随机算法 (Spring 2013)/Random Variables and Expectations

Random Variable

Definition (random variable)

A random variable [math]\displaystyle{ X }[/math] on a sample space [math]\displaystyle{ \Omega }[/math] is a real-valued function [math]\displaystyle{ X:\Omega\rightarrow\mathbb{R} }[/math]. A random variable X is called a discrete random variable if its range is finite or countably infinite.

For a random variable [math]\displaystyle{ X }[/math] and a real value [math]\displaystyle{ x\in\mathbb{R} }[/math], we write "[math]\displaystyle{ X=x }[/math]" for the event [math]\displaystyle{ \{a\in\Omega\mid X(a)=x\} }[/math], and denote the probability of the event by

[math]\displaystyle{ \Pr[X=x]=\Pr(\{a\in\Omega\mid X(a)=x\}) }[/math].

Independent Random Variables

The independence can also be defined for variables:

Definition (Independent variables)

Two random variables [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] are independent if and only if [math]\displaystyle{ \Pr[(X=x)\wedge(Y=y)]=\Pr[X=x]\cdot\Pr[Y=y] }[/math] for all values [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math]. Random variables [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are mutually independent if and only if, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math] and any values [math]\displaystyle{ x_i }[/math], where [math]\displaystyle{ i\in I }[/math], [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i\in I}(X_i=x_i)\right] &= \prod_{i\in I}\Pr[X_i=x_i]. \end{align} }[/math]

Note that in probability theory, the "mutual independence" is not equivalent with "pair-wise independence", which we will learn in the future.

Expectation

Let [math]\displaystyle{ X }[/math] be a discrete random variable. The expectation of [math]\displaystyle{ X }[/math] is defined as follows.

Definition (Expectation)

The expectation of a discrete random variable [math]\displaystyle{ X }[/math], denoted by [math]\displaystyle{ \mathbf{E}[X] }[/math], is given by [math]\displaystyle{ \begin{align} \mathbf{E}[X] &= \sum_{x}x\Pr[X=x], \end{align} }[/math] where the summation is over all values [math]\displaystyle{ x }[/math] in the range of [math]\displaystyle{ X }[/math].

Linearity of Expectation

Perhaps the most useful property of expectation is its linearity.

Theorem (Linearity of Expectations)

For any discrete random variables [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math], and any real constants [math]\displaystyle{ a_1, a_2, \ldots, a_n }[/math], [math]\displaystyle{ \begin{align} \mathbf{E}\left[\sum_{i=1}^n a_iX_i\right] &= \sum_{i=1}^n a_i\cdot\mathbf{E}[X_i]. \end{align} }[/math]

Proof. By the definition of the expectations, it is easy to verify that (try to prove by yourself): for any discrete random variables [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math], and any real constant [math]\displaystyle{ c }[/math], [math]\displaystyle{ \mathbf{E}[X+Y]=\mathbf{E}[X]+\mathbf{E}[Y] }[/math]; [math]\displaystyle{ \mathbf{E}[cX]=c\mathbf{E}[X] }[/math]. The theorem follows by induction. [math]\displaystyle{ \square }[/math]

The linearity of expectation gives an easy way to compute the expectation of a random variable if the variable can be written as a sum.

Example

Supposed that we have a biased coin that the probability of HEADs is [math]\displaystyle{ p }[/math]. Flipping the coin for n times, what is the expectation of number of HEADs?

It looks straightforward that it must be np, but how can we prove it? Surely we can apply the definition of expectation to compute the expectation with brute force. A more convenient way is by the linearity of expectations: Let [math]\displaystyle{ X_i }[/math] indicate whether the [math]\displaystyle{ i }[/math]-th flip is HEADs. Then [math]\displaystyle{ \mathbf{E}[X_i]=1\cdot p+0\cdot(1-p)=p }[/math], and the total number of HEADs after n flips is [math]\displaystyle{ X=\sum_{i=1}^{n}X_i }[/math]. Applying the linearity of expectation, the expected number of HEADs is:

[math]\displaystyle{ \mathbf{E}[X]=\mathbf{E}\left[\sum_{i=1}^{n}X_i\right]=\sum_{i=1}^{n}\mathbf{E}[X_i]=np }[/math].

