Martingales and the Optional Stopping Theorem

This is a guest post by my colleague Adam Lelkes.

The goal of this primer is to introduce an important and beautiful tool from probability theory, a model of fair betting games called martingales. In this post I will assume that the reader is familiar with the basics of probability theory. For those that need to refresh their knowledge, Jeremy’s excellent primers (1, 2) are a good place to start.

The Geometric Distribution and the ABRACADABRA Problem

Before we start playing with martingales, let’s start with an easy exercise. Consider the following experiment: we throw an ordinary die repeatedly until the first time a six appears. How many throws will this take in expectation? The reader might recognize immediately that this exercise can be easily solved using the basic properties of the geometric distribution, which models this experiment exactly. We have independent trials, every trial succeeding with some fixed probability $p$. If $X$ denotes the number of trials needed to get the first success, then clearly $\Pr(X = k) = (1-p)^{k-1} p$ (since first we need $k-1$ failures which occur independently with probability $1-p$, then we need one success which happens with probability $p$). Thus the expected value of $X$ is

$\displaystyle E(X) = \sum_{k=1}^\infty k P(X = k) = \sum_{k=1}^\infty k (1-p)^{k-1} p = \frac1p$

by basic calculus. In particular, if success is defined as getting a six, then $p=1/6$ thus the expected time is $1/p=6$.

Now let us move on to a somewhat similar, but more interesting and difficult problem, the ABRACADABRA problem. Here we need two things for our experiment, a monkey and a typewriter. The monkey is asked to start bashing random keys on a typewriter. For simplicity’s sake, we assume that the typewriter has exactly 26 keys corresponding to the 26 letters of the English alphabet and the monkey hits each key with equal probability. There is a famous theorem in probability, the infinite monkey theorem, that states that given infinite time, our monkey will almost surely type the complete works of William Shakespeare. Unfortunately, according to astronomists the sun will begin to die in a few billion years, and the expected time we need to wait until a monkey types the complete works of William Shakespeare is orders of magnitude longer, so it is not feasible to use monkeys to produce works of literature.

So let’s scale down our goals, and let’s just wait until our monkey types the word ABRACADABRA. What is the expected time we need to wait until this happens? The reader’s first idea might be to use the geometric distribution again. ABRACADABRA is eleven letters long, the probability of getting one letter right is $\frac{1}{26}$, thus the probability of a random eleven-letter word being ABRACADABRA is exactly $\left(\frac{1}{26}\right)^{11}$. So if typing 11 letters is one trial, the expected number of trials is

$\displaystyle \frac1{\left(\frac{1}{26}\right)^{11}}=26^{11}$

which means $11\cdot 26^{11}$ keystrokes, right?

Well, not exactly. The problem is that we broke up our random string into eleven-letter blocks and waited until one block was ABRACADABRA. However, this word can start in the middle of a block. In other words, we considered a string a success only if the starting position of the word ABRACADABRA was divisible by 11. For example, FRZUNWRQXKLABRACADABRA would be recognized as success by this model but the same would not be true for AABRACADABRA. However, it is at least clear from this observation that $11\cdot 26^{11}$ is a strict upper bound for the expected waiting time. To find the exact solution, we need one very clever idea, which is the following:

Let’s Open a Casino!

Do I mean that abandoning our monkey and typewriter and investing our time and money in a casino is a better idea, at least in financial terms? This might indeed be the case, but here we will use a casino to determine the expected wait time for the ABRACADABRA problem. Unfortunately we won’t make any money along the way (in expectation) since our casino will be a fair one.

Let’s do the following thought experiment: let’s open a casino next to our typewriter. Before each keystroke, a new gambler comes to our casino and bets $1 that the next letter will be A. If he loses, he goes home disappointed. If he wins, he bets all the money he won on the event that the next letter will be B. Again, if he loses, he goes home disappointed. (This won’t wreak havoc on his financial situation, though, as he only loses$1 of his own money.) If he wins again, he bets all the money on the event that the next letter will be R, and so on.

