Martingales: a fair game in mathematical form

What makes a random process a fair game?

A process is a martingale when the conditional mean of any later value, given everything seen so far, is the value now. Its paths still wander up and down, but the forecast line drawn from any time is flat.

Before you start · 3 quick checks on conditional expectation, filtrations and the simple random walk, about 4 min

Setup: information, forecasts and the definition

Information. A filtration $(\mathcal{F}_t)_{t \ge 0}$ records what is known at each time: $\mathcal{F}_t$ is the collection of events settled by what has been observed up to time $t$. Nothing observed is forgotten, so $\mathcal{F}_s \subseteq \mathcal{F}_t$ for $s \le t$. For coin games, $\mathcal{F}_k$ is the information in the first $k$ tosses. For a Brownian motion $W$, $\mathcal{F}_t$ is the information in the path $W_u$, $0 \le u \le t$.

Forecasts. For a random variable $Y$ with $\mathbb{E}\lvert Y \rvert \lt \infty$, the conditional expectation $\mathbb{E}[Y \mid \mathcal{F}_s]$ is the forecast of $Y$ from the information at time $s$: a random variable known at time $s$ whose average over each event of $\mathcal{F}_s$ equals the average of $Y$ over that event. On a coin tree it is, at each node, the probability-weighted average of $Y$ over the paths through that node. When $\mathbb{E}[Y^2] \lt \infty$, it is also the best forecast in mean square: among random variables $Z$ known at time $s$, it makes $\mathbb{E}[(Y - Z)^2]$ smallest.

Definition. A process $(X_t)$ is a martingale with respect to $(\mathcal{F}_t)$ when, for every time $t$,

it is adapted: $X_t$ is known at time $t$, that is, $X_t$ is $\mathcal{F}_t$-measurable;
it is integrable: $\mathbb{E}\lvert X_t \rvert \lt \infty$;
it has the martingale property: $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$ for every $s \le t$.

An adapted, integrable process with $\mathbb{E}[X_t \mid \mathcal{F}_s] \le X_s$ for all $s \le t$ is a supermartingale, an unfavorable game. With $\ge$ it is a submartingale, a favorable game.

Rules for forecasts, used on the whole page, for times $s \le t$ and random variables with finite expectations:

Linearity: $\mathbb{E}[aY + bZ \mid \mathcal{F}_s] = a\,\mathbb{E}[Y \mid \mathcal{F}_s] + b\,\mathbb{E}[Z \mid \mathcal{F}_s]$ for numbers $a$ and $b$.
Taking out what is known: if $Y$ is known at time $s$, then $\mathbb{E}[YZ \mid \mathcal{F}_s] = Y\,\mathbb{E}[Z \mid \mathcal{F}_s]$, when $YZ$ has a finite expectation. In particular $\mathbb{E}[Y \mid \mathcal{F}_s] = Y$.
Independence: if $Z$ is independent of $\mathcal{F}_s$, then $\mathbb{E}[Z \mid \mathcal{F}_s] = \mathbb{E}[Z]$, a number.
Tower: $\mathbb{E}\big[\mathbb{E}[Y \mid \mathcal{F}_t] \mid \mathcal{F}_s\big] = \mathbb{E}[Y \mid \mathcal{F}_s]$ and $\mathbb{E}\big[\mathbb{E}[Y \mid \mathcal{F}_s]\big] = \mathbb{E}[Y]$.

Brownian motion. $W$ is a standard Brownian motion, as built in step 9 of the path, From random walk to Brownian motion: $W_0 = 0$, its paths are continuous, and for $0 \le s \lt t$ the increment $W_t - W_s$ is normal with mean 0 and variance $t - s$ and is independent of $\mathcal{F}_s$. The figures draw paths on a partition $0 = t_0 \lt t_1 \lt \dots \lt t_n = T$.

Coin games. Each round a coin is tossed, independently of the other rounds, and lands heads with probability $p$, where $0 \lt p \lt 1$. The player wins $1 on heads and loses $1 on tails. $X_k$ is the net winnings after $k$ rounds, with $X_0 = 0$; it equals $S_k$, the number of heads minus the number of tails so far.

Fine print: "with probability 1", one step in discrete time, and the filtration

Conditional expectations are defined up to events of probability 0, so an equation such as $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$ means that the two random variables are equal with probability 1. On a finite coin tree with $0 \lt p \lt 1$ every path has positive probability, so there the equation holds at every node.

In discrete time it is enough to check one step, $\mathbb{E}[X_{k+1} \mid \mathcal{F}_k] = X_k$ for every $k$: by the tower rule, $\mathbb{E}[X_{k+2} \mid \mathcal{F}_k] = \mathbb{E}\big[\mathbb{E}[X_{k+2} \mid \mathcal{F}_{k+1}] \mid \mathcal{F}_k\big] = \mathbb{E}[X_{k+1} \mid \mathcal{F}_k] = X_k$, and so on for later times.

Being a martingale depends on the filtration. On this page the filtration is always the information generated by the coin tosses or by $W$.

1 See

Sort the processes

Here are six processes. As in the Setup, the information $\mathcal{F}$ is that of the coin tosses for the two games and that of the path of $W$ for the other four. Decide for each one whether it is a martingale, then check your sort. The page then draws the process's paths, and from a time $s$ that you choose it draws the forecast line $t \mapsto \mathbb{E}[X_t \mid \mathcal{F}_s]$ for the later times.

