随机算法 (Spring 2014)/Concentration of Measure

The Bounded Difference Method

Combining Azuma's inequality with the construction of Doob martingales, we have the powerful Bounded Difference Method for concentration of measures.

For arbitrary random variables

Given a sequence of random variables [math]\displaystyle{ X_1,\ldots,X_n }[/math] and a function [math]\displaystyle{ f }[/math]. The Doob sequence constructs a martingale from them. Combining this construction with Azuma's inequality, we can get a very powerful theorem called "the method of averaged bounded differences" which bounds the concentration for arbitrary function on arbitrary random variables (not necessarily a martingale).

Theorem (Method of averaged bounded differences)

Let [math]\displaystyle{ \boldsymbol{X}=(X_1,\ldots, X_n) }[/math] be arbitrary random variables and let [math]\displaystyle{ f }[/math] be a function of [math]\displaystyle{ X_0,X_1,\ldots, X_n }[/math] satisfying that, for all [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ |\mathbf{E}[f(\boldsymbol{X})\mid X_1,\ldots,X_i]-\mathbf{E}[f(\boldsymbol{X})\mid X_1,\ldots,X_{i-1}]|\le c_i, }[/math] Then [math]\displaystyle{ \begin{align} \Pr\left[|f(\boldsymbol{X})-\mathbf{E}[f(\boldsymbol{X})]|\ge t\right]\le 2\exp\left(-\frac{t^2}{2\sum_{i=1}^nc_i^2}\right). \end{align} }[/math]

Proof. Define the Doob Martingale sequence [math]\displaystyle{ Y_0,Y_1,\ldots,Y_n }[/math] by setting [math]\displaystyle{ Y_0=\mathbf{E}[f(X_1,\ldots,X_n)] }[/math] and, for [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ Y_i=\mathbf{E}[f(X_1,\ldots,X_n)\mid X_1,\ldots,X_i] }[/math]. Then the above theorem is a restatement of the Azuma's inequality holding for [math]\displaystyle{ Y_0,Y_1,\ldots,Y_n }[/math]. [math]\displaystyle{ \square }[/math]

For independent random variables

The condition of bounded averaged differences is usually hard to check. This severely limits the usefulness of the method. To overcome this, we introduce a property which is much easier to check, called the Lipschitz condition.

Definition (Lipschitz condition)

A function [math]\displaystyle{ f(x_1,\ldots,x_n) }[/math] satisfies the Lipschitz condition, if for any [math]\displaystyle{ x_1,\ldots,x_n }[/math] and any [math]\displaystyle{ y_i }[/math], [math]\displaystyle{ \begin{align} |f(x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n)-f(x_1,\ldots,x_{i-1},y_i,x_{i+1},\ldots,x_n)|\le 1. \end{align} }[/math]

In other words, the function satisfies the Lipschitz condition if an arbitrary change in the value of any one argument does not change the value of the function by more than 1.

The diference of 1 can be replaced by arbitrary constants, which gives a generalized version of Lipschitz condition.

Definition (Lipschitz condition, general version)

A function [math]\displaystyle{ f(x_1,\ldots,x_n) }[/math] satisfies the Lipschitz condition with constants [math]\displaystyle{ c_i }[/math], [math]\displaystyle{ 1\le i\le n }[/math], if for any [math]\displaystyle{ x_1,\ldots,x_n }[/math] and any [math]\displaystyle{ y_i }[/math], [math]\displaystyle{ \begin{align} |f(x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n)-f(x_1,\ldots,x_{i-1},y_i,x_{i+1},\ldots,x_n)|\le c_i. \end{align} }[/math]

The following "method of bounded differences" can be developed for functions satisfying the Lipschitz condition. Unfortunately, in order to imply the condition of averaged bounded differences from the Lipschitz condition, we have to restrict the method to independent random variables.

