Equivalence of Definitions of Martingale in Discrete Time

Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a filtered probability space.

Let $\sequence {X_n}_{n \mathop \ge 0}$ be an adapted stochastic process.

Definition 1 implies Definition 2
Suppose that:


 * $(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$


 * $(2): \quad \forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} = X_n$.

We prove that:


 * $\forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} = X_n$.

Fix $n \in \Z_{\ge 0}$.

We induct on $m$.

For all $m \in \Z_{\ge 0}$ with $m \ge n$, let $\map P m$ be the proposition:


 * $\expect {X_m \mid \FF_n} = X_n$ almost surely.

Basis for Induction
The case $m = n$ is immediate from Conditional Expectation of Measurable Random Variable.

So $\map P n$ is seen to hold.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P m$ is true, where $m \ge n$, then it logically follows that $\map P {m + 1}$ is true.

So this is our induction hypothesis:
 * $\expect {X_m \mid \FF_n} = X_n$ almost surely.

Then we need to show:
 * $\expect {X_{m + 1} \mid \FF_n} = X_n$ almost surely.

Induction Step
This is our induction step.

From $(2)$, we have:


 * $X_m = \expect {X_{m + 1} \mid \FF_m}$ almost surely.

So we have:


 * $\expect {X_m \mid \FF_n} = \expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n}$ almost surely.

Since $n \le m$, we have $\FF_n \subseteq \FF_m$ since $\sequence {\FF_n}_{n \mathop \ge 0}$ is a filtration.

Then, by the Tower Property of Conditional Expectation, we have:


 * $\expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n} = \expect {X_{m + 1} \mid \FF_n}$ almost surely.

So:


 * $\expect {X_{m + 1} \mid \FF_n} = \expect {X_m \mid \FF_n} = X_n$ almost surely.

So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.

Definition 2 implies Definition 1
Suppose that:


 * $(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$


 * $(2): \quad \forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} = X_n$.

We need to prove that:


 * $\forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} = X_n$

This is immediate from setting $m = n + 1$ in $(2)$.