2 Try

Check the condition node by node

The coin tree

Three rounds of a coin game. Each box is a history of tosses, such as HT, with the value $X_k$ after that history; up is heads. At a node at time $k$, the forecast of the next value is the average of the two values that can follow, weighted by their probabilities: \[\mathbb{E}[X_{k+1} \mid \mathcal{F}_k] = p \cdot X(\text{after heads}) + (1 - p) \cdot X(\text{after tails}).\] Pick a node at time 0, 1 or 2, type your forecast and check it. By the fine print in Start, these one-step checks decide whether $X$ is a martingale.

Your forecast

Key point On a tree the martingale condition is one weighted average per node. A drift shows up as the same gap at every node: $-0.1$ for the unfavorable game, $+1$ for $S_k^2$. Subtracting the accumulated drift, as in $S_k^2 - k$, gives a martingale.

3 Prove

$W_t^2 - t$ is a martingale

Theorem

Assumptions.

$W$ is a standard Brownian motion and $\mathcal{F}_t$ is the information in its path up to time $t$.
For $0 \le s \lt t$, the increment $W_t - W_s$ is independent of $\mathcal{F}_s$ and normal with mean 0 and variance $t - s$.
The rules for forecasts from the Setup.

Conclusion. $X_t = W_t^2 - t$ is a martingale: it is adapted and integrable, and $\mathbb{E}[W_t^2 - t \mid \mathcal{F}_s] = W_s^2 - s$ for all $0 \le s \le t$.

Each equation between random variables below holds with probability 1. The proof fixes $0 \lt s \lt t$; the case $s = t$ is immediate, and at $s = 0$ the forecast is the plain expectation $\mathbb{E}[W_t^2] - t = 0 = X_0$.

Scratch work, not part of the proof.

More results from independent increments

Same assumptions as above. Write $D = W_t - W_s$ for $0 \le s \lt t$.

$W$ is a martingale. $\mathbb{E}[W_t \mid \mathcal{F}_s] = \mathbb{E}[W_s + D \mid \mathcal{F}_s] = W_s + \mathbb{E}[D] = W_s$, by taking out what is known and independence. $W$ is integrable because $\mathbb{E}\lvert W_t \rvert \le \sqrt{\mathbb{E}[W_t^2]} = \sqrt{t}$.
The exponential martingale. For a number $\sigma$, let $X_t = e^{\sigma W_t - \sigma^2 t/2}$. For $D$ normal with mean 0 and variance $t - s$, $\mathbb{E}[e^{\sigma D}] = e^{\sigma^2 (t - s)/2}$. Since $e^{\sigma W_s}$ is known at time $s$ and $e^{\sigma D}$ is independent of $\mathcal{F}_s$, $\mathbb{E}[e^{\sigma W_t} \mid \mathcal{F}_s] = e^{\sigma W_s}\,\mathbb{E}[e^{\sigma D}] = e^{\sigma W_s} e^{\sigma^2 (t - s)/2}$. Multiplying by the number $e^{-\sigma^2 t/2}$ gives $\mathbb{E}[X_t \mid \mathcal{F}_s] = e^{\sigma W_s - \sigma^2 s/2} = X_s$. $X_t \gt 0$ and $\mathbb{E}[X_t] = 1$, so $X$ is integrable.
Coin games. $\mathbb{E}[X_{k+1} \mid \mathcal{F}_k] = X_k + p(1) + (1 - p)(-1) = X_k + (2p - 1)$. The game is a martingale for $p = \tfrac{1}{2}$, a supermartingale for $p \lt \tfrac{1}{2}$ and a submartingale for $p \gt \tfrac{1}{2}$, and $\mathbb{E}[X_k] = k(2p - 1)$.
Constant expectation. For a martingale, the tower rule gives $\mathbb{E}[X_t] = \mathbb{E}\big[\mathbb{E}[X_t \mid \mathcal{F}_s]\big] = \mathbb{E}[X_s]$ for all $s \le t$, so $\mathbb{E}[X_t] = \mathbb{E}[X_0]$. The converse fails. The last toss $\xi_k$ of a fair coin has $\mathbb{E}[\xi_k] = 0$ for every $k \ge 1$, but $\mathbb{E}[\xi_{k+1} \mid \mathcal{F}_k] = 0$, which differs from $\xi_k = \pm 1$ on every path, for every $k \ge 1$.
Stopping. Can a player beat a fair game by choosing when to stop? For stopping rules with a bounded time, no; that is the optional stopping theorem, the subject of a later page.

The next steps of the path use these facts. Step 12, The Itô integral, builds new martingales from $W$, and step 17 of the option pricing path shows that, in the frictionless market model without arbitrage used there, the price of a stock that pays no dividends, divided by the value of a bank account, is a martingale under the risk-neutral probability $Q$. Quadratic variation, step 10 (Quadratic variation and (dW)² = dt), explains why the correction here is exactly $t$.

Key point To prove a martingale property, split $W_t = W_s + (W_t - W_s)$, take out what is known at time $s$, and replace what is independent of $\mathcal{F}_s$ by its mean. For $W_t^2$ the independent part leaves a drift $t - s$, and subtracting $t$ removes it.

4 Apply

Problems

$W$ is a standard Brownian motion with its filtration. Coin tosses and game rounds are independent.

Martingales: a fair game in mathematical form

Sort the processes

Paths and the forecast seen from time \(s\)

Check the condition node by node

The coin tree

\(W_t^2 - t\) is a martingale

Problems

Further reading