If a gambler wins, how much does he win? We said that the casino would be fair, i.e. the expected outcome should be zero. That means that it the gambler bets $1, he should receive$26 if he wins, since the probability of getting the next letter right is exactly $\frac{1}{26}$ (thus the expected value of the change in the gambler’s fortune is $\frac{25}{26}\cdot (-1) + \frac{1}{26}\cdot (+25) = 0$.

Let’s keep playing this game until the word ABRACADABRA first appears and let’s denote the number of keystrokes up to this time as $T$. As soon as we see this word, we close our casino. How much was the revenue of our casino then? Remember that before each keystroke, a new gambler comes in and bets $1, and if he wins, he will only bet the money he has received so far, so our revenue will be exactly $T$ dollars. How much will we have to pay for the winners? Note that the only winners in the last round are the players who bet on A. How many of them are there? There is one that just came in before the last keystroke and this was his first bet. He wins$26. There was one who came three keystrokes earlier and he made four successful bets (ABRA). He wins $\26^4$. Finally there is the luckiest gambler who went through the whole ABRACADABRA sequence, his prize will be $\26^{11}$. Thus our casino will have to give out $26^{11}+26^4+26$ dollars in total, which is just under the price of 200,000 WhatsApp acquisitions.

Now we will make one crucial observation: even at the time when we close the casino, the casino is fair! Thus in expectation our expenses will be equal to our income. Our income is $T$ dollars, the expected value of our expenses is $26^{11}+26^4+26$ dollars, thus $E(T)=26^{11}+26^4+26$. A beautiful solution, isn’t it? So if our monkey types at 150 characters per minute on average, we will have to wait around 47 million years until we see ABRACADABRA. Oh well.

Time to be More Formal

After giving an intuitive outline of the solution, it is time to formalize the concepts that we used, to translate our fairy tales into mathematics. The mathematical model of the fair casino is called a martingale, named after a class of betting strategies that enjoyed popularity in 18th century France. The gambler’s fortune (or the casino’s, depending on our viewpoint) can be modeled with a sequence of random variables. $X_0$ will denote the gambler’s fortune before the game starts, $X_1$ the fortune after one round and so on. Such a sequence of random variables is called a stochastic process. We will require the expected value of the gambler’s fortune to be always finite.

How can we formalize the fairness of the game? Fairness means that the gambler’s fortune does not change in expectation, i.e. the expected value of $X_n$, given $X_1, X_2, \ldots, X_{n-1}$ is the same as $X_{n-1}$. This can be written as $E(X_n | X_1, X_2, \ldots, X_{n-1}) = X_{n-1}$ or, equivalently, $E(X_n - X_{n-1} | X_1, X_2, \ldots, X_{n-1}) = 0$.

The reader might be less comfortable with the first formulation. What does it mean, after all, that the conditional expected value of a random variable is another random variable? Shouldn’t the expected value be a number? The answer is that in order to have solid theoretical foundations for the definition of a martingale, we need a more sophisticated notion of conditional expectations. Such sophistication involves measure theory, which is outside the scope of this post. We will instead naively accept the definition above, and the reader can look up all the formal details in any serious probability text (such as [1]).

Clearly the fair casino we constructed for the ABRACADABRA exercise is an example of a martingale. Another example is the simple symmetric random walk on the number line: we start at 0, toss a coin in each step, and move one step in the positive or negative direction based on the outcome of our coin toss.

The Optional Stopping Theorem

Remember that we closed our casino as soon as the word ABRACADABRA appeared and we claimed that our casino was also fair at that time. In mathematical language, the closed casino is called a stopped martingale. The stopped martingale is constructed as follows: we wait until our martingale X exhibits a certain behaviour (e.g. the word ABRACADABRA is typed by the monkey), and we define a new martingale X’ as follows: let $X'_n = X_n$ if $n < T$ and $X'_n = X_T$ if $n \ge T$ where $T$ denotes the stopping time, i.e. the time at which the desired event occurs. Notice that $T$ itself is a random variable.

