# Probabilistic Bounds — A Primer

Probabilistic arguments are a key tool for the analysis of algorithms in machine learning theory and probability theory. They also assume a prominent role in the analysis of randomized and streaming algorithms, where one imposes a restriction on the amount of storage space an algorithm is allowed to use for its computations (usually sublinear in the size of the input).

While a whole host of probabilistic arguments are used, one theorem in particular (or family of theorems) is ubiquitous: the Chernoff bound. In its simplest form, the Chernoff bound gives an exponential bound on the deviation of sums of random variables from their expected value.

This is perhaps most important to algorithm analysis in the following mindset. Say we have a program whose output is a random variable $X$. Moreover suppose that the expected value of $X$ is the correct output of the algorithm. Then we can run the algorithm multiple times and take a median (or some sort of average) across all runs. The probability that the algorithm gives a wildly incorrect answer is the probability that more than half of the runs give values which are wildly far from their expected value. Chernoff’s bound ensures this will happen with small probability.

So this post is dedicated to presenting the main versions of the Chernoff bound that are used in learning theory and randomized algorithms. Unfortunately the proof of the Chernoff bound in its full glory is beyond the scope of this blog. However, we will give short proofs of weaker, simpler bounds as a straightforward application of this blog’s previous work laying down the theory.

If the reader has not yet intuited it, this post will rely heavily on the mathematical formalisms of probability theory. We will assume our reader is familiar with the material from our first probability theory primer, and it certainly wouldn’t hurt to have read our conditional probability theory primer, though we won’t use conditional probability directly. We will refrain from using measure-theoretic probability theory entirely (some day my colleagues in analysis will like me, but not today).

## Two Easy Bounds of Markov and Chebyshev

The first bound we’ll investigate is almost trivial in nature, but comes in handy. Suppose we have a random variable $X$ which is non-negative (as a function). Markov’s inequality is the statement that, for any constant $a > 0$,

$\displaystyle \textup{P}(X \geq a) \leq \frac{\textup{E}(X)}{a}$

In words, the probability that $X$ grows larger than some fixed constant is bounded by a quantity that is inversely proportional to the constant.

The proof is quite simple. Let $\chi_a$ be the indicator random variable for the event that $X \geq a$ ($\chi_a = 1$ when $X \geq a$ and zero otherwise). As with all indicator random variables, the expected value of $\chi_a$ is the probability that the event happens (if this is mysterious, use the definition of expected value). So $\textup{E}(\chi_a) = \textup{P}(X \geq a)$, and linearity of expectation allows us to include a factor of $a$:

$\textup{E}(a \chi_a) = a \textup{P}(X \geq a)$

The rest of the proof is simply the observation that $\textup{E}(a \chi_a) \leq \textup{E}(X)$. Indeed, as random variables we have the inequality $a \chi_a \leq X$. Whenever $a < X$, the former is equal to zero while the latter is nonnegative. And whenever $a \geq X$, the former is precisely $a$ while the latter is by assumption at least $a$. It follows that $\textup{E}(a \chi_a) \leq \textup{E}(X)$.

This last point is a simple property of expectation we omitted from our first primer. It usually goes by monotonicity of expectation, and we prove it here. First, if $X \geq 0$ then $\textup{E}(X) \geq 0$ (this is trivial). Second, if $0 \leq X \leq Y$, then define a new random variable $Z = Y-X$. Since $Z \geq 0$ and using linearity of expectation, it must be that $\textup{E}(Z) = \textup{E}(Y) - \textup{E}(X) \geq 0$. Hence $\textup{E}(X) \geq \textup{E}(Y)$. Note that we do require that $X$ has a finite expected value for this argument to work, but if this is not the case then Markov’s inequality is nonsensical anyway.

Markov’s inequality by itself is not particularly impressive or useful. For example, if $X$ is the number of heads in a hundred coin flips, Markov’s inequality ensures us that the probability of getting at least 99 heads is at most 50/99, which is about 1/2. Shocking. We know that the true probability is much closer to $2^{-100}$, so Markov’s inequality is a bust.

However, it does give us a more useful bound as a corollary. This bound is known as Chebyshev’s inequality, and its use is sometimes referred to as the second moment method because it gives a bound based on the variance of a random variable (instead of the expected value, the “first moment”).

The statement is as follows.

Chebyshev’s Inequality: Let $X$ be a random variable with finite expected value and positive variance. Then we can bound the probability that $X$ deviates from its expected value by a quantity that is proportional to the variance of $X$. In particular, for any $\lambda > 0$,

$\displaystyle \textup{P}(|X - \textup{E}(X)| \geq \lambda) \leq \frac{\textup{Var}(X)}{\lambda^2}$

And without any additional assumptions on $X$, this bound is sharp.

Proof. The proof is a simple application of Markov’s inequality. Let $Y = (X - \textup{E}(X))^2$, so that $\textup{E}(Y) = \textup{Var}(X)$. Then by Markov’s inequality

$\textup{P}(Y \geq \lambda^2) \leq \frac{\textup{E}(Y)}{\lambda^2}$

Since $Y$ is nonnegative $|X - \textup{E}(X)| = \sqrt(Y)$, and $\textup{P}(Y \geq \lambda^2) = \textup{P}(|X - \textup{E}(X)| \geq \lambda)$. The theorem is proved. $\square$

Chebyshev’s inequality shows up in so many different places (and usually in rather dry, technical bits), that it’s difficult to give a good example application.  Here is one that shows up somewhat often.

Say $X$ is a nonnegative integer-valued random variable, and we want to argue about when $X = 0$ versus when $X > 0$, given that we know $\textup{E}(X)$. No matter how large $\textup{E}(X)$ is, it can still be possible that $\textup{P}(X = 0)$ is arbitrarily close to 1. As a colorful example, let $X$ is the number of alien lifeforms discovered in the next ten years. We might debate that $\textup{E}(X)$ can arbitrarily large: if some unexpected scientific and technological breakthroughs occur tomorrow, we could discover an unbounded number of lifeforms. On the other hand, we are very likely not to discover any, and probability theory allows for such a random variable to exist.

If we know everything about $\textup{Var}(X)$, however, we can get more informed bounds.

Theorem: If $\textup{E}(X) \neq 0$, then $\displaystyle \textup{P}(X = 0) \leq \frac{\textup{Var}(X)}{\textup{E}(X)^2}$.

Proof. Simply choose $\lambda = \textup{E}(X)$ and apply Chebyshev’s inequality.