Corollary (Method of bounded differences)

Let [math]\displaystyle{ \boldsymbol{X}=(X_1,\ldots, X_n) }[/math] be [math]\displaystyle{ n }[/math] independent random variables and let [math]\displaystyle{ f }[/math] be a function satisfying the Lipschitz condition with constants [math]\displaystyle{ c_i }[/math], [math]\displaystyle{ 1\le i\le n }[/math]. Then [math]\displaystyle{ \begin{align} \Pr\left[|f(\boldsymbol{X})-\mathbf{E}[f(\boldsymbol{X})]|\ge t\right]\le 2\exp\left(-\frac{t^2}{2\sum_{i=1}^nc_i^2}\right). \end{align} }[/math]

Proof. For convenience, we denote that [math]\displaystyle{ \boldsymbol{X}_{[i,j]}=(X_i,X_{i+1},\ldots, X_j) }[/math] for any [math]\displaystyle{ 1\le i\le j\le n }[/math]. We first show that the Lipschitz condition with constants [math]\displaystyle{ c_i }[/math], [math]\displaystyle{ 1\le i\le n }[/math], implies another condition called the averaged Lipschitz condition (ALC): for any [math]\displaystyle{ a_i,b_i }[/math], [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ \left|\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a_i\right]-\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=b_i\right]\right|\le c_i. }[/math] And this condition implies the averaged bounded difference condition: for all [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ \left|\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right]-\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]}\right]\right|\le c_i. }[/math] Then by applying the method of averaged bounded differences, the corollary can be proved. For any [math]\displaystyle{ a }[/math], by the law of total expectation, [math]\displaystyle{ \begin{align} &\quad\, \mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a\right]\\ &=\sum_{a_{i+1},\ldots,a_n}\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a, \boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right]\cdot\Pr\left[\boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\mid \boldsymbol{X}_{[1,i-1]},X_i=a\right]\\ &=\sum_{a_{i+1},\ldots,a_n}\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a, \boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right]\cdot\Pr\left[\boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right] \qquad (\mbox{independence})\\ &= \sum_{a_{i+1},\ldots,a_n} f(\boldsymbol{X}_{[1,i-1]},a,\boldsymbol{a}_{[i+1,n]})\cdot\Pr\left[\boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right]. \end{align} }[/math] Let [math]\displaystyle{ a=a_i }[/math] and [math]\displaystyle{ b_i }[/math], and take the diference. Then [math]\displaystyle{ \begin{align} &\quad\, \left|\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a_i\right]-\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=b_i\right]\right|\\ &=\left|\sum_{a_{i+1},\ldots,a_n}\left(f(\boldsymbol{X}_{[1,i-1]},a_i,\boldsymbol{a}_{[i+1,n]})-f(\boldsymbol{X}_{[1,i-1]},b_i,\boldsymbol{a}_{[i+1,n]})\right)\Pr\left[\boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right]\right|\\ &\le \sum_{a_{i+1},\ldots,a_n}\left|f(\boldsymbol{X}_{[1,i-1]},a_i,\boldsymbol{a}_{[i+1,n]})-f(\boldsymbol{X}_{[1,i-1]},b_i,\boldsymbol{a}_{[i+1,n]})\right|\Pr\left[\boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right]\\ &\le \sum_{a_{i+1},\ldots,a_n}c_i\Pr\left[\boldsymbol{X}_{[i+1,n]}=\boldsymbol{a}_{[i+1,n]}\right] \qquad (\mbox{Lipschitz condition})\\ &=c_i. \end{align} }[/math] Thus, the Lipschitz condition is transformed to the ALC. We then deduce the averaged bounded difference condition from ALC. By the law of total expectation, [math]\displaystyle{ \mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]}\right]=\sum_{a}\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a\right]\cdot\Pr[X_i=a\mid \boldsymbol{X}_{[1,i-1]}]. }[/math] We can trivially write [math]\displaystyle{ \mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right] }[/math] as [math]\displaystyle{ \mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right]=\sum_{a}\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right]\cdot\Pr\left[X_i=a\mid \boldsymbol{X}_{[1,i-1]}\right]. }[/math] Hence, the difference is [math]\displaystyle{ \begin{align} &\quad \left|\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right]-\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]}\right]\right|\\ &=\left|\sum_{a}\left(\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right]-\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a\right]\right)\cdot\Pr\left[X_i=a\mid \boldsymbol{X}_{[1,i-1]}\right]\right| \\ &\le \sum_{a}\left|\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i]}\right]-\mathbf{E}\left[f(\boldsymbol{X})\mid \boldsymbol{X}_{[1,i-1]},X_i=a\right]\right|\cdot\Pr\left[X_i=a\mid \boldsymbol{X}_{[1,i-1]}\right] \\ &\le \sum_a c_i\Pr\left[X_i=a\mid \boldsymbol{X}_{[1,i-1]}\right] \qquad (\mbox{due to ALC})\\ &=c_i. \end{align} }[/math] The averaged bounded diference condition is implied. Applying the method of averaged bounded diferences, the corollary follows. [math]\displaystyle{ \square }[/math]