We require our stopping time $T$ to depend only on the past, i.e. that at any time we should be able to decide whether the event that we are waiting for has already happened or not (without looking into the future). This is a very reasonable requirement. If we could look into the future, we could obviously cheat by closing our casino just before some gambler would win a huge prize.

We said that the expected wealth of the casino at the stopping time is the same as the initial wealth. This is guaranteed by Doob’s optional stopping theorem, which states that under certain conditions, the expected value of a martingale at the stopping time is equal to its expected initial value.

Theorem: (Doob’s optional stopping theorem) Let $X_n$ be a martingale stopped at step $T$, and suppose one of the following three conditions hold:

1. The stopping time $T$ is almost surely bounded by some constant;
2. The stopping time $T$ is almost surely finite and every step of the stopped martingale $X_n$ is almost surely bounded by some constant; or
3. The expected stopping time $E(T)$ is finite and the absolute value of the martingale increments $|X_n-X_{n-1}|$ are almost surely bounded by a constant.

Then $E(X_T) = E(X_0).$

We omit the proof because it requires measure theory, but the interested reader can see it in these notes.

For applications, (1) and (2) are the trivial cases. In the ABRACADABRA problem, the third condition holds: the expected stopping time is finite (in fact, we showed using the geometric distribution that it is less than $26^{12}$) and the absolute value of a martingale increment is either 1 or a net payoff which is bounded by $26^{11}+26^4+26$. This shows that our solution is indeed correct.

Gambler’s Ruin

Another famous application of martingales is the gambler’s ruin problem. This problem models the following game: there are two players, the first player has $a$ dollars, the second player has $b$ dollars. In each round they toss a coin and the loser gives one dollar to the winner. The game ends when one of the players runs out of money. There are two obvious questions: (1) what is the probability that the first player wins and (2) how long will the game take in expectation?

Let $X_n$ denote the change in the second player’s fortune, and set $X_0 = 0$. Let $T_k$ denote the first time $s$ when $X_s = k$. Then our first question can be formalized as trying to determine $\Pr(T_{-b} < T_a)$. Let $t = \min \{ T_{-b}, T_a\}$. Clearly $t$ is a stopping time. By the optional stopping theorem we have that

$\displaystyle 0=E(X_0)=E(X_t)=-b\Pr(T_{-b} < T_a)+a(1-\Pr(T_{-b} < T_a))$

thus $\Pr(T_{-b} < T_a)=\frac{a}{a+b}$.

I would like to ask the reader to try to answer the second question. It is a little bit trickier than the first one, though, so here is a hint: $X_n^2-n$ is also a martingale (prove it), and applying the optional stopping theorem to it leads to the answer.

A Randomized Algorithm for 2-SAT

The reader is probably familiar with 3-SAT, the first problem shown to be NP-complete. Recall that 3-SAT is the following problem: given a boolean formula in conjunctive normal form with at most three literals in each clause, decide whether there is a satisfying truth assignment. It is natural to ask if or why 3 is special, i.e. why don’t we work with $k$-SAT for some $k \ne 3$ instead? Clearly the hardness of the problem is monotone increasing in $k$ since $k$-SAT is a special case of $(k+1)$-SAT. On the other hand, SAT (without any bound on the number of literals per clause) is clearly in NP, thus 3-SAT is just as hard as $k$-SAT for any $k>3$. So the only question is: what can we say about 2-SAT?

It turns out that 2-SAT is easier than satisfiability in general: 2-SAT is in P. There are many algorithms for solving 2-SAT. Here is one deterministic algorithm: associate a graph to the 2-SAT instance such that there is one vertex for each variable and each negated variable and the literals $x$ and $y$ are connected by a directed edge if there is a clause $(\bar x \lor y)$. Recall that $\bar x \lor y$ is equivalent to $x \implies y$, so the edges show the implications between the variables. Clearly the 2-SAT instance is not satisfiable if there is a variable x such that there are directed paths $x \to \bar x$ and $\bar x \to x$ (since $x \Leftrightarrow \bar x$ is always false). It can be shown that this is not only a sufficient but also a necessary condition for unsatisfiability, hence the 2-SAT instance is satisfiable if and only if there is are no such path. If there are directed paths from one vertex of a graph to another and vice versa then they are said to belong to the same strongly connected component. There are several graph algorithms for finding strongly connected components of directed graphs, the most well-known algorithms are all based on depth-first search.