$\displaystyle \textup{P}(X = 0) \leq \textup{P}(|X - \textup{E}(X)| \geq \textup{E}(X)) \leq \frac{\textup{Var}(X)}{\textup{E}(X)^2}$

The first inequality follows from the fact that the only time $X$ can ever be zero is when $|X - \textup{E}(X)| = \textup{E}(X)$, and $X=0$ only accounts for one such possibility. $\square$

This theorem says more. If we know that $\textup{Var}(X)$ is significantly smaller than $\textup{E}(X)^2$, then $X > 0$ is more certain to occur. More precisely, and more computationally minded, suppose we have a sequence of random variables $X_n$ so that $\textup{E}(X_n) \to \infty$ as $n \to \infty$. Then the theorem says that if $\textup{Var}(X_n) = o(\textup{E}(X_n)^2)$, then $\textup{P}(X_n > 0) \to 1$. Remembering one of our very early primers on asymptotic notation, $f = o(g)$ means that $f$ grows asymptotically slower than $g$, and in terms of this fraction $\textup{Var}(X) / \textup{E}(X)^2$, this means that the denominator dominates the fraction so that the whole thing tends to zero.

## The Chernoff Bound

The Chernoff bound takes advantage of an additional hypothesis: our random variable is a sum of independent coin flips. We can use this to get exponential bounds on the deviation of the sum. More rigorously,

Theorem: Let $X_1 , \dots, X_n$ be independent random $\left \{ 0,1 \right \}$-valued variables, and let $X = \sum X_i$. Suppose that $\mu = \textup{E}(X)$. Then the probability that $X$ deviates from $\mu$ by more than a factor of $\lambda > 0$ is bounded from above:

$\displaystyle \textup{P}(X > (1+\lambda)\mu) \leq \frac{e^{\lambda \mu}}{(1+\lambda)^{(1+\lambda)\mu}}$

The proof is beyond the scope of this post, but we point the interested reader to these lecture notes.

We can apply the Chernoff bound in an easy example. Say all $X_i$ are fair coin flips, and we’re interested in the probability of getting more than 3/4 of the coins heads. Here $\mu = n/2$ and $\lambda = 1/2$, so the probability is bounded from above by

$\displaystyle \left ( \frac{e}{(3/2)^3} \right )^{n/4} \approx \frac{1}{5^n}$

So as the number of coin flips grows, the probability of seeing such an occurrence diminishes extremely quickly to zero. This is important because if we want to test to see if, say, the coins are biased toward flipping heads, we can simply run an experiment with $n$ sufficiently large. If we observe that more than 3/4 of the flips give heads, then we proclaim the coins are biased and we can be assured we are correct with high probability. Of course, after seeing 3/4 of more heads we’d be really confident that the coin is biased. A more realistic approach is to define some $\varepsilon$ that is small enough so as to say, “if some event occurs whose probability is smaller than $\varepsilon$, then I call shenanigans.” Then decide how many coins and what bound one would need to make the bad event have probability approximately $\varepsilon$. Finding this balance is one of the more difficult aspects of probabilistic algorithms, and as we’ll see later all of these quantities are left as variables and the correct values are discovered in the course of the proof.

## Chernoff-Hoeffding Inequality

The Hoeffding inequality (named after the Finnish statistician, Wassily Høffding) is a variant of the Chernoff bound, but often the bounds are collectively known as Chernoff-Hoeffding inequalities. The form that Hoeffding is known for can be thought of as a simplification and a slight generalization of Chernoff’s bound above.

Theorem: Let $X_1, \dots, X_n$ be independent random variables whose values are within some range $[a,b]$. Call $\mu_i = \textup{E}(X_i)$, $X = \sum_i X_i$, and $\mu \textup{E}(X) = \sum_i \mu_i$. Then for all $t > 0$,

$\displaystyle \textup{P}(|X - \mu| > t) \leq 2e^{(-2t^2)/(b-a)^2}$

For example, if we are interested in the sum of $n$ rolls of a fair six-sided die, then the probability that we deviate from $(7/2)n$ by more than $\sqrt{n}$ is bounded by $2e^{(-2n / 25)}$. Supposing we want to know how many rolls we need to guarantee with probability 0.01 that we don’t deviate too much, we just do the algebra:

$2e^{(-2n / 25)} < 0.01$
$-2n / 25 < \log (0.005)$
$n > \log(0.005^{(-25/2)}) \approx 66$

So with 67 rolls we can be confident that the sum of the rolls will lie between 222 and 240.

Another version of this theorem concerns the average of the $X_i$, and is only a minor modification of the above.

Theorem: If $X_1, \dots, X_n$ are as above, and $X = \frac{1}{n} \sum_i X_i$, with $\mu = \frac{1}{n}(\sum_i \mu_i)$, then for all $t > 0$, we get the following bound

$\displaystyle \textup{P}(|X - \mu| > t) \leq 2e^{(-2nt^2)/(b-a)^2}$

The only difference here is the extra factor of $n$ in the exponent. So the deviation is exponential both in the amount of deviation ($t^2$), and in the number of trials.

This theorem comes up very often in learning theory, in particular to prove Boosting works. Mathematicians will joke about how all theorems in learning theory are just applications of Chernoff-Hoeffding-type bounds. We’ll of course be seeing it again as we investigate boosting and the PAC-learning model in future posts, so we’ll see the theorems applied to their fullest extent then.

Until next time!

# Conditional (Partitioned) Probability — A Primer

One of the main areas of difficulty in elementary probability, and one that requires the highest levels of scrutiny and rigor, is conditional probability. The ideas are simple enough: that we assign probabilities relative to the occurrence of some event. But shrewd applications of conditional probability (and in particular, efficient ways to compute conditional probability) are key to successful applications of this subject. This is the basis for Nate Silver‘s success, the logical flaws of many a political pundit, and the ability for a robot to tell where it is in an environment. As this author usually touts, the best way to avoid the pitfalls of such confusing subjects is to be mathematically rigorous. In doing so we will develop intuition for when conditional probability that experts show off as if it were trivial.

But before we can get to all of that, we will cover a few extra ideas from finite probability theory that were left out of the last post.

Our entire discussion will revolve around a finite probability space, as defined last time. Let’s briefly (and densely) recall some of the notation presented there. We will always denote our probability space by $\Omega$, and the corresponding probability mass function will be $f: \Omega \to [0,1]$. Recall that events are subsets $E \subset \Omega$, and the probability function $P$ accepts as inputs events $E$, and produces as output the sum of the probabilities of members of $E$. We abuse notation by saying $\textup{P}(x) = \textup{P}(\left \{ x \right \}) = f(x)$ and disregarding $f$ for the most part. We really think of $\textup{P}$ as an extension of $f$ to subsets of $\Omega$ instead of just single values of $\Omega$. Further recall that a random variable $X$ is a real-valued function function $\Omega \to \mathbb{R}$.

