ECE595 / STAT598: Machine Learning I Lecture 28 Sample …...E.g., linear in 2D: N = 3 is okay, but...

ECE595 / STAT598: Machine Learning ILecture 28 Sample and Model Complexity

Spring 2020

Stanley Chan

School of Electrical and Computer EngineeringPurdue University

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Outline

Lecture 28 Sample and Model Complexity

Lecture 29 Bias and Variance

Lecture 30 Overfit

Today’s Lecture:

Generalization Bound using VC Dimension

Review of growth function and VC dimensionGeneralization bound

Sample and Model Complexity

Sample complexityModel complexityTrade off

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VC Dimension

Definition (VC Dimension)

The Vapnik-Chervonenkis dimension of a hypothesis set H, denoted by dVC, is the largestvalue of N for which H can shatter all N training samples.

You give me a hypothesis set H, e.g., linear model

You tell me the number of training samples N

Start with a small N

I will be able to shatter for a while, until I hit a bump

E.g., linear in 2D: N = 3 is okay, but N = 4 is not okay

So I find the largest N such that H can shatter N training samples

E.g., linear in 2D: dVC = 3

If H is complex, then expect large dVC

Does not depend on p(x), A and f

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VC Dimension

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VC Dimension

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VC Dimension

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VC Dimension

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VC Dimension

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VC Dimension

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VC Dimension

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VC Dimension

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VC Dimension

Does not depend on p(x), A and f3 / 17

Linking the Growth Function

Theorem (Sauer’s Lemma)

Let dVC be the VC dimension of a hypothesis set H, then

mH(N) ≤dVC∑i=0

). (1)

I skip the proof here. See AML Chapter 2.2 for proof.What is more interesting is this:

dVC∑i=0

)≤ NdVC + 1.

This can be proved by simple induction. Exercise.So together we have

mH(N) ≤ NdVC + 1.

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mH(N) ≤dVC∑i=0

). (1)

I skip the proof here. See AML Chapter 2.2 for proof.

What is more interesting is this:

dVC∑i=0

)≤ NdVC + 1.

mH(N) ≤ NdVC + 1.

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mH(N) ≤dVC∑i=0

). (1)

dVC∑i=0

)≤ NdVC + 1.

This can be proved by simple induction. Exercise.

So together we havemH(N) ≤ NdVC + 1.

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mH(N) ≤dVC∑i=0

). (1)

dVC∑i=0

)≤ NdVC + 1.

This can be proved by simple induction. Exercise.

So together we havemH(N) ≤ NdVC + 1.

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mH(N) ≤dVC∑i=0

). (1)

dVC∑i=0

)≤ NdVC + 1.

mH(N) ≤ NdVC + 1.4 / 17

Difference between VC and Hoeffding

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Generalization Bound Again

Recall the generalization bound

Ein(g)−√

δ≤ Eout(g) ≤ Ein(g) +

Substitute M by mH(N), and then mH(N) ≤ NdVC + 1:

Eout(g) ≤ Ein(g) +

2(NdVC + 1)

Wonderful!

Everything is characterized by δ, N and dVC

dVC tells us the expressiveness of the model

You can also think of dVC as the effective number of parameters

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Ein(g)−√

2(NdVC + 1)

Wonderful!

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Ein(g)−√

2(NdVC + 1)

Wonderful!

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Ein(g)−√

2(NdVC + 1)

Wonderful!

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Ein(g)−√

2(NdVC + 1)

Wonderful!

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Ein(g)−√

2(NdVC + 1)

Wonderful!

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Ein(g)−√

2(NdVC + 1)

Wonderful!

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

If this is the case, then the final hypothesis g ∈ H will generalize.

dVC =∞,

Then His as diverse as it can be

It is not possible to generalize

Message 1: If you choose a complex model, then you need to pay the price of trainingsample

Message 2: If you choose an extremely complex model, then it may not be able togeneralize regardless the number of samples

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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If dVC <∞,

Then as N →∞,

2(NdVC + 1)

δ→ 0.

dVC =∞,

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Generalizing the Generalization Bound

Theorem (Generalization Bound)

For any tolerance δ > 0

4mH(2N)

with probability at least 1− δ.

Some small subtle technical requirements. See AML chapter 2.2

How tight is this generalization bound? Not too tight.

The Hoeffding inequality has a slack. The inequality is too general for all values of Eout

The growth function mH(N) gives the worst case scenario

Bounding mH(N) by a polynomial introduces slack

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4mH(2N)

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4mH(2N)

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4mH(2N)

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4mH(2N)

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4mH(2N)

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Outline

Lecture 28 Sample and Model Complexity

Lecture 29 Bias and Variance

Lecture 30 Overfit

Today’s Lecture:

Generalization Bound using VC Dimension

Review of growth function and VC dimensionGeneralization bound

Sample complexityModel complexityTrade off

9 / 17

Sample Complexity

What is the smallest number of samples required?

Required to ensure training and testing error are close

Close = within certain ε, with confidence 1− δRegardless of what learning algorithm you use

Model Complexity

What is the largest model you can use?

