LECTURE 16: LEARNING THEORY

CS446 Introduction to Machine Learning (Spring 2015) University of Illinois at Urbana-Champaign http://courses.engr.illinois.edu/cs446

Prof. Julia Hockenmaier [email protected]

LECTURE 16: LEARNING THEORY

CS446 Machine Learning

Learning theory questions – Sample complexity:

How many training examples are needed for a learner to converge (with high probability) to a successful hypothesis?

– Computational complexity: How much computational effort is required for a learner to converge (with high probability) to a successful hypothesis?

– Mistake bounds: How many training examples will the learner misclassify before converging to a successful hypothesis?

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PAC learning (Probably Approximately Correct)


Terminology/Assumptions The instance space X is the set of all instances x. Assume each x is of size n. Instances are drawn i.i.d. from an unknown probability distribution D over X: x ~ D A concept c: X → {0,1} is a Boolean function (it identifies a subset of X) A concept class C is a set of concepts The hypothesis space H is the (sub)set of Boolean functions considered by the learner L We evaluate L by its performance on new instances drawn i.i.d. from D

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What can a learner learn? We can’t expect to learn concepts exactly: –  Many concepts may be consistent with the data –  Unseen examples could have any label

We can’t expect to always learn close approximations to the target concept: –  Sometimes the data will not be representative

We can only expect to learn with high probability a close approximation to the target concept.

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PAC Learning – Intuition Recall the conjunction example. We have seen many examples (drawn from D ) If x1 was active in all positive examples we have seen, it is very likely that it will be active in future positive examples In any case, x1 is active only in a small percentage of the examples so our error will be small

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-

True error of a hypothesis

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Instance space X

f h + +

−

−

-

ErrorD = x∈P D[ f (x) ≠ h(x)]

f and h disagree


True error of a hypothesis The true error (errorD(h)) of hypothesis h with respect to target concept c and distribution D is the probability that h will misclassify an instance drawn at random according to D:

errorD(h) = Px~D(c(x) ≠ h(x)) Can we bound the error based on what we know from the training data?

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What is the error for learning conjunctions? p(z): prob. that z = 0 in a positive example Claim: h only makes mistakes on positive examples. A mistake is made only if a literal z that is in h but not in the target f is false in a positive example. In this case, h will say NEG, but the example is POS.

Thus, p(z) is also the probability that z causes h to make a mistake on a randomly drawn example. Hence, Error(h) ≤∑zP(z)

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Learning Conjunctions– Analysis Call a literal z bad if p(z) > ε/n. A bad literal has a significant probability to appear with a positive example but, nevertheless, it has not appeared with one in the training data. Claim: If there are no bad literals, than error(h) < ε. Reason: What if there are bad literals ? –  Let z be a bad literal. –  What is the probability that z will not be eliminated by one example? Pr(z survives one example) = 1- Pr(z is eliminated by one example) = 1 – p(z) < 1- ε/n The probability that z will not be eliminated by m examples is therefore: Pr(z survives m independent examples) = (1 –p(z))m < (1- ε/n)m

There are at most n bad literals, so the probability that some bad literal survives m examples is bounded by n(1- ε/n)m

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Learning Conjunctions– Analysis We want this probability to be small. Say, we want to choose m large enough such that the probability that some bad z survives m examples is less than δ. Pr(z survives m example) = n(1- ε/n)m < δ Using 1-x < e-x it is sufficient to require that n e-mε/n < δ Therefore, we need m examples to guarantee a probability of failure (error > ε) of less than δ. With m examples, the probability that there are no bad literals is > 1-δ With δ=0.1, ε=0.1, and n=100, we need 6907 examples. With δ=0.1, ε=0.1, and n=10, we need 460 example, only 690 for δ=0.01

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)}/1ln(){ln( δε

+> nnm


PAC learnability Consider: –  A concept class C over a set of instances X

(each x is of length n) –  A learner L that uses hypothesis space H C is PAC-learnable by L if for all c∈ C and any distribution D over X, L will output with probability at least (1−δ) and in time that is polynomial in 1/ε, 1/δ, n and size(c), a hypothesis h ∈ H with errorD(h) ≤ ε (for 0 < δ < 0.5 and 0 < ε < 0.5)

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PAC learnability in plain English

– With arbitrarily high probability (p=1−δ), L must output a hypothesis h that has arbitrarily low error ε.

– L must learn h efficiently (using a polynomial amount of time per example, and a polynomial number of examples)

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PAC learning: intuition

A hypothesis h is bad if its true error > ε ∀x ∈ X: PrD(h(x) ≠ h*(x)) > ε

A hypothesis h looks good if it is correct on our training set S ∀s ∈ S : h(s) = h*(s) |S| = N

We want the probability that a bad hypothesis looks good to be smaller than δ

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PAC learning: intuition

We want the probability that a bad hypothesis looks good to be smaller than δ Probability of one bad h getting one x ~ XD correct:

PD(h(x) = h*(x)) ≤ 1-ε

Probability of one bad h getting m x ~ XD correct: PD(h(x) = h*(x)) ≤ (1-ε)m

Prob’ty that any h gets m x~XD correct: ≤ ∣H∣(1-ε)m Exclusive union bound: P(A ∨ B) ≤ Pr(A) + Pr(B) Set ∣H∣(1-ε)m ≤ δ, solve for m

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Sample complexity (for finite hypothesis spaces and consistent learners)

Consistent learner: returns hypotheses that perfectly fit the training data (whenever possible).