## Partitions and Total Probability

A lot of reasoning in probability theory involves decomposing a complicated event into simpler events, or decomposing complicated random variables into simpler ones. Conditional probability is one way to do that, and conditional probability has very nice philosophical interpretations, but it fits into this more general scheme of “decomposing” events and variables into components.

The usual way to break up a set into pieces is via a partition. Recall the following set-theoretic definition.

Definition: partition of a set $X$ is a collection of subsets $X_i \in X$ so that every element $x \in X$ occurs in exactly one of the $X_i$.

Here are a few examples. We can partition the natural numbers $\mathbb{N}$ into even and odd numbers. We can partition the set of people in the world into subsets where each subset corresponds to a country and a person is placed in the subset corresponding to where they were born (an obvious simplification of the real world, but illustrates the point). The avid reader of this blog will remember how we used partitions to define quotient groups and quotient spaces. With a more applied flavor, finding a “good” partition is the ultimate goal of the clustering problem, and we saw a heuristic approach to this in our post on Lloyd’s algorithm.

You should think of a partition as a way to “cut up” a set into pieces. This colorful diagram is an example of a partition of a disc.

In fact, any time we have a canonical way to associate two things in a set, we can create a partition by putting all mutually associated things in the same piece of the partition. The rigorous name for this is an equivalence relation, but we won’t need that for the present discussion (partitions are the same thing as equivalence relations, just viewed in a different way).

Of course, the point is to apply this idea to probability spaces. Points (elements) in our probability space $\Omega$ are outcomes of some random experiment, and subsets $E \subset \Omega$ are events. So we can rephrase a partition for probability spaces as a choice of events $E_i \subset \Omega$ so that every outcome in $\Omega$ is part of exactly one event. Our first observation is quite a trivial one: the probabilities of the events in a partition sum to one. In symbols, if $E_1, \dots, E_m$ form our partition, then

$\displaystyle \sum_{i=1}^m \textup{P}(E_i) = 1$

Indeed, the definition of $\textup{P}$ is to sum over the probabilities of outcomes in an event. Since each outcome occurs exactly once among all the $E_i$, the above sum expands to

$\displaystyle \sum_{\omega \in \Omega} \textup{P}(\omega)$

Which by our axioms for a probability space is just one. We will give this observation the (non-standard) name the Lemma of Total Probability.

This was a nice warmup proof, but we can beef it up to make it more useful. If we have some other event $A$ which is not related to a partition in any way, we can break up $A$ with respect to the partition. Then, assuming this is simpler, we compute the probability that $A$ happens in terms of the probabilities of the pieces.

Theorem: Let $E_1, \dots , E_m$ be a partition of $\Omega$, and let $A$ be an arbitrary event. Then

$\displaystyle \textup{P}(A) = \sum_{i=1}^m \textup{P}(E_i \cap A)$

Proof. The proof is only marginally more complicated than that of the lemma of total probability. The probability of the event $A$ occurring is the sum of the probabilities of each of its outcomes occurring. Each outcome in $A$ occurs in exactly one of the $E_i$, and hence in exactly one of the sets $E_i \cap A$. If $E_i \cap A$ is empty, then its probability of occurring is zero (as per our definitions last time). So the sum on the right expands directly into the definition of $\textup{P}(A)$. $\square$

The area taken up by the set A is the same as the area taken up by the pieces of A which overlap the E’s. That is, the E’s give us a partition of A.

A more useful way of thinking of this is that we can use the $E_i$ to define a partition of $A$ in a natural way. The subsets in the partition will just be the sets $E_i \cap A$, and we will throw out any of these that turn out to be empty. Then we can think of our “new” probability space being $A$, and the theorem is just a special case of the lemma of total probability. Interestingly enough, this special case is often called the Theorem of Total Probability.

The idea to think of the event $A$ as our “new” probability space is extremely useful. It shows its worth most prominently when we interpret the shift as, “gaining the information that $A$ has occurred.” Then the question becomes: given that $A$ occurs, what is the probability that some other event will occur? That is, we’re interested in the probability of some event $B$ relative to $A$. This is called the conditional probability of $B$ with respect to $A$, and is denoted $P(B | A)$ (read “the probability of B given A”).

To compute the conditional probability, simply scale $\textup{P}(A \cap B)$ by the assumed event $\textup{P}(A)$. That is,

$\displaystyle \textup{P}(B | A) = \frac{\textup{P}(A \cap B)}{\textup{P}(A)}$

Wikipedia provides a straightforward derivation of the formula, but the spirit of the proof is exactly what we said above. The denominator is our new sample space, and the numerator is the probability of outcomes that cause $B$ to occur which also cause $A$ to occur. Multiplying both sides of this formula by $\textup{P}(A)$, this identity can be used to arrive at another version of the theorem of total probability:

$\displaystyle \textup{P}(A) = \sum_{i=1}^m \textup{P}(A | E_i) \textup{P}(E_i)$

That is, if we know how to compute the probabilities of the $E_i$, and we know how likely $A$ is to occur in each of those scenarios, then we can compute the total probability of $A$ occurring independently of the $E_i$.

We can come up with loads of more or less trivial examples of the theorem of total probability on simple probability spaces. Say you play a craps-like game where you roll a die twice. If you get a one on the first roll, you lose, and otherwise you have to match your initial roll on the second to win. The probability you win can be analyzed with the theorem on total probability. We partition the sample space into events corresponding to the outcome of the first roll.

$\displaystyle \textup{P}(\textup{Win}) = \sum_{i=1}^6 \textup{P}(\textup{Win } | \textup{ 1st roll }= i) \textup{P}(\textup{1st roll } = i)$

The probability the first roll is $i$ is 1/6, and if the first roll is a 1 then the probability of winning after that is zero. In the other 5 cases the conditional probability is the same regardless of $i$: to match $i$ on the second roll has a 1/6 chance. So the probability of winning is

$\displaystyle 5 \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{5}{36}$

For the working mathematician, these kinds of examples are relatively low-tech, but it illustrates the main way conditional probability is used in practice. We have some process we want to analyze, and we break it up into steps and condition on the results of a given step. We will see in a moment a more complicated example of this.

