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1. ASSIGNMENT 4:
VC DIMENSION
Institute for Machine Learning

2. Contact
Heads:
Markus Holzleitner,
Andreas Radler
————
Institute for Machine Learning
Johannes Kepler University
Altenberger Str. 69
A-4040 Linz
————
E-Mail: theoretical@ml.jku.at
Only mails to this list are answered!
Institute Homepage
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3. Copyright statement:
This material, no matter whether in printed or electronic form,
may be used for personal and non-commercial educational use
only. Any reproduction of this material, no matter whether as a
whole or in parts, no matter whether in printed or in electronic
form, requires explicit prior acceptance of the authors.
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4. Setting
Data points Z = (xi, yi)l
i=1 are sampled iid. from p(x, y)
supported in X × {−1, 1}
Want to learn g : X → {−1, 1} so that expected loss
(according to given loss function) is minimal. We will only
use Lzo in this chapter.
Goal: minimize associated risk/generalization error:
R(g) =
R
X
P
y∈{±1} (L(y, g(x))p(x, y)) dx
Also important: empirical risk:
Remp(g, Z) = Remp(g, l) = 1
l
l
P
i=1
L(yi, g(xi))
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5. Hoeffding’s inequality
Lemma (Hoeffding)
Let X1, ..., Xl be independent random variables drawn accord-
ing to p. Assume further that Xi ∈ [mi, Mi]. Then for t ≥ 0:
p
l
X
i=1
(Xi − E(Xi)) ≥ t
!
≤ exp −
2t2
Pl
i=1(Mi − mi)2
!
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6. Generalization bound: finite function
classes
First step (one single model): Apply Hoeffding to
Xi = L(yi, g(xi)), E(Xi) = R(g) for fixed g ∈ G. Then
mi = 0, Mi = 1 (for all i = 1, ..., l) and for any ε 0:
p (|Remp(g, l) − R(g)| ≥ ε) = p |
l
X
i=1
(Xi − E(Xi))| ≥ lε
!
≤ 2 exp(−2lε2
).
Lemma (Generalization bound: finite model classes)
Let |G| = m. Choose failure probability 0 δ 1. Then with
probability at least 1 − δ for all g ∈ G:
R(g) ≤ Remp(g, l) +
r
ln(2m) + ln(1/δ)
2l
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7. What does this result mean?
Bound the true risk by empirical risk plus capacity term
If function class increases, bound gets worse
If m is small enough compared to l (so that ln m
l is small),
we get a tight bound
The whole bound holds with probability 1 − δ. Decreasing δ
worsens the bound
These arguments break down if |G| = ∞. For this case we
need new ideas.
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8. Shattering coefficient: definition
Definition (Shattering coefficient)
For given sample x1, . . . , xl ∈ X and function class G define
Gx1,...,xl
as set of functions on G that we get when restricting
G to x1, . . . , xl:
Gx1,...,xl
= {g|x1,...,xl
: g ∈ G}
The shattering coefficient N(G, l) of G is defined as maximal
number of functions in Gx1,...,xl
.
N(G, l) = max{|Gx1,...,xl
| : x1, . . . , xl ∈ X}
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9. Shattering coefficient: main result
Theorem (Generalization bound: shattering coefficient)
Let G be an arbitrary function class. Then for 0 1 :
p supg∈G |Remp(g, l) − R(g)| ε
≤ 2N(G, 2l)e
−lε2
4 .
In other words: with probability at least 1−δ all functions g ∈ G
satisfy:
R(g) ≤ Remp(g, l) + 2
r
ln(N(G, 2l)) + ln(1/δ)
l
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10. Symmetrization Lemma
Notation:
Remp(g, l): empirical risk of given sample of l points
R0
emp(g, l): empirical risk of second, independent sample of
l points: Ghost sample
Lemma (Symmetrization)
For ε 2
l :
p(sup
g∈G
|Remp(g, l) − R(g)| ε)
≤ 2p(sup
g∈G
|Remp(g, l) − R0
emp(g, l)|
ε
2
).
Proof can be found e.g. here (Lemma 7.63, see also
notes). 9/14

11. Why symmetrization?
If two g, g̃ coincide on all points of original and ghost
sample: Remp(g, l) = Remp(g̃, l) and R0
emp(g, l) = R0
emp(g̃, l)
→ sup over G in fact only runs over finitely many fcts: all
possible binary fcts on two samples of size l → number of
such fcts bounded by N(G, 2l).
Bound analogous to one with finite function classes, just
replace m by N(G, 2l)
Intuitively: shattering coefficient measures how powerful
fct. class is, how many labelings of dataset it can realize.
For consistency: need ln N(G,2l)
l −
−
−
→
l→∞
0.
However: shattering coefficients difficult to deal with. Need
to now how they grow in l. Study now a tool that helps in
this regard.
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12. Definition: Shattering and VC-dimension
Definition (Shattering)
G shatters a set of points x1, ..., xl, if G can realize all possible
labelings, i.e. |Gx1,...,xl
| = 2l.
Definition (VC-Dimension (from Vapnik-Chervonenkis))
The VC-dimension of G is defined as largest l, so that there
exists a sample of size l that can be shattered by G:
VC(G) = max
n
l ∈ N|∃x1, ..., xl s.t. |Gx1,...,xl
| = 2l
o
.
If max does not exist: VC(G) = ∞.
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13. VC-dimension: examples
X = R, positive class=interior of closed interval, i.e.
G =
1[a,b] : a b ∈ R .
Positive class=interior of right triangles with sides adjacent
to right angle are parallel to aces. Right angle in lower left
corner. X = R2, G = {indicators of right triangles}.
Positive class=interior of convex polygon, X = R2,
G = {indicators of convex polygons with d corners}
X = R, G = {sgn (sin(tx)) : t ∈ R}. Then VC(G) = ∞
X = Rr, G = {area above linear hyperplane}. Show in
exercises: VC(G) = r + 1
X = Rr, ρ 0, G = {hyperplanes with margins at least γ}.
One can prove: if data are restricted to ball of radius R:
VC(G) = min
r, 2R2
γ2
+ 1.
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14. Why VC-dimension? Sauer’s Lemma
Lemma (Vapnik, Chervonenkis, Sauer, Shelah)
Let G be a function class with VC(G) = d. Then:
N(G, l) ≤
Pd
i=0
l
i
for all l ∈ N
In particular, for all l ≥ d: N(G, l) ≤ el
d
d
.
If fct. class has finite VC-dim → shattering coefficient only
grows polynomially.
Infinite VC-dim → exponential growth
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15. VC-dimension: main result
Theorem (Generalization bound: VC-dimension)
Let G a function class with VC(G) = d. Then with probability
at least 1 − δ all functions g ∈ G satisfy
R(g) ≤ Remp(g, l) + 2
s
d ln(2el
d ) + ln(1/δ)
l
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