Now we give a very simple randomized algorithm for 2-SAT (due to Christos Papadimitriou in a ’91 paper): start with an arbitrary truth assignment and while there are unsatisfied clauses, pick one and flip the truth value of a random literal in it. Stop after $O(n^2)$ rounds where $n$ denotes the number of variables. Clearly if the formula is not satisfiable then nothing can go wrong, we will never find a satisfying truth assignment. If the formula is satisfiable, we want to argue that with high probability we will find a satisfying truth assignment in $O(n^2)$ steps.

The idea of the proof is the following: fix an arbitrary satisfying truth assignment and consider the Hamming distance of our current assignment from it. The Hamming distance of two truth assignments (or in general, of two binary vectors) is the number of coordinates in which they differ. Since we flip one bit in every step, this Hamming distance changes by $\pm 1$ in every round. It also easy to see that in every step the distance is at least as likely to be decreased as to be increased (since we pick an unsatisfied clause, which means at least one of the two literals in the clause differs in value from the satisfying assignment).

Thus this is an unfair “gambler’s ruin” problem where the gambler’s fortune is the Hamming distance from the solution, and it decreases with probability at least $\frac{1}{2}$. Such a stochastic process is called a supermartingale — and this is arguably a better model for real-life casinos. (If we flip the inequality, the stochastic process we get is called a submartingale.) Also, in this case the gambler’s fortune (the Hamming distance) cannot increase beyond $n$. We can also think of this process as a random walk on the set of integers: we start at some number and in each round we make one step to the left or to the right with some probability. If we use random walk terminology, 0 is called an absorbing barrier since we stop the process when we reach 0. The number $n$, on the other hand, is called a reflecting barrier: we cannot reach $n+1$, and whenever we get close we always bounce back.

There is an equivalent version of the optimal stopping theorem for supermartingales and submartingales, where the conditions are the same but the consequence holds with an inequality instead of equality. It follows from the optional stopping theorem that the gambler will be ruined (i.e. a satisfying truth assignment will be found) in $O(n^2)$ steps with high probability.

[1] For a reference on stochastic processes and martingales, see the text of Durrett .

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!

Seam Carving for Content-Aware Image Scaling

The Problem with Cropping

Every programmer or graphic designer with some web development experience can attest to the fact that finding good images that have an exactly specified size is a pain. Since the dimensions of the sought picture are usually inflexible, an uncomfortable compromise can come in the form of cropping a large image down to size or scaling the image to have appropriate dimensions.

Both of these solutions are undesirable. In the example below, the caterpillar looks distorted in the scaled versions (top right and bottom left), and in the cropped version (bottom right) it’s more difficult to tell that the caterpillar is on a leaf; we have lost the surrounding context.

In this post we’ll look at a nice heuristic method for rescaling images called seam-carving, which pays attention to the contents of the image as it resacles. In particular, it only removes or adds pixels to the image that the viewer is least-likely to notice. In all but the most extreme cases it will avoid the ugly artifacts introduced by cropping and scaling, and with a bit of additional scaffolding it becomes a very useful addition to a graphic designer’s repertoire. At first we will focus on scaling an image down, and then we will see that the same technique can be used to enlarge an image.

Before we begin, we should motivate the reader with some examples of its use.

It’s clear that the caterpillar is far less distorted in all versions, and even in the harshly rescaled version, parts of the green background are preserved. Although the leaf is warped a little, it is still present, and it’s not obvious that the image was manipulated.