## Partitions via Random Variables

The most common kind of partition is created via a random variable with finitely many values (or countably many, but we haven’t breached infinite probability spaces yet). In this case, we can partition the sample space $\Omega$ based on the values of $X$. That is, for each value $x = X(\omega)$, we will have a subset of the partition $S_x$ be the set of all $\omega$ which map to $x$. In the parlance of functions, it is the preimage of a single value $x$;

$\displaystyle S_x = X^{-1}(x) = \left \{ \omega \in \Omega : X(\omega) = x\right \}$

And as the reader is probably expecting, we can use this to define a “relative” expected value of a random variable. Recall that if the image of $X$ is a finite set $x_1, \dots, x_n$, the expected value of $X$ is a sum

$\displaystyle \textup{E}(X) = \sum_{i=1}^n x_i \textup{P}(X = x_i)$

Suppose $X,Y$ are two such random variables, then the conditional probability of $X$ relative to the event $Y=y$ is the quantity

$\displaystyle \textup{P}(X=x | Y=y) = \frac{\textup{P}(X=x \textup{ and } Y=y)}{\textup{P}(Y=y)}$

And the conditional expectation of $X$ relative to the event $Y = y$, denoted $\textup{E}(X | Y = y)$ is a similar sum

$\displaystyle \textup{E}(X|Y=y) = \sum_{i=1}^n x_i \textup{P}(X = x_i | Y = y)$

Indeed, just as we implicitly “defined” a new sample space when we were partitioning based on events, here we are defining a new random variable (with the odd notation $X | Y=y$) whose domain is the preimage $Y^{-1}(y)$. We can then ask what the probability of it assuming a value $x$ is, and moreover what its expected value is.

Of course there is an analogue to the theorem of total probability lurking here. We want to say something like the true expected value of $X$ is a sum of the conditional expectations over all possible values of $Y$. We have to remember, though, that different values of $y$ can occur with very different probabilities, and the expected values of $X | Y=y$ can change wildly between them. Just as a quick (and morbid) example, if $X$ is the number of people who die on a randomly chosen day, and $Y$ is the number of atomic bombs dropped on that day, it is clear that the probability of $Y$ being positive is quite small, and the expected value of $X = Y=y$ will be dramatically larger if $y$ is positive than if it’s zero. (A few quick calculations based on tragic historic events show it would roughly double, using contemporary non-violent death rate estimates.)

And so instead of simply summing the expectation, we need to take an expectation over the values of $Y$. Thinking again of $X | Y=y$ as a random variable based on values of $Y$, it makes sense mathematically to take expectation. To distinguish between the two types of expectation, we will subscript the variable being “expected,” as in $\textup{E}_X(X|Y)$. That is, we have the following theorem.

TheoremThe expected value of $X$ satisfies

$\textup{E}_X(X) = \textup{E}_Y(\textup{E}_X(X|Y))$

Proof. Expand the definitions of what these values mean, and use the definition of conditional probability $\textup{P}(A \cap B) = \textup{P}(A | B) \textup{P}(B)$. We leave the proof as a trivial exercise to the reader, but if one cannot bear it, see Wikipedia for a full proof. $\square$

Let’s wrap up this post with a non-trivial example of all of this theory in action.

## A Nontrivial Example: the Galton-Watson Branching Process

We are interested (as was the eponymous Sir Francis Galton in the 1800′s) in the survival of surnames through generations of marriage and children. The main tool to study such a generational phenomenon is the Galton-Watson branching process. The idea is quite simple, but its analysis quickly blossoms into a rich and detailed theoretical puzzle and a more general practical tool. Just before we get too deep into things, we should note that these ideas (along with other types of branching processes) are used to analyze a whole host of problems in probability theory and computer science. A few the author has recently been working with are the evolution of random graphs and graph property testing.

The gist is as follows: say we live in a patriarchal society in which surnames are passed down on the male side. We can image a family tree being grown step by step in this way At the root there is a single male, and he has $k$ children, some of which are girls and some of which are boys. They all go on to have some number of children, but only the men pass on the family name to their children, and only their male children pass on the family name further. If we only record the family tree along the male lines, we can ask whether the tree will be finite; that is, whether the family name will die out.

To make this rigorous, let us define an infinite sequence of random variables $X_1 X_2, \dots$ which represent the number of children each person in the tree has, and suppose further that all of these variables are independent and uniformly distributed from $1, \dots, n$ for some fixed $n$. This may be an unrealistic assumption, but it makes the analysis a bit simpler. The number of children more likely follows a Poisson distribution where the mean is a parameter we would estimate from real-world data, but we haven’t spoken of Poisson distributions on this blog yet so we will leave it out.

We further imagine the tree growing step by step: at step $i$ the $i$-th individual in the tree has $X_i$ children and then dies. If the individual is a woman we by default set $X_i = 0$. We can recursively describe the size of the tree at each step by another random variable $Y_i$. Clearly $Y_0 = 1$, and the recursion is $Y_n = Y_{n-1} + X_i - 1$. In words, $Y_i$ represents the current living population with the given surname. We say the tree is finite (the family name dies off), if for some $i$ we get $Y_i = 0$. The first time at which this happens is when the family name dies off, but abstractly we can imagine the sequence of random variables continuing forever. This is sometimes called fictitious continuation.

At last, we assume that the probability of having a boy or girl is a split 1/2. Now we can start asking questions. What is the probability that the surname dies off? What is the expected size of the tree in terms of $n$?

For the first question we use the theorem of total probability. In particular, suppose the first person has two boys. Then the whole tree is finite precisely when both boys’ sub-trees are finite. Indeed, the two boys’ sub-trees are independent of one another, and so the probability of both being finite is the product of the probabilities of each being finite. That is, more generally

$\displaystyle \textup{P}(\textup{finite } | k \textup{ boys}) = \textup{P}(\textup{finite})^k \textup{P}(\textup{two boys})$

Setting $z = \textup{P}(\textup{the tree is finite})$, we can compute $z$ directly by conditioning on all possibilities of the first person’s children. Notice how we must condition twice here.

$\displaystyle z = \sum_{i=0}^n \sum_{k=0}^i \textup{P}(k \textup{ boys } | i \textup{ children}) \textup{P}(i \textup{ children}) z^k$

The probability of getting $k$ boys is the same as flipping $i$ coins and getting $k$ heads, which is just

$\displaystyle \textup{P}(k \textup{ boys } | i \textup{ children}) = \binom{i}{k}\frac{1}{2^i}$

So the equation is

$\displaystyle z = \sum_{i=0}^n \sum_{k=0}^i \binom{i}{k} \frac{1}{2^i} \cdot \frac{1}{n} z^k$

From here, we’ve reduced the problem down to picking the correct root of a polynomial. For example, when $n=4$, the polynomial equation to solve is

$\displaystyle 64z = 5 + 10z + 10z^2 + 5z^3 + z^4$

We have to be a bit careful, here though. Not all solutions to this equation are valid answers. For instance, the roots must be between 0 and 1 (inclusive), and if there are multiple then one must rule out the irrelevant roots by some additional argument. Moreover, we would need to use a calculus argument to prove there is always a solution between 0 and 1 in the first place. But after all that is done, we can estimate the correct root computationally (or solve for exactly when our polynomials have small degree). Here for $n=4$, the probability of being finite is about 0.094.

