Martingale difference sequence

The information is part of the claim

• $D_t$ is $\mathcal{F}_t$-measurable: it is an adapted process, observable by time $t$.
• $\mathbb{E}^{\mathbb{P}}[\lvert D_t\rvert]<\infty$: the required means exist.
• Almost surely,

Zero mean, changing risk

Consider a synthetic two-step model. Let $\xi_1,\xi_2$ be independent signs, each equally likely to be $+1$ or $-1$. Define two changes, measured in USD:

Use the trivial initial information $\mathcal{F}_0=\{\varnothing,\Omega\}$, then $\mathcal{F}_1=\sigma(\xi_1)$ and $\mathcal{F}_2=\sigma(\xi_1,\xi_2)$. Each change is observable at its own time and bounded, so adaptedness and integrability hold. The first change has mean zero. For the second,

The unconditional variance of $D_2$ is the average $(9+1)/2=5\,\text{USD}^2$. Its distribution is plainly different from that of $D_1$, whose variance is $1\,\text{USD}^2$. Neither identical distributions nor constant conditional variance is required. Normality is not required either: this example has only four equally likely paths.

The accumulated process $M_0=0$, $M_1=D_1$, $M_2=D_1+D_2$ is a martingale. Its possible terminal values are USD $4,-2,0,-2$, indexed by sign paths $++,+-,-+,--$. After the first up step, the two terminal values average to USD 1; after the first down step, they average to USD $-1$.

Uncorrelated can still mean dependent

In our example, $\text{Cov}(D_1,D_2)=0$, yet the changes are dependent. Seeing $\lvert D_2\rvert=\text{USD }3$ reveals that $D_1=\text{USD }1$. Before seeing it, that first up step had probability $1/2$; afterwards its conditional probability is 1. Zero covariance misses this dependence in magnitude.

Forecasting the mean and forecasting the risk

For any integrable adapted outcome $Y_t$, subtracting its true conditional mean gives

a martingale difference sequence. An estimated model's residuals need not have this property: its forecast may omit relevant past information or estimate the conditional mean poorly.

This distinction matters for financial returns. A conditional-mean model can leave no forecastable mean in its errors while a volatility model still predicts their scale. The martingale-difference assumption therefore leaves room for time-varying conditional risk; it does not establish that raw returns have zero conditional mean, nor does it specify a complete trading-profit model.

Practice: is zero covariance enough?

Let $Z$ take values $-1,0,1$ with equal probability. Set $A=2Z\,\text{USD}$ and $B=(3Z^2-2)\,\text{USD}$. First reveal $A$, then $B$, using the trivial initial information and the natural filtration of these two changes. Both changes have mean zero and $\text{Cov}(A,B)=0$. Do they form a martingale difference sequence?

Sources

• Peter Bartlett, Statistical Learning Theory, Lecture 5, slide 17: definition, adaptation, integrability and martingale differences.
• Steven Lalley, Conditional Expectation and Martingales, §1.2: martingale increments, orthogonality under square integrability and partial-sum second moments.
• Alexander J. McNeil, Rüdiger Frey and Paul Embrechts, Quantitative Risk Management: Concepts, Techniques and Tools, revised edition, Princeton University Press, 2015, §4.1.1, Definition 4.6: information-relative martingale differences, zero autocovariance and the additional constant-variance requirement for weak white noise.