The real power of the linearity of expectations is that it does not require the random variables to be independent, thus can be applied to any set of random variables. For example:

[math]\displaystyle{ \mathbf{E}\left[\alpha X+\beta X^2+\gamma X^3\right] = \alpha\cdot\mathbf{E}[X]+\beta\cdot\mathbf{E}\left[X^2\right]+\gamma\cdot\mathbf{E}\left[X^3\right]. }[/math]

However, do not exaggerate this power!

• For an arbitrary function [math]\displaystyle{ f }[/math] (not necessarily linear), the equation [math]\displaystyle{ \mathbf{E}[f(X)]=f(\mathbf{E}[X]) }[/math] does not hold generally.
• For variances, the equation [math]\displaystyle{ var(X+Y)=var(X)+var(Y) }[/math] does not hold without further assumption of the independence of [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math].

Conditional Expectation

Conditional expectation can be accordingly defined:

Definition (conditional expectation)

For random variables [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math], [math]\displaystyle{ \mathbf{E}[X\mid Y=y]=\sum_{x}x\Pr[X=x\mid Y=y], }[/math] where the summation is taken over the range of [math]\displaystyle{ X }[/math].

The Law of Total Expectation

There is also a law of total expectation.

Theorem (law of total expectation)

Let [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] be two random variables. Then [math]\displaystyle{ \mathbf{E}[X]=\sum_{y}\mathbf{E}[X\mid Y=y]\cdot\Pr[Y=y]. }[/math]

Random Quicksort

Given as input a set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] numbers, we want to sort the numbers in [math]\displaystyle{ S }[/math] in increasing order. One of the most famous algorithm for this problem is the Quicksort algorithm.

• if [math]\displaystyle{ |S|\gt 1 }[/math] do:
  • pick an [math]\displaystyle{ x\in S }[/math] as the pivot;
  • partition [math]\displaystyle{ S }[/math] into [math]\displaystyle{ S_1 }[/math], [math]\displaystyle{ \{x\} }[/math], and [math]\displaystyle{ S_2 }[/math], where all numbers in [math]\displaystyle{ S_1 }[/math] are smaller than [math]\displaystyle{ x }[/math] and all numbers in [math]\displaystyle{ S_2 }[/math] are larger than [math]\displaystyle{ x }[/math];
  • recursively sort [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math];

The time complexity of this sorting algorithm is measured by the number of comparisons.

For the deterministic quicksort algorithm, the pivot is picked from a fixed position (e.g. the first number in the array). The worst-case time complexity in terms of number of comparisons is [math]\displaystyle{ \Theta(n^2) }[/math].

We consider the following randomized version of the quicksort.

• if [math]\displaystyle{ |S|\gt 1 }[/math] do:
  • uniformly pick a random [math]\displaystyle{ x\in S }[/math] as the pivot;
  • partition [math]\displaystyle{ S }[/math] into [math]\displaystyle{ S_1 }[/math], [math]\displaystyle{ \{x\} }[/math], and [math]\displaystyle{ S_2 }[/math], where all numbers in [math]\displaystyle{ S_1 }[/math] are smaller than [math]\displaystyle{ x }[/math] and all numbers in [math]\displaystyle{ S_2 }[/math] are larger than [math]\displaystyle{ x }[/math];
  • recursively sort [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math];

Analysis

Our goal is to analyze the expected number of comparisons during an execution of RandQSort with an arbitrary input [math]\displaystyle{ S }[/math]. We achieve this by measuring the chance that each pair of elements are compared, and summing all of them up due to Linearity of Expectation.

Let [math]\displaystyle{ a_i }[/math] denote the [math]\displaystyle{ i }[/math]th smallest element in [math]\displaystyle{ S }[/math]. Let [math]\displaystyle{ X_{ij}\in\{0,1\} }[/math] be the random variable which indicates whether [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared during the execution of RandQSort. That is:

[math]\displaystyle{ \begin{align} X_{ij} &= \begin{cases} 1 & a_i\mbox{ and }a_j\mbox{ are compared}\\ 0 & \mbox{otherwise} \end{cases}. \end{align} }[/math]

Elements [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared only if one of them is chosen as pivot. After comparison they are separated (thus are never compared again). So we have the following observation:

Claim 1: Every pair of [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared at most once.

