Martingales

A single random variable models one uncertain quantity. To model a quantity that evolves, like a gambler’s fortune or the price of an asset, we need a whole family of random variables indexed by time, together with a bookkeeping of what is known at each instant. Those two ingredients are the stochastic process and the filtration.

A martingale is the special case where, given everything known so far, the expected next value equals the current one, i.e., the mathematical model of a fair game.

Stochastic processes

Definition: Stochastic Process

A stochastic process is a family of random variables $\{X_t\}_{t \in T}$ defined on a common probability space $(\Omega, \cF, \Pr)$, indexed by a set $T$ interpreted as time.

Most commonly the index set is $T = \{0, 1, 2, \ldots\}$, giving a discrete-time process, or $T = [0, \infty)$, giving a continuous-time process. Everything in this section is discrete-time, so $T = \{0, 1, 2, \ldots\}$ throughout.

The second ingredient records how information accumulates.

Definition: Filtration

A filtration is a non-decreasing sequence of sub-$\sigma$-fields of $\cF$,

$$\cF_0 \subseteq \cF_1 \subseteq \cF_2 \subseteq \cdots,$$

where $\cF_n$ is interpreted as the information available by time $n$. The set of events we can resolve only ever grows with time.

The canonical example is the filtration generated by the process itself,

$$\cF_n = \sigma(X_0, \ldots, X_n),$$

the $\sigma$-field carrying exactly the information in the first $n+1$ observations: everything that can be decided once $X_0, \ldots, X_n$ have been seen, and nothing more.

For a process and a filtration to fit together, the process must not look into the future.

Definition: Adapted Process

A stochastic process $\{X_t\}_{t \in T}$ is adapted to the filtration $\{\cF_t\}_{t \in T}$ if $X_t$ is $\cF_t$-measurable for every $t \in T$.

Measurability with respect to $\cF_t$ means $X_t$ is determined by the information available at time $t$. So adaptedness says that the value of the process at time $t$ is known at time $t$. A process is always adapted to the filtration it generates, since $X_n$ is $\sigma(X_0, \ldots, X_n)$-measurable by construction.

To see what adaptedness rules out, here is a process that fails it.

Example (a process that is not adapted). Let $\{X_n\}_{n \ge 0}$ be any process with generated filtration $\cF_n = \sigma(X_0, \ldots, X_n)$, and define the running three-term average

$$Y_n = \tfrac{1}{3}\big(X_{n-1} + X_n + X_{n+1}\big), \qquad Y_0 = X_0.$$

Then $Y_n$ depends on $X_{n+1}$, which $\cF_n$ knows nothing about, so $Y_n$ is not $\cF_n$-measurable. Hence $\{Y_n\}_{n \ge 0}$ is not adapted: computing $Y_n$ requires peeking one step into the future.

Adaptedness is the compatibility condition that ties a stochastic process to a filtration, and every definition that follows assumes it.

Martingales

Definition: Martingale

Let $(\Omega, \cF, \Pr)$ be a probability space with a filtration $\{\cF_n\}_{n \ge 0}$. An $\{\cF_n\}$-adapted process $\{X_n\}_{n \ge 0}$ is a (discrete-time) martingale with respect to $\{\cF_n\}$ if:

1. $\E|X_n| < \infty$ for every $n = 0, 1, 2, \ldots$
2. $\E(X_{n+1} \mid \cF_n) = X_n$ for every $n = 0, 1, 2, \ldots$

Condition (1) keeps the conditional expectation in (2) well-defined. Condition (2) is the martingale property, and it carries the whole idea: the expected future value, given all the information up to the present, is exactly the present value. The best forecast of tomorrow is today.