Applications

Occupancy problem

Throwing [math]\displaystyle{ m }[/math] balls uniformly and independently at random to [math]\displaystyle{ n }[/math] bins, we ask for the occupancies of bins by the balls. In particular, we are interested in the number of empty bins.

This problem can be described equivalently as follows. Let [math]\displaystyle{ f:[m]\rightarrow[n] }[/math] be a uniform random function from [math]\displaystyle{ [m]\rightarrow[n] }[/math]. We ask for the number of [math]\displaystyle{ i\in[n] }[/math] that [math]\displaystyle{ f^{-1}(i) }[/math] is empty.

For any [math]\displaystyle{ i\in[n] }[/math], let [math]\displaystyle{ X_i }[/math] indicate the emptiness of bin [math]\displaystyle{ i }[/math]. Let [math]\displaystyle{ X=\sum_{i=1}^nX_i }[/math] be the number of empty bins.

[math]\displaystyle{ \mathbf{E}[X_i]=\Pr[\mbox{bin }i\mbox{ is empty}]=\left(1-\frac{1}{n}\right)^m. }[/math]

By the linearity of expectation,

[math]\displaystyle{ \mathbf{E}[X]=\sum_{i=1}^n\mathbf{E}[X_i]=n\left(1-\frac{1}{n}\right)^m. }[/math]

We want to know how [math]\displaystyle{ X }[/math] deviates from this expectation. The complication here is that [math]\displaystyle{ X_i }[/math] are not independent. So we alternatively look at a sequence of independent random variables [math]\displaystyle{ Y_1,\ldots, Y_m }[/math], where [math]\displaystyle{ Y_j\in[n] }[/math] represents the bin into which the [math]\displaystyle{ j }[/math]th ball falls. Clearly [math]\displaystyle{ X }[/math] is function of [math]\displaystyle{ Y_1,\ldots, Y_m }[/math].

We than observe that changing the value of any [math]\displaystyle{ Y_i }[/math] can change the value of [math]\displaystyle{ X }[/math] by at most 1, because one ball can affect the emptiness of at most one bin. Thus as a function of independent random variables [math]\displaystyle{ Y_1,\ldots, Y_m }[/math], [math]\displaystyle{ X }[/math] satisfies the Lipschitz condition. Apply the method of bounded differences, it holds that

[math]\displaystyle{ \Pr\left[\left|X-n\left(1-\frac{1}{n}\right)^m\right|\ge t\sqrt{m}\right]=\Pr[|X-\mathbf{E}[X]|\ge t\sqrt{m}]\le 2e^{-t^2/2} }[/math]

Thus, for sufficiently large [math]\displaystyle{ n }[/math] and [math]\displaystyle{ m }[/math], the number of empty bins is tightly concentrated around [math]\displaystyle{ n\left(1-\frac{1}{n}\right)^m\approx \frac{n}{e^{m/n}} }[/math]