Now that the reader’s appetite has been whet, let’s jump into the mathematics of it. This method was pioneered by Avidan and Shamir, and the impatient reader can jump straight to their paper (which contains many more examples). In this post we hope to fill in the background and show a working implementation.

Images as Functions

One common way to view an image is as an approximation to a function of two real variables. Suppose we have an $n \times m$-pixel image ($n$ rows and $m$ columns of pixels). For simplicity (during the next few paragraphs), we will also assume that the pixel values of an image are grayscale intensity values between 0 and 255. Then we can imagine the pixel values as known integer values of a function $f: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}$. That is, if we take two integers $0 \leq x < n$ and $0 \leq y < m$ then we know the value $f(x,y)$; it’s just the intensity value at the corresponding pixel. For values outside these ranges, we can impose arbitrary values for $I$ (we don’t care what’s happening outside the image).

Moreover, it makes sense to assume that $f$ is a well-behaved function in between the pixels (i.e. it is differentiable). And so we can make reasonable guessed as to the true derivative of $f$ by looking at the differences between adjacent pixels. There are many ways to get a good approximation of the derivative of an image function, but we should pause a moment to realize why this is important to nail down for the purpose of resizing images.

A good rule of thumb with images is that regions of an image which are most important to the viewer are those which contain drastic changes in intensity or color. For instance, consider this portrait of Albert Einstein.

Which parts of this image first catch the eye? The unkempt hair, the wrinkled eyes, the bushy mustache? Certainly not the misty background, or the subtle shadows on his chin.

Indeed, one could even claim that an image having a large derivative at a certain pixel corresponds to high information content there (of course this is not true of all images, but perhaps it’s reasonable to claim this for photographs). And if we want to scale an image down in size, we are interested in eliminating those regions which have the smallest information content. Of course we cannot avoid losing some information: the image after resizing is smaller than the original, and a reasonable algorithm should not add any new information. But we can minimize the damage by intelligently picking which parts to remove; our naive assumption is that a small derivative at a pixel implies a small amount of information.

Of course we can’t just remove “regions” of an image to change its proportions. We have to remove the same number of pixels in each row or column to reduce the corresponding dimension (width or height, resp.). Before we get to that, though, let’s write a program to compute the gradient. For this program and the rest of the post we will use the Processing programming language, and our demonstrations will use the Javascript cross-compiler processing.js. The nice thing about Processing is that if you know Java then you know processing. All the basic language features are the same, and it’s just got an extra few native types and libraries to make graphics rendering and image displaying easier. As usual, all of the code used in this blog post is available on this blog’s Github page.

Let’s compute the gradient of this picture, and call the picture $I$:

A very nice picture whose gradient we can compute. It was taken by the artist Ria Czichotzki.

Since this is a color image, we will call it a function $I: \mathbb{R}^2 \to \mathbb{R}^3$, in the sense that the input is a plane coordinate $(x,y)$, and the output $I(x,y) = (r,g,b)$ is a triple of color intensity values. We will approximate the image’s partial derivative $\left \langle \partial I / \partial x, \partial I / \partial y \right \rangle$ at $(x,y)$ by inspecting values of $I$ in a neighborhood of the point:

$I(x-1,y), I(x+1, y), I(x,y-1), I(x,y+1)$.

For each pixel we call the value $|I(x+1,y) - I(x-1,y)| / 2$ the partial derivative in the $x$ direction, and $|I(x,y+1) - I(x,y-1)| / 2$ the partial in the $y$ direction. Note that the values $I(x,y)$ are vectors, so the norm signs here are really computing the distance between the two values of $I$.

There are two ways to see why this makes sense as an approximation. The first is analytic: by definition, the partial derivative $\partial I / \partial x$ is a limit:

$\displaystyle \lim_{h \to 0} \frac{|I(x+h,y) - I(x,y)|}{h}$

It turns out that this limit is equivalent to

$\displaystyle \lim_{h \to 0} \frac{|I(x+h,y) - I(x-h,y)|}{2h}$

And the closer $h$ gets to zero the better the approximation of the limit is. Since the closest we can make $h$ is $h=1$ (we don’t know any other values of $I$ with nonzero $h$), we plug in the corresponding values for neighboring pixels. The partial $\partial I / \partial y$ is similar.

