The canonical example
Take the symmetric random walk: independent fair steps
\xi_i = \pm 1 (each sign with probability
\tfrac12, so \mathbb{E}[\xi_i] = \tfrac12(+1) + \tfrac12(-1) = 0),
and partial sums S_n = \xi_1 + \xi_2 + \cdots + \xi_n on the natural
filtration \mathcal{F}_n = \sigma(\xi_1, \dots, \xi_n). We verify the
martingale property \mathbb{E}[S_{n+1} \mid \mathcal{F}_n] = S_n line by
line, naming the rule that licenses each step.
Split tomorrow into today plus the new step. By definition
S_{n+1} = S_n + \xi_{n+1}, so
\mathbb{E}[S_{n+1} \mid \mathcal{F}_n] = \mathbb{E}[\,S_n + \xi_{n+1} \mid \mathcal{F}_n\,].
Linearity of conditional expectation separates the two pieces:
\mathbb{E}[\,S_n + \xi_{n+1} \mid \mathcal{F}_n\,] = \mathbb{E}[S_n \mid \mathcal{F}_n] + \mathbb{E}[\xi_{n+1} \mid \mathcal{F}_n].
First term — S_n is already known. Since
S_n is \mathcal{F}_n-measurable (it is built from
\xi_1, \dots, \xi_n), conditioning on
\mathcal{F}_n leaves it untouched — "taking out what is known":
\mathbb{E}[S_n \mid \mathcal{F}_n] = S_n.
Second term — the new step is independent of the past. Because
\xi_{n+1} is independent of
\mathcal{F}_n = \sigma(\xi_1, \dots, \xi_n), the conditioning information is
useless and the conditional expectation collapses to the ordinary one:
\mathbb{E}[\xi_{n+1} \mid \mathcal{F}_n] = \mathbb{E}[\xi_{n+1}] = 0.
Recombine. Adding the two results,
\mathbb{E}[S_{n+1} \mid \mathcal{F}_n] = S_n + 0 = S_n,
which is exactly the martingale property. More generally the same three moves — split off the new
increment, keep the known part, and use mean-zero independence on the rest — show that
any running sum of independent mean-zero increments is a martingale.
Refresh the figure to draw a fresh path. It wanders, but it has no preferred direction: it is just
as likely to drift above the dashed mean level M_0 = 0 as below it.
Constant mean, and why finance cares
The
tower property
makes the mean stand perfectly still. The cleanest statement is that
\mathbb{E}[M_n] = \mathbb{E}[M_0] for every
n, and we prove it by induction.
One-step result. Start from the defining equation
\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n and take an ordinary
(unconditional) expectation of both sides. On the left, the tower property
\mathbb{E}[\mathbb{E}[\,\cdot \mid \mathcal{F}_n]] = \mathbb{E}[\,\cdot\,]
applies:
\mathbb{E}[M_{n+1}] = \mathbb{E}\!\big[\,\mathbb{E}[M_{n+1} \mid \mathcal{F}_n]\,\big].
Now substitute the martingale property inside the outer expectation,
\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n:
\mathbb{E}\!\big[\,\mathbb{E}[M_{n+1} \mid \mathcal{F}_n]\,\big] = \mathbb{E}[M_n].
Chaining the last two displays gives the single-step conclusion
\mathbb{E}[M_{n+1}] = \mathbb{E}[M_n] \qquad \text{for every } n \ge 0.
Induction. This says consecutive means are equal. Take the base case
\mathbb{E}[M_0] = \mathbb{E}[M_0] (trivially true), and assume as the
inductive hypothesis that \mathbb{E}[M_n] = \mathbb{E}[M_0]. The one-step
result just proved then advances the hypothesis by one index:
\mathbb{E}[M_{n+1}] = \mathbb{E}[M_n] = \mathbb{E}[M_0].
By induction the equality holds for all n, which we may display as the
telescoping chain
\mathbb{E}[M_n] = \mathbb{E}[M_{n-1}] = \cdots = \mathbb{E}[M_1] = \mathbb{E}[M_0].
A fair game is fair forever: your expected fortune at any future time equals your starting stake.
This is exactly why martingales sit at the heart of pricing. Under the risk-neutral measure,
a properly discounted, arbitrage-free asset price is a martingale — its expected discounted value
tomorrow is its price today — and that single condition is what makes option prices well defined.
Let (\mathcal{F}_n) be a filtration. An adapted, integrable process
(M_n) is a martingale when
\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n \qquad \text{for every } n \ge 0,
(a submartingale if \ge M_n, a
supermartingale if \le M_n). For a martingale the
unconditional mean is constant in time,
\mathbb{E}[M_n] = \mathbb{E}[M_0] \qquad \text{for all } n,
obtained by taking expectations of the defining equation (tower property) to get
\mathbb{E}[M_{n+1}] = \mathbb{E}[M_n], then inducting on
n.