Pattern Matching

Let [math]\displaystyle{ \boldsymbol{X}=(X_1,\ldots,X_n) }[/math] be a sequence of characters chosen independently and uniformly at random from an alphabet [math]\displaystyle{ \Sigma }[/math], where [math]\displaystyle{ m=|\Sigma| }[/math]. Let [math]\displaystyle{ \pi\in\Sigma^k }[/math] be an arbitrarily fixed string of [math]\displaystyle{ k }[/math] characters from [math]\displaystyle{ \Sigma }[/math], called a pattern. Let [math]\displaystyle{ Y }[/math] be the number of occurrences of the pattern [math]\displaystyle{ \pi }[/math] as a substring of the random string [math]\displaystyle{ X }[/math].

By the linearity of expectation, it is obvious that

[math]\displaystyle{ \mathbf{E}[Y]=(n-k+1)\left(\frac{1}{m}\right)^k. }[/math]

We now look at the concentration of [math]\displaystyle{ Y }[/math]. The complication again lies in the dependencies between the matches. Yet we will see that [math]\displaystyle{ Y }[/math] is well tightly concentrated around its expectation if [math]\displaystyle{ k }[/math] is relatively small compared to [math]\displaystyle{ n }[/math].

For a fixed pattern [math]\displaystyle{ \pi }[/math], the random variable [math]\displaystyle{ Y }[/math] is a function of the independent random variables [math]\displaystyle{ (X_1,\ldots,X_n) }[/math]. Any character [math]\displaystyle{ X_i }[/math] participates in no more than [math]\displaystyle{ k }[/math] matches, thus changing the value of any [math]\displaystyle{ X_i }[/math] can affect the value of [math]\displaystyle{ Y }[/math] by at most [math]\displaystyle{ k }[/math]. [math]\displaystyle{ Y }[/math] satisfies the Lipschitz condition with constant [math]\displaystyle{ k }[/math]. Apply the method of bounded differences,

[math]\displaystyle{ \Pr\left[\left|Y-\frac{n-k+1}{m^k}\right|\ge tk\sqrt{n}\right]=\Pr\left[\left|Y-\mathbf{E}[Y]\right|\ge tk\sqrt{n}\right]\le 2e^{-t^2/2} }[/math]

Combining unit vectors

Let [math]\displaystyle{ u_1,\ldots,u_n }[/math] be [math]\displaystyle{ n }[/math] unit vectors from some normed space. That is, [math]\displaystyle{ \|u_i\|=1 }[/math] for any [math]\displaystyle{ 1\le i\le n }[/math], where [math]\displaystyle{ \|\cdot\| }[/math] denote the vector norm (e.g. [math]\displaystyle{ \ell_1,\ell_2,\ell_\infty }[/math]) of the space.

Let [math]\displaystyle{ \epsilon_1,\ldots,\epsilon_n\in\{-1,+1\} }[/math] be independently chosen and [math]\displaystyle{ \Pr[\epsilon_i=-1]=\Pr[\epsilon_i=1]=1/2 }[/math].

Let

[math]\displaystyle{ v=\epsilon_1u_1+\cdots+\epsilon_nu_n, }[/math]

and

[math]\displaystyle{ X=\|v\|. }[/math]

This kind of construction is very useful in combinatorial proofs of metric problems. We will show that by this construction, the random variable [math]\displaystyle{ X }[/math] is well concentrated around its mean.

[math]\displaystyle{ X }[/math] is a function of independent random variables [math]\displaystyle{ \epsilon_1,\ldots,\epsilon_n }[/math]. By the triangle inequality for norms, it is easy to verify that changing the sign of a unit vector [math]\displaystyle{ u_i }[/math] can only change the value of [math]\displaystyle{ X }[/math] for at most 2, thus [math]\displaystyle{ X }[/math] satisfies the Lipschitz condition with constant 2. The concentration result follows by applying the method of bounded differences:

[math]\displaystyle{ \Pr[|X-\mathbf{E}[X]|\ge 2t\sqrt{n}]\le 2e^{-t^2/2}. }[/math]