The second way to view it is geometric.

The slope of the blue secant line is not a bad approximation to the derivative at x, provided the resolution is fine enough.

The salient fact here is that a nicely-behaved curve at $x$ will have a derivative close to the secant line between the points $(x-1, f(x-1))$ and $(x+1, f(x+1))$. Indeed, this idea inspires the original definition of the derivative. The slope of the secant line is just $(f(x+1) - f(x-1)) / 2$. As we saw in our post on numerical integration, we can do much better than a linear guess (specifically, we can use do any order of polynomial interpolation we wish), but for the purposes of displaying the concept of seam-carving, a linear guess will suffice.

And so with this intuitive understanding of how to approximate the gradient, the algorithm to actually do it is a straightforward loop. Here we compute the horizontal gradient (that is, the derivative $\partial I / \partial x$).

PImage horizontalGradient(PImage img) {
color left, right;
int center;
PImage newImage = createImage(img.width, img.height, RGB);

for (int x = 0; x < img.width; x++) {
for (int y = 0; y < img.height; y++) {
center = x + y*img.width;

left = x == 0 ? img.pixels[center] : img.pixels[(x-1) + y*img.width];
right = x == img.width-1 ? img.pixels[center] : img.pixels[(x+1) + y*img.width];

newImage.pixels[center] = color(colorDistance(left, right));
}
}

return newImage;
}


The details are a bit nit-picky, but the idea is simple. If we’re inspecting a non-edge pixel, then we can use the formula directly and compute the values of the neighboring left and right pixels. Otherwise, the “left” pixel or the “right” pixel will be outside the bounds of the image, and so we replace it with the pixel we’re inspecting. Mathematically, we’d be computing the difference $|I(x, y) - I(x+1, y)|$ and $|I(x-1,y) - I(x, y)|$. Additionally, since we’ll later only be interested in the relative sizes of the gradient, we can ignore the factor of 1/2 in the formula we derived.

The parts of this code that are specific to Processing also deserve some attention. Specifically, we use the built-in types PImage and color, for representing images and colors, respectively. The “createImage” function creates an empty image of the specified size. And peculiarly, the pixels of a PImage are stored as a one-dimensional array. So as we’re iterating through the rows and columns, we must compute the correct location of the sought pixel in the pixel array (this is why we have a variable called “center”). Finally, as in Java, the ternary if notation is used to keep the syntax short, and those two lines simply check for the boundary conditions we stated above.

The last unexplained bit of the above code is the “colorDistance” function. As our image function $I(x,y)$ has triples of numbers as values, we need to compute the distance between two values via the standard distance formula. We have encapsulated this in a separate function. Note that because (in this section of the blog) we are displaying the results in an image, we have to convert to an integer at the end.

int colorDistance(color c1, color c2) {
float r = red(c1) - red(c2);
float g = green(c1) - green(c2);
float b = blue(c1) - blue(c2);
return (int)sqrt(r*r + g*g + b*b);
}


Let’s see this in action on the picture we introduced earlier.

The reader who is interested in comparing the two more closely may visit this interactive page. Note that we only compute the horizontal gradient, so certain locations in the image have a large derivative but are still dark in this image. For instance, the top of the door in the background and the wooden bars supporting the bottom of the chair are dark despite the vertical color variations.

The vertical gradient computation is entirely analogous, and is left as an exercise to the reader.

Since we want to inspect both vertical and horizontal gradients, we will call the total gradient matrix $G$ the matrix whose entries $g_{i,j}$ are the sums of the magnitudes of the horizontal and vertical gradients at $i,j$:

$\displaystyle g_{i,j} = \left | \frac{\partial I}{\partial x} (i,j) \right | + \left | \frac{\partial I}{\partial y} (i,j) \right |$

The function $e(x,y) = g_{x,y}$ is often called an energy function for $I$. We will mention now that there are other energy functions one can consider, and use this energy function for the remainder of this post.