Relaxing the equality in (2) to an inequality gives the two one-sided variants:

$$\E(X_{n+1} \mid \cF_n) \le X_n \;\implies\; \{X_n\} \text{ is a } \textbf{supermartingale},$$ $$\E(X_{n+1} \mid \cF_n) \ge X_n \;\implies\; \{X_n\} \text{ is a } \textbf{submartingale}.$$

The gambling reading makes the names concrete. Taking expectations on both sides of the martingale property and using the tower property, $\E[X_{n+1}] = \E[\E(X_{n+1} \mid \cF_n)] = \E[X_n]$, so the mean is constant in time:

$$\E[X_n] = \E[X_0] \qquad \text{for every } n.$$

A martingale is a fair game: on average you neither gain nor lose. The same computation on the one-sided versions makes the sequence of means monotone:

• Supermartingale: $\E[X_0] \ge \E[X_1] \ge \E[X_2] \ge \cdots$. You do not gain on average.
• Submartingale: $\E[X_0] \le \E[X_1] \le \E[X_2] \le \cdots$. You do not lose on average.

The terminology runs opposite to the everyday sense of the prefixes: a supermartingale is the one whose expectation drifts down. A useful way to remember it is that a supermartingale sits above where it is heading, while a submartingale sits below.

Example: the symmetric random walk

The cleanest first example of a martingale is the simple random walk.

Let $\{X_n\}_{n \ge 1}$ be independent and identically distributed with

$$\Pr(X_n = 1) = \Pr(X_n = -1) = \tfrac{1}{2},$$

and define the partial sums

$$S_0 = 0, \qquad S_n = X_1 + \cdots + X_n \quad (n \ge 1).$$

At each step the walk moves one unit forward or backward with equal probability. The two adjectives name these two features: symmetric because $+1$ and $-1$ are equally likely, and simple because every step has size exactly one.

Take the generated filtration $\cF_n = \sigma(X_1, \ldots, X_n)$ for $n \ge 1$, with $\cF_0 = \{\emptyset, \Omega\}$ the trivial $\sigma$-field. Then $\{S_n\}_{n \ge 0}$ is an $\{\cF_n\}$-martingale.

1. Integrability. Since each $X_i \in \{-1, +1\}$, the walk after $n$ steps satisfies $|S_n| \le n$, so $\E|S_n| \le n < \infty$.

2. Martingale property. Split off the latest increment and apply linearity:

   $$\E[S_{n+1} \mid \cF_n] = \E[S_n + X_{n+1} \mid \cF_n] = \E[S_n \mid \cF_n] + \E[X_{n+1} \mid \cF_n].$$

   The first term simplifies because $S_n$ is $\cF_n$-measurable, so it passes through conditioning untouched: $\E[S_n \mid \cF_n] = S_n$. The second term simplifies because $X_{n+1}$ is independent of $\cF_n = \sigma(X_1, \ldots, X_n)$, so conditioning has no effect: $\E[X_{n+1} \mid \cF_n] = \E[X_{n+1}] = 0$. Therefore

   $$\E[S_{n+1} \mid \cF_n] = S_n + 0 = S_n.$$

Remark

The only fact about the increments that entered the martingale property is $\E[X_{n+1}] = 0$ (together with independence and integrability). The exact distribution played no role: any independent, integrable, mean-zero increments produce a martingale, not just the $\pm 1$ steps.

This makes the generalization to a biased walk immediate. Suppose the increments are i.i.d. and integrable with $\E[X_1] = a \neq 0$. The same split gives $\E[S_{n+1} \mid \cF_n] = S_n + a$, so

$$\{S_n\} \text{ is a } \begin{cases} \textbf{supermartingale} & \text{if } a < 0, \\ \textbf{submartingale} & \text{if } a > 0. \end{cases}$$

To turn a biased walk back into a martingale, subtract the accumulated mean. The centered walk

$$M_n = S_n - n\,\E[X_1] = X_1 + \cdots + X_n - n\,\E[X_1]$$

satisfies

$$\E[M_{n+1} \mid \cF_n] = (S_n + a) - (n+1)a = S_n - na = M_n,$$

so $\{M_n\}_{n \ge 0}$ is an $\{\cF_n\}$-martingale. All of these processes, symmetric or biased, centered or not, are still called random walks.