We leave the second question, on the expected size of the tree, for the reader to ponder. Next time we’ll devote an entire post to Bayes Theorem (a trivial consequence of the definition of conditional probability), and see how it helps us compute probabilities for use in programs.

Until then!

# Probability Theory — A Primer

It is a wonder that we have yet to officially write about probability theory on this blog. Probability theory underlies a huge portion of artificial intelligence, machine learning, and statistics, and a number of our future posts will rely on the ideas and terminology we lay out in this post. Our first formal theory of machine learning will be deeply ingrained in probability theory, we will derive and analyze probabilistic learning algorithms, and our entire treatment of mathematical finance will be framed in terms of random variables.

And so it’s about time we got to the bottom of probability theory. In this post, we will begin with a naive version of probability theory. That is, everything will be finite and framed in terms of naive set theory without the aid of measure theory. This has the benefit of making the analysis and definitions simple. The downside is that we are restricted in what kinds of probability we are allowed to speak of. For instance, we aren’t allowed to work with probabilities defined on all real numbers. But for the majority of our purposes on this blog, this treatment will be enough. Indeed, most programming applications restrict infinite problems to finite subproblems or approximations (although in their analysis we often appeal to the infinite).

We should make a quick disclaimer before we get into the thick of things: this primer is not meant to connect probability theory to the real world. Indeed, to do so would be decidedly unmathematical. We are primarily concerned with the mathematical formalisms involved in the theory of probability, and we will leave the philosophical concerns and applications to  future posts. The point of this primer is simply to lay down the terminology and basic results needed to discuss such topics to begin with.

So let us begin with probability spaces and random variables.

## Finite Probability Spaces

We begin by defining probability as a set with an associated function. The intuitive idea is that the set consists of the outcomes of some experiment, and the function gives the probability of each event happening. For example, a set $\left \{ 0,1 \right \}$ might represent heads and tails outcomes of a coin flip, while the function assigns a probability of one half (or some other numbers) to the outcomes. As usual, this is just intuition and not rigorous mathematics. And so the following definition will lay out the necessary condition for this probability to make sense.

Definition: A finite set $\Omega$ equipped with a function $f: \Omega \to [0,1]$ is a probability space if the function $f$ satisfies the property

$\displaystyle \sum_{\omega \in \Omega} f(\omega) = 1$

That is, the sum of all the values of $f$ must be 1.

Sometimes the set $\Omega$ is called the sample space, and the act of choosing an element of $\Omega$ according to the probabilities given by $f$ is called drawing an example. The function $f$ is usually called the probability mass function. Despite being part of our first definition, the probability mass function is relatively useless except to build what follows. Because we don’t really care about the probability of a single outcome as much as we do the probability of an event.

Definition: An event $E \subset \Omega$ is a subset of a sample space.

For instance, suppose our probability space is $\Omega = \left \{ 1, 2, 3, 4, 5, 6 \right \}$ and $f$ is defined by setting $f(x) = 1/6$ for all $x \in \Omega$ (here the “experiment” is rolling a single die). Then we are likely interested in more exquisite kinds of outcomes; instead of asking the probability that the outcome is 4, we might ask what is the probability that the outcome is even? This event would be the subset $\left \{ 2, 4, 6 \right \}$, and if any of these are the outcome of the experiment, the event is said to occur. In this case we would expect the probability of the die roll being even to be 1/2 (but we have not yet formalized why this is the case).

As a quick exercise, the reader should formulate a two-dice experiment in terms of sets. What would the probability space consist of as a set? What would the probability mass function look like? What are some interesting events one might consider (if playing a game of craps)?

Of course, we want to extend the probability mass function $f$ (which is only defined on single outcomes) to all possible events of our probability space. That is, we want to define a probability measure $\textup{P}: 2^\Omega \to \mathbb{R}$, where $2^\Omega$ denotes the set of all subsets of $\Omega$. The example of a die roll guides our intuition: the probability of any event should be the sum of the probabilities of the outcomes contained in it. i.e. we define

$\displaystyle \textup{P}(E) = \sum_{e \in E} f(e)$

where by convention the empty sum has value zero. Note that the function $\textup{P}$ is often denoted $\textup{Pr}$.

So for example, the coin flip experiment can’t have zero probability for both of the two outcomes 0 and 1; the sum of the probabilities of all outcomes must sum to 1. More coherently: $\textup{P}(\Omega) = \sum_{\omega \in \Omega} f(\omega) = 1$ by the defining property of a probability space. And so if there are only two outcomes of the experiment, then they must have probabilities $p$ and $1-p$ for some $p$. Such a probability space is often called a Bernoulli trial.

Now that the function $\textup{P}$ is defined on all events, we can simplify our notation considerably. Because the probability mass function $f$ uniquely determines $\textup{P}$ and because $\textup{P}$ contains all information about $f$ in it ($\textup{P}(\left \{ \omega \right \}) = f(\omega)$), we may speak of $\textup{P}$ as the probability measure of $\Omega$, and leave $f$ out of the picture. Of course, when we define a probability measure, we will allow ourselves to just define the probability mass function and the definition of $\textup{P}$ is understood as above.

There are some other quick properties we can state or prove about probability measures: $\textup{P}(\left \{ \right \}) = 0$ by convention, if $E, F$ are disjoint then $\textup{P}(E \cup F) = \textup{P}(E) + \textup{P}(F)$, and if $E \subset F \subset \Omega$ then $\textup{P}(E) \leq \textup{P}(F)$. The proofs of these facts are trivial, but a good exercise for the uncomfortable reader to work out.

## Random Variables

The next definition is crucial to the entire theory. In general, we want to investigate many different kinds of random quantities on the same probability space. For instance, suppose we have the experiment of rolling two dice. The probability space would be

$\displaystyle \Omega = \left \{ (1,1), (1,2), (1,3), \dots, (6,4), (6,5), (6,6) \right \}$

Where the probability measure is defined uniformly by setting all single outcomes to have probability 1/36. Now this probability space is very general, but rarely are we interested only in its events. If this probability space were interpreted as part of a game of craps, we would likely be more interested in the sum of the two dice than the actual numbers on the dice. In fact, we are really more interested in the payoff determined by our roll.

Sums of numbers on dice are certainly predictable, but a payoff can conceivably be any function of the outcomes. In particular, it should be a function of $\Omega$ because all of the randomness inherent in the game comes from the generation of an output in $\Omega$ (otherwise we would define a different probability space to begin with).

