Uniform convergence in probability

Uniform convergence in probability is a form of convergence in probability in statistical asymptotic theory and probability theory. It means that, under certain conditions, the ''empirical frequencies'' of all events in a certain event-family converge to their ''theoretical probabilities''. Uniform convergence in probability has applications to statistics as well as machine learning as part of statistical learning theory.

The law of large numbers says that, for each ''single'' event $A$, its empirical frequency in a sequence of independent trials converges (with high probability) to its theoretical probability. In many application however, the need arises to judge simultaneously the probabilities of events of an entire class $S$ from one and the same sample. Moreover it, is required that the relative frequency of the events converge to the probability uniformly over the entire class of events $S$ The Uniform Convergence Theorem gives a sufficient condition for this convergence to hold. Roughly, if the event-family is sufficiently simple (its VC dimension is sufficiently small) then uniform convergence holds.

Definitions

For a class of predicates $H$ defined on a set $X$ and a set of samples $x=(x_1,x_2,\dots,x_m)$, where $x_i\in X$, the ''empirical frequency'' of $h\in H$ on $x$ is

: $\widehat_x(h)=\frac 1 m , \, .$

The ''theoretical probability'' of $h\in H$ is defined as $Q_P(h) = P\.$

The Uniform Convergence Theorem states, roughly, that if $H$ is "simple" and we draw samples independently (with replacement) from $X$ according to any distribution $P$, then with high probability, the empirical frequency will be close to its expected value, which is the theoretical probability.

Here "simple" means that the Vapnik–Chervonenkis dimension of the class $H$ is small relative to the size of the sample. In other words, a sufficiently simple collection of functions behaves roughly the same on a small random sample as it does on the distribution as a whole.

The Uniform Convergence Theorem was first proved by Vapnik and Chervonenkis using the concept of growth function.

Uniform convergence theorem

The statement of the uniform convergence theorem is as follows:Martin Anthony Peter, l. Bartlett. Neural Network Learning: Theoretical Foundations, pages 46–50. First Edition, 1999. Cambridge University Press

If $H$ is a set of $\$-valued functions defined on a set $X$ and $P$ is a probability distribution on $X$ then for $\varepsilon>0$ and $m$ a positive integer, we have:
: $P^m\\leq 4\Pi_H(2m)e^.$
: where, for any $x\in X^m,$,
: $Q_P(h)=P\,$
: $\widehat_x(h)=\frac 1 m , \,$
: and $, x, =m$. $P^m$ indicates that the probability is taken over $x$ consisting of $m$ i.i.d. draws from the distribution $P$.

: $\Pi_H$ is defined as: For any $\$-valued functions $H$ over $X$ and $D\subseteq X$,
: $\Pi_H(D)=\.$

And for any natural number $m$, the shattering number $\Pi_H(m)$ is defined as:
: $\Pi_H(m)=\max, \, .$

From the point of Learning Theory one can consider $H$ to be the Concept/Hypothesis class defined over the instance set $X$. Before getting into the details of the proof of the theorem we will state Sauer's Lemma which we will need in our proof.

Sauer–Shelah lemma

The Sauer–Shelah lemmaSham Kakade and Ambuj Tewari, CMSC 35900 (Spring 2008) Learning Theory, Lecture 11
/ref> relates the shattering number $\Pi_(m)$ to the VC Dimension.

Lemma: $\Pi_(m)\leq\left( \frac\right)^$, where $d$ is the VC Dimension of the concept class $H$.

Corollary: $\Pi_(m)\leq m^$.

Proof of uniform convergence theorem

and are the sources of the proof below. Before we get into the details of the proof of the ''Uniform Convergence Theorem'' we will present a high level overview of the proof.

#''Symmetrization:'' We transform the problem of analyzing $, Q_(h)-\widehat_(h), \geq\varepsilon$ into the problem of analyzing $, \widehat_(h)-\widehat_(h), \geq\varepsilon/2$, where $r$ and $s$ are i.i.d samples of size $m$ drawn according to the distribution $P$. One can view $r$ as the original randomly drawn sample of length $m$, while $s$ may be thought as the testing sample which is used to estimate $Q_(h)$.
#''Permutation:'' Since $r$ and $s$ are picked identically and independently, so swapping elements between them will not change the probability distribution on $r$ and $s$. So, we will try to bound the probability of $, \widehat_(h)-\widehat_(h), \geq\varepsilon/2$ for some $h \in H$ by considering the effect of a specific collection of permutations of the joint sample $x=r, , s$. Specifically, we consider permutations $\sigma(x)$ which swap $x_i$ and $x_$ in some subset of $$. The symbol $r, , s$ means the concatenation of $r$ and $s$.
#''Reduction to a finite class:'' We can now restrict the function class $H$ to a fixed joint sample and hence, if $H$ has finite VC Dimension, it reduces to the problem to one involving a finite function class.