Seams, and Dynamic Programming

Back to the problem of resizing, we want a way to remove only those regions of an image that have low total gradient across all of the pixels in the region removed. But of course when resizing an image we must maintain the rectangular shape, and so we have to add or remove the same number of pixels in each column or row.

For the purpose of scaling an image down in width (and the other cases are similar), we have a few options. We could find the pixel in each row with minimal total gradient and remove it. More conservatively, we could remove those columns with minimal gradient (as a sum of the total gradient of each pixel in the column). More brashly, we could just remove pixels of lowest gradient willy-nilly from the image, and slide the rows left.

If none of these ideas sound like they would work, it’s because they don’t. We encourage the unpersuaded reader to try out each possibility on a variety of images to see just how poorly they perform. But of these options, removing an entire column happens to distort the image less than the others. Indeed, the idea of a “seam” in an image is just a slight generalization of a column. Intuitively, a seam $s_i$ is a trail of pixels traversing the image from the bottom to the top, and at each step the pixel trail can veer to the right or left by at most one pixel.

Definition: Let $I$ be an $n \times m$ image with nonnegative integer coordinates indexed from zero. A vertical seam in $I$ is a list of coordinates $s_i = (x_i, y_i)$ with the following properties:

• $y_0 = 0$ is at the bottom of the image.
• $y_{n-1} = n-1$ is at the top of the image.
• $y_i$ is strictly increasing.
• $|x_i - x_{i+1}| \leq 1$ for all $0 \leq i < n-1$.

These conditions simply formalize what we mean by a seam. The first and second impose that the seam traverses from top to bottom. The third requires the seam to always “go up,” so that there is only one pixel in each row. The last requires the seam to be “connected” in the sense that it doesn’t veer too far at any given step.

Here are some examples of some vertical seams. One can easily define horizontal seams by swapping the placement of $x, y$ in the above list of conditions.

So the goal is now to remove the seams of lowest total gradient. Here the total gradient of a seam is just the sum of the energy values of the pixels in the seam.

Unfortunately there are many more seams to choose from than columns (or even individual pixels). It might seem difficult at first to find the seam with the minimal total gradient. Luckily, if we’re only interested in minima, we can use dynamic programming to compute the minimal seam ending at any given pixel in linear time.

We point the reader unfamiliar with dynamic programming to our Python primer on this topic. In this case, the sub-problem we’re working with is the minimal total gradient value of all seams from the bottom of the image to a fixed pixel. Let’s call this value $v(a,b)$. If we know $v(a,b)$ for all pixels below, say, row $i$, then we can compute the $v(i+1,b)$ for the entire row $i+1$ by taking pixel $(i+1,j)$, and adding its gradient value to the minimum of the values of possible predecessors in a seam, $v(i,j-1), v(i,j), v(i,j+1)$ (respecting the appropriate boundary conditions).

Once we’ve computed $v(a,b)$ for the entire matrix, we can look at the minimal value at the top of the image $\min_j v(n,j)$, and work backwards down the image to compute which seam gave us this minimum.

Let’s make this concrete and compute the function $v$ as a two-dimensional array called “seamFitness.”

void computeVerticalSeams() {
seamFitness = new float[img.width][img.height];
for (int i = 0; i < img.width; i++) {
}

for (int y = 1; y < img.height; y++) {
for (int x = 0; x < img.width; x++) {

if (x == 0) {
seamFitness[x][y] += min(seamFitness[x][y-1], seamFitness[x+1][y-1]);
} else if (x == img.width-1) {
seamFitness[x][y] += min(seamFitness[x][y-1], seamFitness[x-1][y-1]);
} else {
seamFitness[x][y] += min(seamFitness[x-1][y-1], seamFitness[x][y-1], seamFitness[x+1][y-1]);
}
}
}
}


We have two global variables at work here (global is bad, I know, but it’s Processing; it’s made for prototyping). The seamFitness array, and the gradientMagnitude array. We assume at the start of this function that the gradientMagnitude array is filled with sensible values.

