this case [math]\mathbf{I}(h(X_i)\neq Y_i) = 1[/math] when [math]h(X_i)\neq Y_i[/math] i.e. when misclassifications happens.
e.g., 100 new data points with known (true) labels
[math]y_1 = h(x_1)[/math]
...
[math]y_{100} = h(x_{100})[/math]
To calculate the empirical error we count how many labels our function [math]h(x)[/math] assigned incorrectly and divide by n=100
Bayes Classifier
First recall Bayes' Rule, in the format [math]P(Y|X) = \frac{P(X|Y) P(Y)} {P(X)} [/math]
P(Y|X) : posterior , probability of [math]Y[/math] given [math]X[/math]
P(X|Y) : likelihood, probability of [math]X[/math] generating [math]Y[/math] or probability of [math]Y[/math] being generated by [math]X[/math]
P(Y) : prior, probability of [math]Y[/math] being selected
P(X) : marginal, probability of obtaining [math]X[/math]
We will start with the simplest case: [math]\mathcal{Y} = \{0,1\}[/math]
[math] r(x) = P(Y=1|X=x) = \frac{P(X=x|Y=1) P(Y=1)} {P(X=x)} = \frac{P(X=x|Y=1) P(Y=1)} {P(X=x|Y=1) P(Y=1) + P(X=x|Y=0) P(Y=0)}[/math]
Bayes' rule can be approached by computing either:
- the posterior: P(Y=1|X=x) and P(Y=0|X=x) or
- the likelihood: P(X=x|Y=1) and P(X=x|Y=0)
The former reflects a Bayesian approach - based on the degree of belief. The latter reflects a Frequentist approach - based on the frequency of observation.
There was some class discussion on which approach should be used. Both the ease of computation and the validity of both approaches were discussed. Overall, frequentists consider X to be a random variable. But they do not consider Y to be a random variable, because it has to take on one of the values from a fixed set (in the above case it would be either 0 or 1 and there is only one correct label for a given value X=x). Thus, from a frequentist's perspective it does not make sense to talk about the probability of Y.
This is actually a grey area and sometimes Bayesians and Frequentists use each others' approaches. So using Bayes' rule doesn't necessarily mean you're a Bayesian. Overall, the question remains unresolved.
The Bayes Classifier uses [math]P(Y=1|X=x)[/math]
[math] P(Y=1|X=x) = \frac{P(X=x|Y=1) P(Y=1)} {P(X=x|Y=1) P(Y=1) + P(X=x|Y=0) P(Y=0)}[/math]
P(Y=1) : the prior, based on belief/evidence beforehand
denominator : marginalized by summation
[math]h(x) = \begin{cases} 1 \ \ \hat{r}(x) \gt 1/2 \\ 0 \ \ otherwise \end{cases} [/math]
The set [math]\mathcal{D}(h) = \{ x : P(Y=1|X=x) = P(Y=0|X=x)... \} [/math]
which defines a decision boundary.
Theorem: Bayes rule is optimal. I.e., if h is any other classification rule,
then [math]L(h^*) \lt = L(h)[/math]
(This is to be proved in homework.)
Therefore, Why do we need any other method? A: Because X densities are often/typically unknown. I.e., [math]f_k(x)[/math] and/or [math]\pi_k[/math] unknown.
[math]P(Y=k|X=x) = \frac{P(X=x|Y=k)P(Y=k)} {P(X=x)} = \frac{f_k(x) \pi_k} {\sum_k f_k(x) \pi_k}[/math] f_k(x) is referred to as the class conditional distribution (~likelihood).
Therefore, we rely on some data to estimate quantities.
Three Main Approaches
1. Empirical Risk Minimization: Choose a set of classifier H (e.g., line, neural network) and find [math]h^* \in H[/math] that minimizes (some estimate of) L(h).
2. Regression: Find an estimate ([math]\hat{r}[/math]) of function [math]r[/math] and define [math]h(x) = \begin{cases} 1 \ \ \hat{r}(x) \gt 1/2 \\ 0 \ \ otherwise \end{cases} [/math]
The [math] 1/2 [/math] in the expression above is a threshold set for the regression prediction output.
In general regression refers to finding a continuous, real valued y. The problem here is more difficult, because of the restricted domain (y is a set of discrete label values).