We present the technical details of the proof.

Symmetrization

Lemma: Let $V=\$ and
: $R=\.$

Then for $m\geq\frac 2$, $P^m(V)\leq 2P^(R)$.

Proof:
By the triangle inequality,

if $, Q_(h)-\widehat_r(h), \geq\varepsilon$ and $, Q_P(h)-\widehat_s (h), \leq\varepsilon /2$ then $, \widehat_r(h)-\widehat_s (h), \geq\varepsilon /2$.

Therefore,

:
\begin
& P^(R) \\ pt\geq & P^\ \\ pt= & \int_V P^m\ \, dP^m(r) \\ pt= & A
\end

since $r$ and $s$ are independent.

Now for $r\in V$ fix an $h\in H$ such that $, Q_P(h)-\widehat_r(h), \geq\varepsilon$. For this $h$, we shall show that

: $P^m \left\ \geq\frac 1 2.$

Thus for any $r\in V$, $A\geq\frac2$ and hence $P^(R)\geq\frac2$. And hence we perform the first step of our high level idea.

Notice, $m\cdot \widehat_s(h)$ is a binomial random variable with expectation $m\cdot Q_(h)$ and variance $m\cdot Q_P(h)(1-Q_P(h))$. By Chebyshev's inequality we get

: $P^m \left\ \leq \frac \leq \frac 1 \leq\frac 1 2$

for the mentioned bound on $m$. Here we use the fact that $x(1-x)\leq 1/4$ for $x$.

Permutations

Let $\Gamma_$ be the set of all permutations of $\$ that swaps $i$ and $m+i$ $\forall i$ in some subset of $\$.

Lemma: Let $R$ be any subset of $X^$ and $P$ any probability distribution on $X$. Then,

: P^(R)=E Pr[\sigma(x)\in_R\leq_\max_(\Pr Pr[\sigma(x)\in_R\leq_\max_(\Pr[\sigma(x)\in_R">sigma(x)\in_R.html"_;"title="Pr[\sigma(x)\in_R">Pr[\sigma(x)\in_R\leq_\max_(\Pr[\sigma(x)\in_R,_

where_the_expectation_is_over_$x$_chosen_according_to_$P^$,_and_the_probability_is_over_$\sigma$_chosen_uniformly_from_$\Gamma_$.

Proof:_
For_any_$\sigma\in\Gamma_m,$

:_$_P^(R)_=_P^\_$

(since_coordinate_permutations_preserve_the_product_distribution_$_P^$.)

:_
\begin
\therefore_P^(R)_=__&_\int_1_(x)_\,_dP^(x)_\\_pt=__&_\frac\sum__\int__1_R(\sigma(x))_\,_dP^(x)_\\_pt=__&_\int__\frac_1_\sum__1_R_(\sigma(x))_\,_dP^(x)_\\_pt&_\text_.html" ;"title="sigma(x)\in_R.html" ;"title="sigma(x)\in_R.html" ;"title="Pr[\sigma(x)\in R">Pr[\sigma(x)\in R\leq \max_(\Pr[\sigma(x)\in R">sigma(x)\in_R.html" ;"title="Pr[\sigma(x)\in R">Pr[\sigma(x)\in R\leq \max_(\Pr[\sigma(x)\in R,

where the expectation is over $x$ chosen according to $P^$, and the probability is over $\sigma$ chosen uniformly from $\Gamma_$.

Proof:
For any $\sigma\in\Gamma_m,$

: $P^(R) = P^\$

(since coordinate permutations preserve the product distribution $P^$.)

:
\begin
\therefore P^(R) = & \int_1_(x) \, dP^(x) \\ pt= & \frac\sum_ \int_ 1_R(\sigma(x)) \, dP^(x) \\ pt= & \int_ \frac 1 \sum_ 1_R (\sigma(x)) \, dP^(x) \\ pt& \text ">\Gamma_, \text \\ pt= & \int_ \Pr[\sigma(x)\in R\, dP^(x) \quad \text \\ pt\leq & \max_(\Pr sigma(x)\in R.
\end

The maximum is guaranteed to exist since there is only a finite set of values that probability under a random permutation can take.