Here we first initialize the zero’th row of the seamFitness array to have the same values as the gradient of the image. This is simply because a seam of length 1 has only one gradient value. Note here the coordinates are a bit backwards: the first coordinate represents the choice of a column, and the second represents the choice of a row. We can think of the coordinate axes of our image function having the origin in the bottom-left, the same as we might do mathematically.

Then we iterate over the rows in the matrix, and in each column we compute the fitness based on the fitness of the previous row. That’s it

To actually remove a seam, we need to create a new image of the right size, and shift the pixels to the right (or left) of the image into place. The details are technically important, but tedious to describe fully. So we leave the inspection of the code as an exercise to the reader. We provide the Processing code on this blog’s Github page, and show an example of its use below. Note each the image resizes every time the user clicks within the image.

Photograph by Raphael Goetter.

It’s interesting (and indeed the goal) to see how at first nothing is warped, and then the lines on the walls curve around the woman’s foot, and then finally the woman’s body is distorted before she gets smushed into a tiny box by the oppressive mouse.

As a quick side note, we attempted to provide an interactive version of this Processing program online in the same way we did for the gradient computation example. Processing is quite nice in that any Processing program (which doesn’t use any fancy Java libraries) can be cross-compiled to Javascript via the processing.js library. This is what we did for the gradient example. But in doing so for the (admittedly inefficient and memory-leaky) seam-carving program, it appeared to run an order of magnitude slower in the browser than locally. This was this author’s first time using Processing, so the reason for the drastic jump in runtime is unclear. If any readers are familiar with processing.js, a clarification would be very welcome in the comments.

Inserting Seams, Removing Objects, and Videos

In addition to removing seams to scale an image down, one can just as easily insert seams to make an image larger. To insert a seam, just double each pixel in the seam and push the rest of the pixels on the row to the right. The process is not hard, but it requires avoiding one pitfall: if we just add a single seam at a time, then the seam with minimum total energy will never change! So we’ll just add the same seam over and over again. Instead, if we want to add $k$ seams, one should compute the minimum $k$ seams and insert them all. If the desired resize is too large, then the programmer should pick an appropriate batch size and add seams in batches.

Another nice technique that comes from the seam-carving algorithm is to intelligently protect or destroy specific regions in the image. To do this requires a minor modification of the gradient computation, but the rest of the algorithm is identical. To protect a region, provide some way of user input specifying which pixels in the image are important, and give those pixels an artificially large gradient value (e.g., the maximum value of an integer). If the down-scaling is not too extreme, the seam computations will be guaranteed not to use any of those pixels, and inserted seams will never repeat those pixels. To remove a region, we just give the desired pixels an arbitrarily low gradient value. Then these pixels will be guaranteed to occur in the minimal seams, and will be removed from the picture.

The technique of seam-carving is a very nice tool, and as we just saw it can be extended to a variety of other techniques. In fact, seam-carving and its applications to object removal and image resizing are implemented in all of the recent versions of Photoshop. The techniques are used to adapt applications to environments with limited screen space, such as a mobile phone or tablet. Seam carving can even be adapted for use in videos. This involves an extension of the dynamic program to work across multiple frames, formally finding a minimal graph cut between two frames so that each piece of the cut is a seam in the corresponding frame. Of course there is a lot more detail to it (and the paper linked above uses this detail to improve the basic image-resizing algorithm), but that’s the rough idea.

We’ve done precious little on this blog with images, but we’d like to get more into graphics programming. There’s a wealth of linear algebra, computational geometry, and artificial intelligence hiding behind most of the computer games we like to play, and it would be fun to dive deeper into these topics. Of course, with every new post this author suggests ten new directions for this blog to go. It’s a curse and a blessing.

Until next time!

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!