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Delve into the realm of Vapnik-Chervonenkis dimension to explore lower and upper bounds in PAC learning models, dealing with hypotheses and concept projections.
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Vapnik-Chervonenkis Dimension Part II: Lower and Upper bounds
PAC Learning model • There exists a distribution D over domain X • Examples: <x, c(x)> • Goal: • With high probability (1-d) • find h in H such that • error(h,c ) < e
Definitions: Projection • Given a concept c over X • associate it with a set (all positive examples) • Projection (sets) • For a concept class C and subset S • PC(S) = { c S | c C} • Projection (vectors) • For a concept class C and S = {x1, … , xm} • PC(S) = {<c(x1), … , c(xm)> | c C}
Definition: VC-dim • Clearly |PC(S) | 2m • C shatters S if |PC(S) | =2m • VC dimension of a class C: • The size d of the largest set S that shatters C. • Can be infinite. • For a finite class C • VC-dim(C) log |C|
Lower bounds: Setting • Static learning algorithm: • asks for a sample S of size m(e,d) • Based on S selects a hypothesis
Lower bounds: Setting • Theorem: • If VC-dim(C) = then C is not learnable. • Proof: • Let m = m(0.1,0.1) • Find 2m points which are shattered (set T) • Let D be the uniform distribution on T • Set ct(xi)=1 with probability ½. • Expected error ¼. • Finish proof!
Lower Bound: Feasible • Theorem • VC-dim(C)=d+1, then m(e,d)=W(d/e) • Proof: • Let T be a set of d+1 points which is shattered. • Let the distribution D be: • z0 with prob. 1-8e • zi with prob. 8e/d
Continue • Set ct(z0)=1 and ct(zi)=1 with probability ½ • Expected error 2e • Bound confidence • for accuracy e
Lower Bound: Non-Feasible • Theorem • For two hypotheses m(e,d)=W((log 1/d)/e2) • Proof: • Let H={h0, h1}, where hb(x)=b • Two distributions: • D0: Pr[<x,1>]= ½ - g and Pr[<y,0>]= ½ + g • D1: Pr[<x,1>]= ½ + g and Pr[<y,0>]= ½ - g
Epsilon net • Epsilon bad concepts • Be ( c ) = { h | error(h,c) >e } • A set of points S is an e-net w.r.t. D if • for every hin Be ( c ) • there exists a point x in S • such that h(x) c(x)
Sample size • Event A: • The sample S1 is not an epsilon net, |S1|=m. • Assume A holds • Let h be a epsilon-bad consistent hypothesis. • Sample an additional sample S2 • with probability at least 1/2 • the errors of h on S2 is em/2 • for m=|S2|= O(1/e)
continues • Event B • There exists h in Be ( c ) • and h consistent with S1 • h has e m/2 errors on S2 • Pr[ B | A ] 1/2 • 2 Pr[B] P[A] • Let F be the projection of C to S1 S2 • F=PC(S1 S2 )
Error set • ER(h)={ x : x S1 S2 and c(x)=h(x)} • |ER(h)| em/2 • Event A: • ER(h) S1 = • Event B: • ER(h) S1 = • ER(h) S2= ER(h)
Combinatorial problem • 2m black and white balls • exactly l black balls • Consider a random partition to S1 and S2 • The probability that all the black balls in S2
Completing the proof • Probability of B • Pr[B] |F| 2-l |F| 2-em/2 • Probability of A • Pr[A] Pr[B] |F| 2-em/2 • Confidence d Pr[A] • Sample • m=O( (1/e) log 1/ d + (1/e) log |F| ) • Need to bound |F| !!!
Bounding |F| • Define: • J(m,d)=J(m-1,d) + J(m-1,d-1) • J(m,0)=1 and J(0,d)=1 • Solving the recursion • Claim: • Let VC-dim(C)=d and |S|=m, • then |PC(S)| J(m,d)