How to Prove a Process Is a Martingale

Fix a probability measure and filtration. Verify that the process is adapted and that each value has finite absolute expectation. Then compare

E[X_t\mid\mathcal F_s]

with

X_s

for every

s\le t

. Equality gives a martingale. A conditional expectation at least as large as

X_s

gives a submartingale, and one at most as large as

X_s

gives a supermartingale. These comparisons hold almost surely.

The three lines a martingale proof needs

Use this order for a proof.

Adaptedness. For every $t$ , show that $X_t$ is a measurable function of information in $\mathcal F_t$ . For example, $B_t^2-t$ is measurable in the Brownian filtration because $B_t$ is known at time $t$ .
Integrability. Establish $E|X_t|<\infty$ for each fixed $t$ . This makes the conditional expectations well-defined.
Conditional drift. Decompose $X_t$ into information known at $s$ and future terms. Pull measurable factors outside the conditional expectation when the products are integrable. For an independent future increment, replace its conditional expectation with its unconditional expectation.

In discrete time, it is enough to show the one-step comparison because the tower property propagates it. For example, if $E[X_{n+2}\mid\mathcal F_{n+1}]\ge X_{n+1}$ , then conditioning both sides on $\mathcal F_n$ gives $E[X_{n+2}\mid\mathcal F_n]\ge E[X_{n+1}\mid\mathcal F_n]\ge X_n$ . For continuous-time exercises, a direct calculation for arbitrary $s<t$ is often simplest.

Classify the Brownian square for every drift correction

Let

B_t

be standard Brownian motion and

\mathcal F_t

its natural filtration. Consider

X_t=B_t^2-ct

, where

c

is a fixed real number. We want to know which way its conditional mean moves after time

s

Check what is known and finite. The value of $B_t$ is known at time $t$ , so $X_t$ is adapted. Also $E|X_t|\le E[B_t^2]+|c|t=(1+|c|)t$ . This makes the conditional expectations well-defined.
Separate the old value from the new shock. Write $B_t=B_s+D$ , with $D=B_t-B_s$ . The increment is independent of $\mathcal F_s$ , so its conditional first and second moments are its ordinary moments, $0$ and $t-s$ .
Expand the square before conditioning. The cross term $2B_sD$ has conditional mean zero because the known factor $2B_s$ can be taken outside. The $D^2$ term contributes $t-s$ , which is the variance of the new shock.

\begin{aligned} E[X_t\mid\mathcal F_s] &=E[B_s^2+2B_sD+D^2-ct\mid\mathcal F_s]\\ &=B_s^2+(t-s)-ct\\ &=X_s+(1-c)(t-s). \end{aligned}

For every $s<t$ , the sign is the sign of $1-c$ . The process is a submartingale if $c<1$ , a martingale if $c=1$ , and a supermartingale if $c>1$ . In an exam, the line separating $E[D\mid\mathcal F_s]$ from $E[D^2\mid\mathcal F_s]$ is worth writing out. It explains why the square creates a drift that the linear correction can exactly offset.

The same approach works in reverse when you want to construct a martingale from a terminal payoff. Let

Y

be integrable and set

M_t=E[Y\mid\mathcal F_t]

. Adaptedness holds by construction, conditional Jensen gives

E|M_t|\le E|Y|

, and conditioning on

\mathcal F_s

again gives

M_s

by the tower property. That is the pattern behind a discounted claim value under a pricing measure.

How to Prove a Process Is a Martingale

The three lines a martingale proof needs

Classify the Brownian square for every drift correction

Find the drift correction for squared Brownian motion

References and next steps