Reduction to a finite class

Lemma: Basing on the previous lemma,
: \max_(\Pr sigma(x)\in R\leq 4\Pi_H(2m)e^ .

Proof:
Let us define $x=(x_1,x_2,\ldots,x_)$ and $t=, H, _x,$ which is at most $\Pi_H(2m)$. This means there are functions $h_1,h_2,\ldots,h_t\in H$ such that for any $h\in H,\exists i$ between $1$ and $t$ with $h_i(x_k)=h(x_k)$ for $1\leq k\leq 2m.$

We see that $\sigma(x)\in R$ iff for some $h$ in $H$ satisfies,
$, \frac, \, -\frac, \, , \geq\frac$.
Hence if we define $w^_=1$ if $h_(x_)=1$ and $w^_=0$ otherwise.

For $1\leq i\leq m$ and $1\leq j\leq t$, we have that $\sigma(x)\in R$ iff for some $j$ in $$ satisfies $, \frac 1 m \left(\sum_i w^j_-\sum_i w^j_\right), \geq\frac \varepsilon 2$. By union bound we get

: \Pr sigma(x)\in Rleq t\cdot \max\left(\Pr \left, _\frac_1_m_\left(\sum_i_w^j_-\sum_i_w^j_\right)\_\geq_\frac_\varepsilon_2_\right\right).

Since,_the_distribution_over_the_permutations_$\sigma$_is_uniform_for_each_$i$,_so_$w^j_-w^j_$_equals_$\pm_, w^j_i-w^j_,$,_with_equal_probability.

Thus,

:_\Pr\left \frac_1_m_\left(\sum_i_\left(w^j_-w^j_\right)\right)\\geq\frac_\varepsilon_2\right=_\Pr\left _\frac_1_m_\left(_\sum_i, w^j_i-w^j_, \beta_i\right)\\geq\frac_\varepsilon_2\right

where_the_probability_on_the_right_is_over_$\beta_$_and_both_the_possibilities_are_equally_likely._By_ Hoeffding's_inequality,_this_is_at_most_$2e^$.

Finally,_combining_all_the_three_parts_of_the_proof_we_get_the_Uniform_Convergence_Theorem.

_References

{{Reflist

Combinatorics
Machine_learning
Articles_containing_proofs
Probability_theoremshtml" ;"title="\frac 1 m \left(\sum_i w^j_ - \sum_i w^j_\right), \geq \frac \varepsilon 2]\right)

:\leq \Pi_(2m)\cdot \max\left(\Pr\left \left, \frac 1 m \left(\sum_i w^j_-\sum_i w^j_\right)\ \geq \frac \varepsilon 2 \right\right).

Since, the distribution over the permutations $\sigma$ is uniform for each $i$, so $w^j_-w^j_$ equals $\pm , w^j_i-w^j_,$, with equal probability.

Thus,

: \Pr\left \frac 1 m \left(\sum_i \left(w^j_-w^j_\right)\right)\\geq\frac \varepsilon 2\right= \Pr\left \frac 1 m \left( \sum_i, w^j_i-w^j_, \beta_i\right)\\geq\frac \varepsilon 2\right

where the probability on the right is over $\beta_$ and both the possibilities are equally likely. By Hoeffding's inequality, this is at most $2e^$.

Finally, combining all the three parts of the proof we get the Uniform Convergence Theorem.

References

{{Reflist

Combinatorics
Machine learning
Articles containing proofs
Probability theorems>\frac 1 m \left(\sum_i w^j_ - \sum_i w^j_\right), \geq \frac \varepsilon 2right)

:\leq \Pi_(2m)\cdot \max\left(\Pr\left \left, \frac 1 m \left(\sum_i w^j_-\sum_i w^j_\right)\ \geq \frac \varepsilon 2 \right\right).

Since, the distribution over the permutations $\sigma$ is uniform for each $i$, so $w^j_-w^j_$ equals $\pm , w^j_i-w^j_,$, with equal probability.

Thus,

: \Pr\left \frac 1 m \left(\sum_i \left(w^j_-w^j_\right)\right)\\geq\frac \varepsilon 2\right= \Pr\left \frac 1 m \left( \sum_i, w^j_i-w^j_, \beta_i\right)\\geq\frac \varepsilon 2\right

where the probability on the right is over $\beta_$ and both the possibilities are equally likely. By Hoeffding's inequality, this is at most $2e^$.

Finally, combining all the three parts of the proof we get the Uniform Convergence Theorem.

References

{{Reflist

Combinatorics
Machine learning
Articles containing proofs
Probability theorems