And of course, we can compare these two different quantities (the amount of money and the sum of the two dice) within the framework of the same probability space. This “quantity” we speak of goes by the name of a random variable.

Definition: random variable $X$ is a real-valued function on the sample space $\Omega \to \mathbb{R}$.

So for example the random variable for the sum of the two dice would be $X(a,b) = a+b$. We will slowly phase out the function notation as we go, reverting to it when we need to avoid ambiguity.

We can further define the set of all random variables $\textup{RV}(\Omega)$. It is important to note that this forms a vector space. For those readers unfamiliar with linear algebra, the salient fact is that we can add two random variables together and multiply them by arbitrary constants, and the result is another random variable. That is, if $X, Y$ are two random variables, so is $aX + bY$ for real numbers $a, b$. This function operates linearly, in the sense that its value is $(aX + bY)(\omega) = aX(\omega) + bY(\omega)$. We will use this property quite heavily, because in most applications the analysis of a random variable begins by decomposing it into a combination of simpler random variables.

Of course, there are plenty of other things one can do to functions. For example, $XY$ is the product of two random variables (defined by $XY(\omega) = X(\omega)Y(\omega)$) and one can imagine such awkward constructions as $X/Y$ or $X^Y$. We will see in a bit why it these last two aren’t often used (it is difficult to say anything about them).

The simplest possible kind of random variable is one which identifies events as either occurring or not. That is, for an event $E$, we can define a random variable which is 0 or 1 depending on whether the input is a member of $E$. That is,

Definition: An indicator random variable $1_E$ is defined by setting $1_E(\omega) = 1$ when $\omega \in E$ and 0 otherwise. A common abuse of notation for singleton sets is to denote $1_{\left \{ \omega \right \} }$ by $1_\omega$.

This is what we intuitively do when we compute probabilities: to get a ten when rolling two dice, one can either get a six, a five, or a four on the first die, and then the second die must match it to add to ten.

The most important thing about breaking up random variables into simpler random variables will make itself clear when we see that expected value is a linear functional. That is, probabilistic computations of linear combinations of random variables can be computed by finding the values of the simpler pieces. We can’t yet make that rigorous though, because we don’t yet know what it means to speak of the probability of a random variable’s outcome.

Definition: Denote by $\left \{ X = k \right \}$ the set of outcomes $\omega \in \Omega$ for which $X(\omega) = k$. With the function notation, $\left \{ X = k \right \} = X^{-1}(k)$.

This definition extends to constructing ranges of outcomes of a random variable. i.e., we can define $\left \{ X < 5 \right \}$ or $\left \{ X \textup{ is even} \right \}$ just as we would naively construct sets. It works in general for any subset of $S \subset \mathbb{R}$. The notation is $\left \{ X \in S \right \} = X^{-1}(S)$, and we will also call these sets events. The notation becomes useful and elegant when we combine it with the probability measure $\textup{P}$. That is, we want to write things like $\textup{P}(X \textup{ is even})$ and read it in our head “the probability that $X$ is even”.

This is made rigorous by simply setting

$\displaystyle \textup{P}(X \in S) = \sum_{\omega \in X^{-1}(S)} \textup{P}(\omega)$

In words, it is just the sum of the probabilities that individual outcomes will have a value under $X$ that lands in $S$. We will also use for $\textup{P}(\left \{ X \in S \right \} \cap \left \{ Y \in T \right \})$ the shorthand notation $\textup{P}(X \in S, Y \in T)$ or $\textup{P}(X \in S \textup{ and } Y \in T)$.

Often times $\left \{ X \in S \right \}$ will be smaller than $\Omega$ itself, even if $S$ is large. For instance, let the probability space be the set of possible lottery numbers for one week’s draw of the lottery (with uniform probabilities), let $X$ be the profit function. Then $\textup{P}(X > 0)$ is very small indeed.

We should also note that because our probability spaces are finite, the image of the random variable $\textup{im}(X)$ is a finite subset of real numbers. In other words, the set of all events of the form $\left \{ X = x_i \right \}$ where $x_i \in \textup{im}(X)$ form a partition of $\Omega$. As such, we get the following immediate identity:

$\displaystyle 1 = \sum_{x_i \in \textup{im} (X)} P(X = x_i)$

The set of such events is called the probability distribution of the random variable $X$.

The final definition we will give in this section is that of independence. There are two separate but nearly identical notions of independence here. The first is that of two events. We say that two events $E,F \subset \Omega$ are independent if the probability of both $E, F$ occurring is the product of the probabilities of each event occurring. That is, $\textup{P}(E \cap F) = \textup{P}(E)\textup{P}(F)$. There are multiple ways to realize this formally, but without the aid of conditional probability (more on that next time) this is the easiest way. One should note that this is distinct from $E,F$ being disjoint as sets, because there may be a zero-probability outcome in both sets.

The second notion of independence is that of random variables. The definition is the same idea, but implemented using events of random variables instead of regular events. In particular, $X,Y$ are independent random variables if

$\displaystyle \textup{P}(X = x, Y = y) = \textup{P}(X=x)\textup{P}(Y=y)$

for all $x,y \in \mathbb{R}$.

## Expectation

We now turn to notions of expected value and variation, which form the cornerstone of the applications of probability theory.

Definition: Let $X$ be a random variable on a finite probability space $\Omega$. The expected value of $X$, denoted $\textup{E}(X)$, is the quantity

$\displaystyle \textup{E}(X) = \sum_{\omega \in \Omega} X(\omega) \textup{P}(\omega)$

Note that if we label the image of $X$ by $x_1, \dots, x_n$ then this is equivalent to

$\displaystyle \textup{E}(X) = \sum_{i=1}^n x_i \textup{P}(X = x_i)$

The most important fact about expectation is that it is a linear functional on random variables. That is,

Theorem: If $X,Y$ are random variables on a finite probability space and $a,b \in \mathbb{R}$, then

$\displaystyle \textup{E}(aX + bY) = a\textup{E}(X) + b\textup{E}(Y)$

Proof. The only real step in the proof is to note that for each possible pair of values $x, y$ in the images of $X,Y$ resp., the events $E_{x,y} = \left \{ X = x, Y=y \right \}$ form a partition of the sample space $\Omega$. That is, because $aX + bY$ has a constant value on $E_{x,y}$, the second definition of expected value gives

$\displaystyle \textup{E}(aX + bY) = \sum_{x \in \textup{im} (X)} \sum_{y \in \textup{im} (Y)} (ax + by) \textup{P}(X = x, Y = y)$

and a little bit of algebraic elbow grease reduces this expression to $a\textup{E}(X) + b\textup{E}(Y)$. We leave this as an exercise to the reader, with the additional note that the sum $\sum_{y \in \textup{im}(Y)} \textup{P}(X = x, Y = y)$ is identical to $\textup{P}(X = x)$. $\square$

If we additionally know that $X,Y$ are independent random variables, then the same technique used above allows one to say something about the expectation of the product $\textup{E}(XY)$ (again by definition, $XY(\omega) = X(\omega)Y(\omega)$). In this case $\textup{E}(XY) = \textup{E}(X)\textup{E}(Y)$. We leave the proof as an exercise to the reader.