3. Density Estimation: Estimate [math]P(X=x|Y=0)[/math] from [math]X_i[/math]'s for which [math]Y_i = 0[/math] Estimate [math]P(X=x|Y=1)[/math] from [math]X_i[/math]'s for which [math]Y_i = 1[/math] and let [math]P(Y=?) = (1/n) \sum_{i=1}^{n} Y_i[/math]
Define [math]\hat{r}(x) = \hat{P}(Y=1|X=x)[/math] and [math]h(x) = \begin{cases} 1 \ \ \hat{r}(x) \gt 1/2 \\ 0 \ \ otherwise \end{cases} [/math]
It is possible that there may not be enough data to estimate from for density estimation. But the main problem lies with high dimensional spaces, as the estimation results may not be good (high error rate) and sometimes even infeasible. The term curse of dimensionality was coined by Bellman <ref>R. E. Bellman, Dynamic Programming. Princeton University Press, 1957</ref> to describe this problem.
As the dimension of the space goes up, the learning requirements go up exponentially.
Multi-Class Classification
Generalize to case Y takes on k>2 values.
Theorem: [math]Y \in \mathcal{Y} = {1,2,..., k}[/math] optimal rule
[math]h*(x) = argmax_k P[/math]
where [math]P(Y=k|X=x) = \frac{f_k(x) \pi_k} {\sum_r f_r \pi_r}[/math]
LDA and QDA
(linear discriminant analysis, and quadratic discriminant analysis)
Simplest: Use approach 3 (above) and assume a parametric model for densities. Assume class conditional is Gaussian.
LDA (also known as FDA (Fisher's), which is in fact not really the same thing)
[math]\mathcal{Y} = \{ 0,1 \}[/math] assumed (i.e., 2 labels)
[math]h(x) = \begin{cases} 1 \ \ P(Y=1|X=x) \gt P(Y=0|X=x) \\ 0 \ \ otherwise \end{cases} [/math]
[math]P(Y=1|X=x) = \frac{f_1(x) \pi_1} {\sum_k f_k \pi_k} \ \ [/math] (denom = P(x))
1) Assume Gaussian distributions
[math]f_k(x) = \frac{1}{(2\pi)^{-d/2} |\Sigma_k|^{-1/2}} exp(-(1/2)(\mathbf{x} - \mathbf{\mu_k}) \Sigma_k^{-1}(\mathbf{x}-\mathbf{\mu_k}) )[/math]
must compare [math]\frac{f_1(x) \pi_1} {p(x)}[/math] with [math]\frac{f_0(x) \pi_0} {p(x)}[/math] Note that the p(x) denom can be ignored: [math]f_1(x) \pi_1[/math] with [math]f_0(x) \pi_0 [/math]
To find the decision boundary, set [math]f_1(x) \pi_1 = f_0(x) \pi_0 [/math]
Because we are assuming [math]\Sigma_1 = \Sigma_0[/math], we can use [math]\Sigma = \Sigma_0 = \Sigma_1[/math].
Cancel [math](2\pi)^{-d/2} |\Sigma_k|^{-1/2}[/math] from both sides.
Take log of both sides.
Subtract one side from both sides, leaving zero on one side.
[math]-(1/2)(\mathbf{x} - \mathbf{\mu_1})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu_1}) + log(\pi_1) - [-(1/2)(\mathbf{x} - \mathbf{\mu_0})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu_0}) + log(\pi_0)] = 0 [/math]
[math](1/2)[-\mathbf{x}^T \Sigma^{-1}\mathbf{x} - \mathbf{\mu_1}^T \Sigma^{-1} \mathbf{\mu_1} + 2\mathbf{\mu_1}^T \Sigma^{-1} \mathbf{x}
+ \mathbf{x}^T \Sigma^{-1}\mathbf{x} + \mathbf{\mu_0}^T \Sigma^{-1} \mathbf{\mu_0} - 2\mathbf{\mu_0}^T \Sigma^{-1} \mathbf{x} ]
+ log(\pi_1/\pi_0) = 0 [/math]
Cancelling out the terms quadratic in [math]\mathbf{x}[/math] and rearranging results in
[math](1/2)[-\mathbf{\mu_1}^T \Sigma^{-1} \mathbf{\mu_1} + \mathbf{\mu_0}^T \Sigma^{-1} \mathbf{\mu_0} + (2\mathbf{\mu_1}^T \Sigma^{-1} - 2\mathbf{\mu_0}^T \Sigma^{-1}) \mathbf{x}] + log(\pi_1/\pi_0) = 0 [/math]
And we see that the first pair of terms is constant, and the second pair is linear on x.
So that we end up with something of the form
[math]ax + b = 0[/math].
References
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Jgpitt - 2011/09/21