stat841f11

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STAT 441/841 / CM 463/763 - Tuesday, 2011/09/20

Wiki Course Notes

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Classification

definitions

classification: Predict a discrete random variable [math]Y[/math] by using another random variable [math]X[/math] iid

[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 make new [math]Y[/math] prediction [math]h(X)[/math].

Example: an image (2-d), which can be represented by a long vector [math]\mathbf{X}[/math]


true error rate for classifier [math]h[/math]

[math]L(h) = (P(h(X) \neq Y )[/math]


empirical error rate (or training error rate)

[math]\hat{L}_n(h) = (1/n) \sum_{i=1}^{n} I(h(X_i) \neq Y_i)[/math]

where [math]I()[/math] indicates misclassification, i.e.,

[math]I = 1 \ \rightarrow \ h()[/math] misclassifies this [math]X[/math]

[math]I = 0 \ \rightarrow \ h()[/math] does not misclassify this [math]X[/math]

e.g., 100 new data points with known (true) labels

[math]y_1 = h(x_1)[/math]

...

[math]y_{100} = h(x_{100})[/math]

compare the known-true against the classified-true, count, 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

P(X|Y)  : likelihood

P(Y)  : prior

P(X)  : marginal

We will start with the simplest case: [math]\mathcal{Y} = \{0,1\}[/math]

[math]r(x) = P(Y=1|X=x)[/math] [math] = \frac{P(X=x|Y=1) P(Y=1)} {P(X=x)}[/math] [math] = \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]

Which of the following do we prefer to work with,

  • P(Y=1|X=x) or
  • P(X=x|Y=1) ?

The former reflects a Bayesian approach - degree of belief. So, Y is not a RV. The latter reflects a frequentist approach - frequency (of observation). (Ease of computation and validity were proposed and discussed by students.)

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) ...} {...} = \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 problem is more difficult, because of restricted domain (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]

Problems?

not enough data to estimate? - possibly

high error rate?

big one: not good in high-dimensional space


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 ...}[/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) = [(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)} with \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{\mu_1}^T \Sigma^{-1} \mathbf{\mu_1} + 2\mathbf{\mu_1}^T \Sigma^{-1} \mathbf{x} + \mathbf{x}^T \Sigma^{-1} + \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]


Which reduces to [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_1}^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].

Jgpitt - 2011/09/21


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