Now intuitively the expected value of a random variable is the “center” of the values assumed by the random variable. It is important, however, to note that the expected value need not be a value assumed by the random variable itself; that is, it might not be true that $\textup{E}(X) \in \textup{im}(X)$. For instance, in an experiment where we pick a number uniformly at random between 1 and 4 (the random variable is the identity function), the expected value would be:

$\displaystyle 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{4} = \frac{5}{2}$

But the random variable never achieves this value. Nevertheless, it would not make intuitive sense to call either 2 or 3 the “center” of the random variable (for both 2 and 3, there are two outcomes on one side and one on the other).

Let’s see a nice application of the linearity of expectation to a purely mathematical problem. The power of this example lies in the method: after a shrewd decomposition of a random variable $X$ into simpler (usually indicator) random variables, the computation of $\textup{E}(X)$ becomes trivial.

tournament  $T$ is a directed graph in which every pair of distinct vertices has exactly one edge between them (going one direction or the other). We can ask whether such a graph has a Hamiltonian path, that is, a path through the graph which visits each vertex exactly once. The datum of such a path is a list of numbers $(v_1, \dots, v_n)$, where we visit vertex $v_i$ at stage $i$ of the traversal. The condition for this to be a valid Hamiltonian path is that $(v_i, v_{i+1})$ is an edge in $T$ for all $i$.

Now if we construct a tournament on $n$ vertices by choosing the direction of each edges independently with equal probability 1/2, then we have a very nice probability space and we can ask what is the expected number of Hamiltonian paths. That is, $X$ is the random variable giving the number of Hamiltonian paths in such a randomly generated tournament, and we are interested in $\textup{E}(X)$.

To compute this, simply note that we can break $X = \sum_p X_p$, where $p$ ranges over all possible lists of the vertices. Then $\textup{E}(X) = \sum_p \textup{E}(X_p)$, and it suffices to compute the number of possible paths and the expected value of any given path. It isn’t hard to see the number of paths is $n!$ as this is the number of possible lists of $n$ items. Because each edge direction is chosen with probability 1/2 and they are all chosen independently of one another, the probability that any given path forms a Hamiltonian path depends on whether each edge was chosen with the correct orientation. That’s just

$\textup{P}(\textup{first edge and second edge and } \dots \textup{ and last edge})$

which by independence is

$\displaystyle \prod_{i = 1}^n \textup{P}(i^\textup{th} \textup{ edge is chosen}) = \frac{1}{2^{n-1}}$

That is, the expected number of Hamiltonian paths is $n!2^{-(n-1)}$.

## Variance and Covariance

Just as expectation is a measure of center, variance is a measure of spread. That is, variance measures how thinly distributed the values of a random variable $X$ are throughout the real line.

Definition: The variance of a random variable $X$ is the quantity $\textup{E}((X - \textup{E}(X))^2)$.

That is, $\textup{E}(X)$ is a number, and so $X - \textup{E}(X)$ is the random variable defined by $(X - \textup{E}(X))(\omega) = X(\omega) - \textup{E}(X)$. It is the expectation of the square of the deviation of $X$ from its expected value.

One often denotes the variance by $\textup{Var}(X)$ or $\sigma^2$. The square is for silly reasons: the standard deviation, denoted $\sigma$ and equivalent to $\sqrt{\textup{Var}(X)}$ has the same “units” as the outcomes of the experiment and so it’s preferred as the “base” frame of reference by some. We won’t bother with such physical nonsense here, but we will have to deal with the notation.

The variance operator has a few properties that make it quite different from expectation, but nonetheless fall our directly from the definition. We encourage the reader to prove a few:

• $\textup{Var}(X) = \textup{E}(X^2) - \textup{E}(X)^2$.
• $\textup{Var}(aX) = a^2\textup{Var}(X)$.
• When $X,Y$ are independent then variance is additive: $\textup{Var}(X+Y) = \textup{Var}(X) + \textup{Var}(Y)$.
• Variance is invariant under constant additives: $\textup{Var}(X+c) = \textup{Var}(X)$.

In addition, the quantity $\textup{Var}(aX + bY)$ is more complicated than one might first expect. In fact, to fully understand this quantity one must create a notion of correlation between two random variables. The formal name for this is covariance.

Definition: Let $X,Y$ be random variables. The covariance of $X$ and $Y$, denoted $\textup{Cov}(X,Y)$, is the quantity $\textup{E}((X - \textup{E}(X))(Y - \textup{E}(Y)))$.

Note the similarities between the variance definition and this one: if $X=Y$ then the two quantities coincide. That is, $\textup{Cov}(X,X) = \textup{Var}(X)$.

There is a nice interpretation to covariance that should accompany every treatment of probability: it measures the extent to which one random variable “follows” another. To make this rigorous, we need to derive a special property of the covariance.

Theorem: Let $X,Y$ be random variables with variances $\sigma_X^2, \sigma_Y^2$. Then their covariance is at most the product of the standard deviations in magnitude:

$|\textup{Cov}(X,Y)| \leq \sigma_X \sigma_Y$

Proof. Take any two non-constant random variables $X$ and $Y$ (we will replace these later with $X - \textup{E}(X), Y - \textup{E}(Y)$). Construct a new random variable $(tX + Y)^2$ where $t$ is a real variable and inspect its expected value. Because the function is squared, its values are all nonnegative, and hence its expected value is nonnegative. That is, $\textup{E}((tX + Y)^2)$. Expanding this and using linearity gives

$\displaystyle f(t) = t^2 \textup{E}(X^2) + 2t \textup{E}(XY) + \textup{E}(Y^2) \geq 0$

This is a quadratic function of a single variable $t$ which is nonnegative. From elementary algebra this means the discriminant is at most zero. i.e.