The same random walk hides a second martingale. With
S_n = \xi_1 + \cdots + \xi_n and
\xi_i = \pm 1 fair and independent, note first that
\xi_{n+1}^2 = 1 always (a sign squared is
1), so \mathbb{E}[\xi_{n+1}^2] = 1. Expand
the square of tomorrow's position:
S_{n+1}^2 = (S_n + \xi_{n+1})^2 = S_n^2 + 2 S_n \xi_{n+1} + \xi_{n+1}^2.
Condition on \mathcal{F}_n and take the three terms in turn.
S_n^2 is \mathcal{F}_n-measurable, so it stays;
in the cross term S_n comes out as known and
\mathbb{E}[\xi_{n+1} \mid \mathcal{F}_n] = \mathbb{E}[\xi_{n+1}] = 0
kills it; and \mathbb{E}[\xi_{n+1}^2 \mid \mathcal{F}_n] = \mathbb{E}[\xi_{n+1}^2] = 1:
\mathbb{E}[S_{n+1}^2 \mid \mathcal{F}_n] = S_n^2 + 2 S_n \cdot 0 + 1 = S_n^2 + 1.
Now subtract (n+1) from both sides so the extra
+1 is absorbed by the bookkeeping:
\mathbb{E}[\,S_{n+1}^2 - (n+1) \mid \mathcal{F}_n\,] = S_n^2 + 1 - (n+1) = S_n^2 - n.
So M_n = S_n^2 - n is a martingale. Its constant mean
\mathbb{E}[S_n^2 - n] = \mathbb{E}[S_0^2 - 0] = 0 immediately gives
\mathbb{E}[S_n^2] = n — the variance of the walk grows linearly in time,
the discrete seed of the \sqrt{t} scaling of Brownian motion.
Can a clever betting rule turn a fair game into a winning one? Let
(M_n) be a martingale and let (H_n) be a
predictable strategy — H_n, your stake on step
n, is fixed using only \mathcal{F}_{n-1}, the
information available before the step. Your accumulated winnings are the
martingale transform
(H \cdot M)_n = \sum_{k=1}^{n} H_k\,(M_k - M_{k-1}).
Look at the conditional expectation of the next gain. The increment is
H_{n+1}(M_{n+1} - M_n); since H_{n+1} is
\mathcal{F}_n-measurable it comes out as known, and the martingale
property makes the bracket's conditional mean vanish:
\mathbb{E}\big[H_{n+1}(M_{n+1} - M_n) \mid \mathcal{F}_n\big] = H_{n+1}\big(\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] - M_n\big) = H_{n+1}\,(M_n - M_n) = 0.
Hence \mathbb{E}[(H \cdot M)_{n+1} \mid \mathcal{F}_n] = (H \cdot M)_n:
the transformed process is itself a martingale, so
\mathbb{E}[(H \cdot M)_n] = 0 for every strategy. No predictable rule can
give you a positive expected profit — the precise sense in which "you can't beat a fair game". This
is the combinatorial heart of arbitrage-free pricing: if discounted prices are a martingale under
the risk-neutral measure, then every self-financing trading strategy has zero expected
discounted gain, so no strategy manufactures a riskless profit.
It is tempting to picture a martingale as a process that stays constant, or one that
drifts back toward its mean. Both pictures are wrong. The defining property constrains only
the expected future value given the past:
\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n,
not any single realisation. An individual path may fluctuate violently — the symmetric random walk
above is a martingale yet is unbounded, wandering arbitrarily far from
0 with probability one. What is pinned down is the lack of predictable
drift: no information available today lets you forecast an up-move or a down-move on average. A
martingale is a fair game, not a still one.
Two further traps. First, don't confuse the three siblings: a martingale has zero
expected drift, a submartingale
(\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] \ge M_n) tends to rise, and a
supermartingale (\le M_n) tends to fall — the inequality
direction, not the intuition of the words, is what counts.
Second, the martingale property is never absolute: it is always relative to a
specified filtration (\mathcal{F}_n) and a probability measure
\mathbb{P}. Change the information structure or the measure and the same
process can stop being a martingale — or start being one. That is not a technicality but the entire
engine of risk-neutral pricing:
a discounted stock price is generally not a martingale under the real-world measure
(it drifts upward as a reward for risk), yet becomes one under the equivalent risk-neutral measure
\mathbb{Q}, and that switch is exactly what makes arbitrage-free option
prices well defined.