Properties

This section collects the structural consequences of the martingale definition: how the one-step property extends across many steps, what convex functions do to a martingale, and what happens when you bet on a martingale with a predictable strategy.

Martingale properties

Conditioning across multiple steps

The defining property relates consecutive times, $\E(X_{n+1} \mid \cF_n) = X_n$. It extends to any pair of times without change in form: the best forecast of a far-future value, given the present information, is still the present value.

Theorem: Conditioning Across Steps

Statement

Let $\{X_n\}$ be a martingale and let $m < n$. Then

$$\E(X_n \mid \cF_m) = X_m.$$

If $\{X_n\}$ is a supermartingale the equality weakens to $\E(X_n \mid \cF_m) \le X_m$, and if it is a submartingale, to $\E(X_n \mid \cF_m) \ge X_m$.

Proof

We give the supermartingale case; the martingale and submartingale cases are identical with $\le$ replaced by $=$ or $\ge$. The argument is induction on $n$ with $m$ fixed.

1. Base case $n = m + 1$. This is exactly the defining inequality $\E(X_{m+1} \mid \cF_m) \le X_m$.

2. Inductive step. Assume the claim holds at $n - 1$, that is, $\E(X_{n-1} \mid \cF_m) \le X_m$. Since $\cF_m \subseteq \cF_{n-1}$, the tower property lets us condition on the larger field first:

   $$\E(X_n \mid \cF_m) = \E\!\big( \E(X_n \mid \cF_{n-1}) \mid \cF_m \big).$$

   The inner term satisfies $\E(X_n \mid \cF_{n-1}) \le X_{n-1}$ by the supermartingale property, so by monotonicity,

   $$\E(X_n \mid \cF_m) \le \E(X_{n-1} \mid \cF_m) \le X_m,$$

   the final inequality being the inductive hypothesis.

Convex functions of a martingale

Applying a convex function to a martingale cannot produce another martingale in general, but it always produces a submartingale. This is the process-level shadow of Jensen’s inequality.

Theorem: Convex Functions Give Submartingales

Statement

Let $\{X_n\}$ be an $\{\cF_n\}$-martingale and let $\varphi : \R \to \R$ be convex with $\E|\varphi(X_n)| < \infty$ for every $n$. Then $\{\varphi(X_n)\}$ is a submartingale with respect to $\{\cF_n\}$.

Proof

Since $X_n$ is $\cF_n$-measurable and a convex function is continuous, hence Borel measurable, $\varphi(X_n)$ is $\cF_n$-measurable: the process is adapted, and integrability is assumed. It remains to verify the submartingale inequality. Apply conditional Jensen’s inequality, then the martingale property $\E(X_{n+1} \mid \cF_n) = X_n$:

$$\E\!\big( \varphi(X_{n+1}) \mid \cF_n \big) \;\ge\; \varphi\!\big( \E(X_{n+1} \mid \cF_n) \big) \;=\; \varphi(X_n).$$

This is exactly the defining inequality of a submartingale for $\{\varphi(X_n)\}$.

Two important special cases:

1. taking $\varphi(x) = |x|$ shows that $\{|X_n|\}$ is a submartingale, and
2. taking $\varphi(x) = x^2$ (when $X_n \in L^2$) shows that $\{X_n^2\}$ is a submartingale.

Both feed into the maximal and convergence theorems for martingales.

Discrete stochastic integral

The construction

Definition: Discrete Stochastic Integral

Given two processes $\{X_n\}$ and $\{H_n\}$, the discrete stochastic integral of $H$ with respect to $X$ is the process

$$(H \cdot X)_n = \sum_{m=1}^n H_m (X_m - X_{m-1}), \qquad n \ge 1,$$

with $(H \cdot X)_0 = 0$.

The construction has a direct financial reading. Take $X_n$ to be the price of a stock at time $n$ and $H_n$ the number of shares held over the step from time $n-1$ to time $n$. Then $H_m (X_m - X_{m-1})$ is the profit earned on that step, and the sum

$$(H \cdot X)_n = \text{total profit up to time } n.$$

For this reason the discrete stochastic integral is also called the martingale transform: the gain of the trading strategy $\{H_n\}$ played against the price process $\{X_n\}$.

Predictable processes

A trading strategy cannot use information it does not yet have. The position held over the step to time $n$ must be chosen before the price move on that step is revealed. This is the content of predictability.

Definition: Predictable Process

A process $\{H_n\}_{n \ge 1}$ is predictable if $H_n$ is $\cF_{n-1}$-measurable for every $n \ge 1$.

Since $\cF_{n-1} \subseteq \cF_n$, a predictable process is in particular adapted. Predictability is the stronger requirement: $H_n$ is fixed using only the information available one step earlier, so you cannot bet on an increment you have already seen.

The transform of a martingale

Betting against a fair game is again a fair game. No predictable, bounded strategy can tilt the expected outcome.

Theorem: The Martingale Transform

Statement

Let $\{H_n\}$ be a predictable process with each $H_n$ bounded.

• If $\{X_n\}$ is a martingale, then $(H \cdot X)_n$ is a martingale.
• If $\{X_n\}$ is a super- or submartingale and $\{H_n\}$ is in addition non-negative, then $(H \cdot X)_n$ is a super- or submartingale respectively.

Boundedness of each $H_n$ guarantees that $(H \cdot X)_n$ is integrable, so integrability need not be checked separately.

Proof

We check the one-step property. Separating the last increment,

$$\E\!\big[ (H \cdot X)_{n+1} \mid \cF_n \big] = \E\!\big[ (H \cdot X)_n + H_{n+1}(X_{n+1} - X_n) \mid \cF_n \big].$$

Both $(H \cdot X)_n$ and $H_{n+1}$ are $\cF_n$-measurable, the latter by predictability, so by linearity and taking out what is known,

$$= (H \cdot X)_n + H_{n+1}\big( \E[X_{n+1} \mid \cF_n] - X_n \big).$$

The bracket is

$$\E[X_{n+1} \mid \cF_n] - X_n \;=\; \begin{cases} = 0 & \text{if } \{X_n\} \text{ is a martingale}, \\ \le 0 & \text{if a supermartingale}, \\ \ge 0 & \text{if a submartingale}. \end{cases}$$

In the martingale case the bracket vanishes, so $(H \cdot X)_n$ is a martingale regardless of the sign of $H_{n+1}$. In the super- or submartingale case, multiplying the signed bracket by $H_{n+1} \ge 0$ preserves its sign, which is exactly the corresponding (super/sub)martingale property for $(H \cdot X)_n$.

The non-negativity hypothesis is only needed for the one-sided cases. For a martingale the increment has zero conditional mean, so the strategy’s sign is irrelevant and any bounded predictable $\{H_n\}$ works.

Remark

On average you cannot make money off a fair game: if the stock price is a martingale, every bounded predictable strategy has zero expected gain.