$\displaystyle 4 \textup{E}(XY)^2 - 4 \textup{E}(X^2) \textup{E}(Y^2) \leq 0$

and so dividing by 4 and replacing $X,Y$ with $X - \textup{E}(X), Y - \textup{E}(Y)$, resp., gives

$\textup{Cov}(X,Y)^2 \leq \sigma_X^2 \sigma_Y^2$

and the result follows. $\square$

Note that equality holds in the discriminant formula precisely when $Y = -tX$ (the discriminant is zero), and after the replacement this translates to $Y - \textup{E}(Y) = -t(X - \textup{E}(X))$ for some fixed value of $t$. In other words, for some real numbers $a,b$ we have $Y = aX + b$.

This has important consequences even in English: the covariance is maximized when $Y$ is a linear function of $X$, and otherwise is bounded from above and below. By dividing both sides of the inequality by $\sigma_X \sigma_Y$ we get the following definition:

Definition: The Pearson correlation coefficient of two random variables $X,Y$ is defined by

$\displaystyle r= \frac{\textup{Cov}(X,Y)}{\sigma_X \sigma_Y}$

If $r$ is close to 1, we call $X$ and $Y$ positively correlated. If $r$ is close to -1 we call them negatively correlated, and if $r$ is close to zero we call them uncorrelated.

The idea is that if two random variables are positively correlated, then a higher value for one variable (with respect to its expected value) corresponds to a higher value for the other. Likewise, negatively correlated variables have an inverse correspondence: a higher value for one correlates to a lower value for the other. The picture is as follows:

The  horizontal axis plots a sample of values of the random variable $X$ and the vertical plots a sample of $Y$. The linear correspondence is clear. Of course, all of this must be taken with a grain of salt: this correlation coefficient is only appropriate for analyzing random variables which have a linear correlation. There are plenty of interesting examples of random variables with non-linear correlation, and the Pearson correlation coefficient fails miserably at detecting them.

Here are some more examples of Pearson correlation coefficients applied to samples drawn from the sample spaces of various (continuous, but the issue still applies to the finite case) probability distributions:

Various examples of the Pearson correlation coefficient, credit Wikipedia.

Though we will not discuss it here, there is still a nice precedent for using the Pearson correlation coefficient. In one sense, the closer that the correlation coefficient is to 1, the better a linear predictor will perform in “guessing” values of $Y$ given values of $X$ (same goes for -1, but the predictor has negative slope).

But this strays a bit far from our original point: we still want to find a formula for $\textup{Var}(aX + bY)$. Expanding the definition, it is not hard to see that this amounts to the following proposition:

Proposition: The variance operator satisfies

$\displaystyle \textup{Var}(aX+bY) = a^2\textup{Var}(X) + b^2\textup{Var}(Y) + 2ab \textup{Cov}(X,Y)$

And using induction we get a general formula:

$\displaystyle \textup{Var} \left ( \sum_{i=1}^n a_i X_i \right ) = \sum_{i=1}^n \sum_{j = 1}^n a_i a_j \textup{Cov}(X_i,X_j)$

Note that in the general sum, we get a bunch of terms $\textup{Cov}(X_i,X_i) = \textup{Var}(X_i)$.

Another way to look at the linear relationships between a collection of random variables is via a covariance matrix.

Definition: The covariance matrix of a collection of random variables $X_1, \dots, X_n$ is the matrix whose $(i,j)$ entry is $\textup{Cov}(X_i,X_j)$.

As we have already seen on this blog in our post on eigenfaces, one can manipulate this matrix in interesting ways. In particular (and we may be busting out an unhealthy dose of new terminology here), the covariance matrix is symmetric and nonnegative, and so by the spectral theorem it has an orthonormal basis of eigenvectors, which allows us to diagonalize it. In more direct words: we can form a new collection of random variables $Y_j$ (which are linear combinations of the original variables $X_i$) such that the covariance of distinct pairs $Y_j, Y_k$ are all zero. In one sense, this is the “best perspective” with which to analyze the random variables. We gave a general algorithm to do this in our program gallery, and the technique is called principal component analysis.

## Next Up

So far in this primer we’ve seen a good chunk of the kinds of theorems one can prove in probability theory. Fortunately, much of what we’ve said for finite probability spaces holds for infinite (discrete) probability spaces and has natural analogues for continuous probability spaces.

Next time, we’ll investigate how things change for discrete probability spaces, and should we need it, we’ll follow that up with a primer on continuous probability. This will get our toes wet with some basic measure theory, but as every mathematician knows: analysis builds character.

Until then!

# Ramsey Number Lower Bound

Define the Ramsey number $R(k,m)$ to be the minimum number $n$ of vertices required of the complete graph $K_n$ so that for any two-coloring (red, blue) of the edges of $K_n$ one of two things will happen:

• There is a red $k$-clique; that is, a complete subgraph of $k$ vertices for which all edges are red.
• There is a blue $m$-clique.

It is known that these numbers are always finite, but it is very difficult to compute them exactly.

Problem: Prove that the Ramsey number $R(m,m) > n$ whenever $n,m$ satisfy

$\displaystyle \binom{n}{m}2^{1-\binom{m}{2}} < 1$

Solution: Color the edges of $K_n$ uniformly at random (that is, each edge has probability 1/2 of being colored red). For any complete subgraph $G = K_m$, define by event $A_G$ the event that $G$ is monochromatic (its edges are either all red or all blue).

Now the probability that $A_G$ occurs (where $G$ is fixed ahead of time) is easy to compute:

$\displaystyle \textup{Pr}(A_G) = \left (\frac{1}{2} \right)^{\binom{m}{2} - 1} = 2^{1-\binom{m}{2}}$

Since there are $\binom{n}{m}$ possible subgraphs with $m$ vertices, The probability that for some $G$ the event $A_G$ occurs is at most

$\displaystyle \binom{n}{m}2^{1-\binom{m}{2}}$

Whenever this quantity is strictly less than 1 (by assumption) then there is a positive probability that no event $A_G$ will occur. That is, there is a positive probability that a random coloring will have no monochromatic subgraph $K_m$. So there must exist such a coloring, and the Ramsey number $R(m,m)$ must be larger than $n$. $\square$

Discussion: This proof (originally due to Erdős) is a classic example of the so-called probabilistic method. In particular, we create a probability space from the object we wish to study, and then we make claims about the probability of joint events.

While it seems quite simple in nature, the probabilistic method has been successfully applied to a wide variety of problems in mathematics. For instance, there is an elegant proof in complexity theory that $\textup{BPP} \subset \textup{P/poly}$ which uses this same method. The probabilistic method has been applied to loads of problems in combinatorics, number theory, and graph theory, and it forms the foundation of the area of random graph theory (which is the setting in which one studies social networks). Perhaps unsurprisingly, there is also a proof of the fundamental theorem of algebra that uses the probabilistic method.