Refers to the hypothesis set

With respect to the number of training samples

Largest = measured in terms of VC dimension

Can use = within certain ε, with confidence 1− δRegardless of what learning algorithm you use

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Sample Complexity

Model Complexity

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Sample Complexity

Model Complexity

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Sample Complexity

Close = within certain ε, with confidence 1− δ

Regardless of what learning algorithm you use

Model Complexity

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Sample Complexity

Model Complexity

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Sample Complexity

Model Complexity

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Sample Complexity

Model Complexity

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Sample Complexity

Model Complexity

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Sample Complexity

Model Complexity

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Sample Complexity

Model Complexity

Can use = within certain ε, with confidence 1− δ

Regardless of what learning algorithm you use

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Sample Complexity

Model Complexity

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Sample Complexity

The generalization bound is

4mH(2N)

If you want the generalization error to be at most ε, then√8

4mH(2N)

δ≤ ε.

Rearrange terms and use VC dimension,

N ≥ 8

ε2log

(4(2N)dVC + 1

Example. dVC = 3. ε = 0.1. δ = 0.1 (90% confidence). Then the number of samples weneed is

N ≥ 8

0.12log

(4(2N)3 + 4

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Sample Complexity

4mH(2N)

δ≤ ε.

N ≥ 8

ε2log

(4(2N)dVC + 1

N ≥ 8

0.12log

(4(2N)3 + 4

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Sample Complexity

4mH(2N)

δ≤ ε.

N ≥ 8

ε2log

(4(2N)dVC + 1

N ≥ 8

0.12log

(4(2N)3 + 4

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Sample Complexity

4mH(2N)

δ≤ ε.

N ≥ 8

ε2log

(4(2N)dVC + 1

N ≥ 8

0.12log

(4(2N)3 + 4

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Sample Complexity

4mH(2N)

δ≤ ε.

N ≥ 8

ε2log

(4(2N)dVC + 1

N ≥ 8

0.12log

(4(2N)3 + 4

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Sample Complexity

How to solve for N in this equation?

N ≥ 8

0.12log

(4(2N)3 + 4

Put N = 1000 to the right hand side

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

Not enough. So put N = 21, 193 to the right hand side. Iterate.

Then we get N ≈ 30, 000.

So we need at least 30,000 samples.

However, generalization bound is not tight. So our estimate is over-estimate.

Rule of thumb, 10× dVC.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

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Sample Complexity

N ≥ 8

0.12log

(4(2N)3 + 4

N ≥ 8

0.12log

(4(2× 1000)3 + 4

)≈ 21, 193.

Rule of thumb, 10× dVC.12 / 17

Error Bar

(4 ((2N)dVC + 1)

What error bar can we offer?

Example. N = 100. δ = 0.1 (90% confidence). dVC = 1.

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

If we use N = 1000, thenEout(g) ≤ Ein(g) + 0.301.

Somewhat more respectable estimate.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

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Error Bar

(4 ((2N)dVC + 1)

100log

(4 ((2× 100) + 1)

)≈ Ein(g) + 0.848.

Close to useless.

Somewhat more respectable estimate.13 / 17

Model Complexity

4 ((2N)dVC + 1)

δ︸︷︷︸=ε(N,H,δ)

ε(N,H, δ) = penalty of the model complexity

If dVC is large, then ε(N,H, δ) is big

So the generalization error is large

There is a trade-off curve

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Model Complexity

4 ((2N)dVC + 1)

δ︸︷︷︸=ε(N,H,δ)

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Model Complexity

4 ((2N)dVC + 1)

δ︸︷︷︸=ε(N,H,δ)

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Model Complexity

4 ((2N)dVC + 1)

δ︸︷︷︸=ε(N,H,δ)

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Model Complexity

4 ((2N)dVC + 1)

δ︸︷︷︸=ε(N,H,δ)

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Model Complexity

4 ((2N)dVC + 1)

δ︸︷︷︸=ε(N,H,δ)

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Trade-off Curve

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Generalization Bound for Testing

Testing Set: Dtest = {x1, . . . , xL}.The final hypothesis g is already determined. So no need to use Union bound.

The Hoeffding is as simple as

P{|Ein(g)− Eout(g)| > ε

}≤ 2e−2ε

If you have a lot of testing samples, then Ein(g) will be good estimate of Eout(g)

Independent of model complexity

Only δ and L

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Testing Set: Dtest = {x1, . . . , xL}.

The final hypothesis g is already determined. So no need to use Union bound.

}≤ 2e−2ε

Only δ and L

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}≤ 2e−2ε

Only δ and L

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}≤ 2e−2ε

Only δ and L

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}≤ 2e−2ε

Only δ and L

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}≤ 2e−2ε

Only δ and L

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}≤ 2e−2ε

Only δ and L

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}≤ 2e−2ε

Only δ and L16 / 17

Reading List

Yasar Abu-Mostafa, Learning from Data, chapter 2.1

Mehrya Mohri, Foundations of Machine Learning, Chapter 3.2

Stanford Note http://cs229.stanford.edu/notes/cs229-notes4.pdf

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ECE595 / STAT598: Machine Learning I Lecture 28 Sample …...E.g., linear in 2D: N = 3 is okay, but...

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