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Version space VSH,D

The version space VSH,D is the set of all hypotheses h∈ H that correctly classify the training data D: VSH,D = { h∈ H | ∀〈x,c(x)〉∈ D: h(x) = c(x) } Every consistent learner outputs a hypothesis h belonging to the version space. We need to only bound the number of examples needed to assure that the version space does not contain any unacceptable hypotheses

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Sample complexity (finite H) –  The version space VSH,D is said to be ε-exhausted with

respect to concept c and distribution D if every h∈ VSH,D has true error < ε with respect to c and distribution D

–  If H is finite, and the data D is a sequence of m i.i.d. samples of c, then for any 0 ≤ ε ≤ 1, the probability that VSH,D is not ε-exhausted with respect to c is ≤ |H|e-εm

–  #training examples required to reduce probability of failure below δ: Find m such that |H|e-εm < δ

–  So, a consistent learner needs m ≥ 1/ε (ln |H| + ln(1/δ)) examples to get an error below δ (often an overestimate; |H| can be very large)

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Occam’s Razor (1) Claim: The probability that there exists a hypothesis h ∈ H that (1) is consistent with m examples and (2) satisfies error(h) > ε ( ErrorD(h) = Prx 2 D [f(x) !=h(x)] ) is less than |H|(1- ε )m .

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Occam’s Razor (1) Proof: Let h be a bad hypothesis. Probability that h is consistent with one example f(x): Px~D(f(x) = h(x)) < 1−ε Since the m examples are drawn independently, the probability that h is consistent with all m examples is less than (1−ε)m The probability that some hypothesis in H is consistent with m examples is smaller than |H|(1−ε)m 20

Note that we don’t need a true f for this argument; it can be done with h, rela:ve to a distribu:on over X £ Y.


Consistent learners Immediately from the definition, we get the following general scheme for PAC learning: Given a sample D of m examples –  Find some h ∈ H that is consistent with all m examples

We showed that if m is large enough, a consistent hypothesis must be close enough to f Check that m is not too large (polynomial in the relevant parameters) : we showed that the “closeness” guarantee requires that

m > 1/² (ln |H| + ln 1/±) –  Show that the consistent hypothesis h ∈ H can be computed efficiently In the case of conjunctions –  We used the Elimination algorithm to find a hypothesis h that is consistent

with the training set (easy to compute) –  We can show directly that if we have sufficiently many examples (polynomial

in the number of variables), than h is close to the target function.

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Vapnik-Chervonenkis (VC) dimension

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Infinite Hypothesis Space The previous analysis was restricted to finite hypothesis spaces Some infinite hypothesis spaces are more expressive than others –  E.g., Rectangles, vs. 17- sides convex polygons vs.

general convex polygons –  Linear threshold function vs. a conjunction of LTUs Need a measure of the expressiveness of an infinite hypothesis space other than its size The Vapnik-Chervonenkis dimension (VC dimension) provides such a measure. Analogous to |H|, there are bounds for sample complexity using VC(H)


VC dimension (basic idea) The VC dimension of a hypothesis space H measures the complexity of H by the number of distinct instances from X that can be completely discriminated (‘shattered’) using H, not by the number of distinct hypotheses (|H|).

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VC dimension (basic idea) An unbiased hypothesis space H shatters the entire instance space X (is able to induce every possible partition on the set of all possible instances) The larger the subset X that can be shattered, the more expressive a hypothesis space is, i.e., the less biased.

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Shattering a set of instances A set of instances S is shattered by the hypothesis space H if and only if for every dichotomy of S there is a hypothesis h in H that is consistent with this dichotomy. –  dichotomy: partition instances in S into + and – –  one dichotomy = label all instances in a subset P⊆ S as +, and instances in the complement of P, S\P as –

The ability of H to shatter S is a measure of its capacity to represent concepts over S

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Instances: Real numbers Hypothesis space: Intervals [a, b] defines the positive instances We can shatter any dataset of two reals, but we cannot shatter datasets of three reals


Shattering

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- - + a b


VC dimension of H The VC dimension of the hypothesis space H, VC(H), is the size of the largest finite subset of the instance space X that can be shattered by H. If arbitrarily large finite subsets of X can be shattered by X then VC(H) = ∞

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VC Dimension The VC dimension of hypothesis space H over instance space X is the size of the largest finite subset of X that is shattered by H. – If there exists one (or more) subsets of size

d that can be shattered, then VC(H) ≥ d – If no subset of size d can be shattered,

then VC(H) < d

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|S| = 4

|S| = 2 |S| = 1

VC Dimension of linear classifiers in 2 dimensions

The VC dimension of a 2-d linear classifier is 3: The largest set of points that can be labeled arbitrarily

Note that |H| is infinite, but expressiveness is quite low. 30

|S| = 3

✔

✗

✔ ✔


VC dimension if H is finite If H is finite: VC(H) ≤ log2|H| – A set S with d instances has 2d distinct

subsets/dichotomies. – Hence, H requires 2d distinct hypotheses

to shatter d instances. – If VC(H) = d: 2d ≤ |H|

hence: VC(H) = d ≤ log2|H|

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COLT Conclusions The PAC framework provides a reasonable model for theoretically analyzing the effectiveness of learning algorithms. The sample complexity for any consistent learner using the hypothesis space, H, can be determined from a measure of H’s expressiveness (|H|, VC(H)) If the sample complexity is tractable, then the computational complexity of finding a consistent hypothesis governs the complexity of the problem. Sample complexity bounds given here are far from being tight, but separates learnable classes from non-learnable classes (and show what’s important). Computational complexity results exhibit cases where information theoretic learning is feasible, but finding good hypothesis is intractable. The theoretical framework allows for a concrete analysis of the complexity of learning as a function of various assumptions (e.g., relevant variables)

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LECTURE 16: LEARNING THEORY