Therefore the sum of [math]\displaystyle{ X_{ij} }[/math] for all pair [math]\displaystyle{ \{i, j\} }[/math] gives the total number of comparisons. The expected number of comparisons is [math]\displaystyle{ \mathbf{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] }[/math]. Due to Linearity of Expectation, [math]\displaystyle{ \mathbf{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] = \sum_{i=1}^n\sum_{j\gt i}\mathbf{E}\left[X_{ij}\right] }[/math]. Our next step is to analyze [math]\displaystyle{ \mathbf{E}\left[X_{ij}\right] }[/math] for each [math]\displaystyle{ \{i, j\} }[/math].

By the definition of expectation and [math]\displaystyle{ X_{ij} }[/math],

[math]\displaystyle{ \begin{align} \mathbf{E}\left[X_{ij}\right] &= 1\cdot \Pr[a_i\mbox{ and }a_j\mbox{ are compared}] + 0\cdot \Pr[a_i\mbox{ and }a_j\mbox{ are not compared}]\\ &= \Pr[a_i\mbox{ and }a_j\mbox{ are compared}]. \end{align} }[/math]

We are going to bound this probability.

Claim 2: [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared if and only if one of them is chosen as pivot when they are still in the same subset.

This is easy to verify: just check the algorithm. The next one is a bit complicated.

Claim 3: If [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are still in the same subset then all [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] are in the same subset.

We can verify this by induction. Initially, [math]\displaystyle{ S }[/math] itself has the property described above; and partitioning any [math]\displaystyle{ S }[/math] with the property into [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math] will preserve the property for both [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math]. Therefore Claim 3 holds.

Combining Claim 2 and 3, we have:

Claim 4: [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared only if one of [math]\displaystyle{ \{a_i, a_j\} }[/math] is chosen from [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math].

And apparently,

Claim 5: Every one of [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] is chosen equal-probably.

This is because our RandQSort chooses the pivot uniformly at random.

Claim 4 and Claim 5 together imply:

[math]\displaystyle{ \begin{align} \Pr[a_i\mbox{ and }a_j\mbox{ are compared}] &\le \frac{2}{j-i+1}. \end{align} }[/math]

Remark: Perhaps you feel confused about the above argument. You may ask: "The algorithm chooses pivots for many times during the execution. Why in the above argument, it looks like the pivot is chosen only once?" Good question! Let's see what really happens by looking closely. For any pair [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math], initially [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] are all in the same set [math]\displaystyle{ S }[/math] (obviously!). During the execution of the algorithm, the set which containing [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] are shrinking (due to the pivoting), until one of [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] is chosen, and the set is partitioned into different subsets. We ask for the probability that the chosen one is among [math]\displaystyle{ \{a_i, a_j\} }[/math]. So we really care about "the last" pivoting before [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] is split. Formally, let [math]\displaystyle{ Y }[/math] be the random variable denoting the pivot element. We know that for each [math]\displaystyle{ a_k\in\{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math], [math]\displaystyle{ Y=a_k }[/math] with the same probability, and [math]\displaystyle{ Y\not\in\{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] with an unknown probability (remember that there might be other elements in the same subset with [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math]). The probability we are looking for is actually [math]\displaystyle{ \Pr[Y\in \{a_i, a_j\}\mid Y\in\{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\}] }[/math], which is always [math]\displaystyle{ \frac{2}{j-i+1} }[/math], provided that [math]\displaystyle{ Y }[/math] is uniform over [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math]. The conditional probability rules out the irrelevant events in a probabilistic argument.

Summing all up:

[math]\displaystyle{ \begin{align} \mathbf{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] &= \sum_{i=1}^n\sum_{j\gt i}\mathbf{E}\left[X_{ij}\right]\\ &\le \sum_{i=1}^n\sum_{j\gt i}\frac{2}{j-i+1}\\ &= \sum_{i=1}^n\sum_{k=2}^{n-i+1}\frac{2}{k} & & (\mbox{Let }k=j-i+1)\\ &\le \sum_{i=1}^n\sum_{k=1}^{n}\frac{2}{k}\\ &= 2n\sum_{k=1}^{n}\frac{1}{k}\\ &= 2n H(n). \end{align} }[/math]

[math]\displaystyle{ H(n) }[/math] is the [math]\displaystyle{ n }[/math]th Harmonic number. It holds that

[math]\displaystyle{ \begin{align}H(n) = \ln n+O(1)\end{align} }[/math].

Therefore, for an arbitrary input [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] numbers, the expected number of comparisons taken by RandQSort to sort [math]\displaystyle{ S }[/math] is [math]\displaystyle{ \mathrm{O}(n\log n) }[/math].