Centered Sums of Independent Random Variables

Why Centered Independent Sums Are Martingales

Suppose observations arrive one at a time, and each new observation has mean zero and is independent of the observations already seen. The accumulated sum has no predictable drift: once the past is known, the expected contribution from the next observation is still zero. This simple fact is a basic way to construct martingales and to check them by conditional expectation.

All processes below are in discrete time, even when their summands have continuous distributions. Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge1}\) be independent, integrable random variables, where integrable means \(\mathbb{E}|X_n|<\infty\). Define their partial sums by \(M_0=0\) and \(M_n=\sum_{k=1}^{n}X_k\) for \(n\ge1\). Their natural filtration is \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(X_1,\ldots,X_n)\); it records exactly the values observed through time \(n\).

The filtration matters: the martingale condition asks for a conditional expectation given the information available before the next summand is revealed. Independence gives the key conditional-expectation calculation. If an integrable random variable \(X\) is independent of a sigma-algebra \(\mathcal{G}\), then \(\mathbb{E}[X\mid\mathcal{G}]=\mathbb{E}X\) almost surely. In the centered case this conditional expectation is zero. Indeed, the constant \(\mathbb{E}X\) is \(\mathcal{G}\)-measurable, and for every \(A\in\mathcal{G}\), independence gives \(\mathbb{E}[X\mathbf{1}_A]=\mathbb{E}X\,\mathbb{P}(A)\), which is the defining integral identity for conditional expectation.

Proof. Each \(M_n\) is \(\mathcal{F}_n\)-measurable because it is a sum of variables \(X_1,\ldots,X_n\), each measurable with respect to \(\mathcal{F}_n\). It is integrable because \(|M_n|\le\sum_{k=1}^n|X_k|\), so \(\mathbb{E}|M_n|\le\sum_{k=1}^n\mathbb{E}|X_k|<\infty\). For \(n\ge1\), \(X_n\) is independent of \(\mathcal{F}_{n-1}=\sigma(X_1,\ldots,X_{n-1})\), since the sequence is independent. The conditional-expectation fact above and centering give \(\mathbb{E}[X_n\mid\mathcal{F}_{n-1}]=0\) almost surely. Since \(M_n=M_{n-1}+X_n\), linearity of conditional expectation yields \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}+0=M_{n-1}\) almost surely. These are all the martingale requirements. \(\square\)

The calculation relies on the new summand being independent of the past, not merely on its unconditional mean being zero. For independent variables the centering condition is also necessary, as the next result makes precise.

Proof. If every \(\mathbb{E}X_n=0\), the theorem just proved shows that \((M_n)\) is a martingale. Conversely, suppose \((M_n)\) is a martingale. For each \(n\ge1\), its defining conditional-expectation identity and \(M_n=M_{n-1}+X_n\) imply \(\mathbb{E}[X_n\mid\mathcal{F}_{n-1}]=0\) almost surely. Taking expectations on both sides gives \(\mathbb{E}X_n=0\); the conditional expectation has the same expectation as the integrable variable \(X_n\). This holds for every \(n\), proving the reverse implication. \(\square\)

Working with Discrete-Valued Summands

For a centered-sum process, it is often clearest to write the next conditional expectation in two stages: first express the new sum as the old sum plus one increment, then condition on the past. In a discrete example, one can also check the centering directly by summing the possible values times their probabilities.

Working with Continuous-Valued Summands

The same reasoning applies when the summands have densities rather than probability masses. The conditional calculation is unchanged: independence makes the conditional mean of a new summand equal to its ordinary mean. The only additional computation is to verify that the density gives a finite mean and that centering subtracts that mean.

What the Assumptions Do—and Do Not—Say

Centering is a statement about an unconditional mean; a martingale requires a conditional-mean statement given the past. Independence bridges the two by ensuring that the past does not change the mean of the next summand. Without independence, even individually centered summands need not produce a martingale.

Common mistake: conditioning on information that already includes the summand. In the Bernoulli example, it would be wrong to write \(\mathbb{E}[X_n\mid\mathcal{F}_n]=0\). Since \(X_n\) is known given \(\mathcal{F}_n\), the correct identity is \(\mathbb{E}[X_n\mid\mathcal{F}_n]=X_n\). For example, if \(B_n=1\), then \(X_n=\tfrac23\), so its conditional expectation given \(\mathcal{F}_n\) is \(\tfrac23\), not zero. The martingale calculation conditions on \(\mathcal{F}_{n-1}\), which does not yet contain \(X_n\).

Here is a separate failure of independence. Let \(X_1\) take \(1\) and \(-1\) with probability \(\tfrac12\) each, and set \(X_2=X_1\). Both variables are integrable and centered, but they are not independent. With \(\mathcal{F}_1=\sigma(X_1)\), one has \(\mathbb{E}[X_2\mid\mathcal{F}_1]=X_1\), not zero. Thus \(M_2=X_1+X_2=2X_1\) has conditional expectation \(2X_1\) given \(\mathcal{F}_1\), whereas \(M_1=X_1\); the martingale identity fails whenever \(X_1\ne0\), which here occurs with probability one.

For independent summands with nonzero means, the raw partial sums are generally not martingales. If each \(\mathbb{E}X_k=a_k\), the conditional increment has mean \(a_k\), not zero. One can instead center each increment by subtracting its mean: \(X_k-a_k\) is integrable, centered, and the resulting partial sums form a martingale by the theorem. This adjustment changes the process being studied; it does not make the original sums martingales.

The practical test is therefore precise: identify the information \(\mathcal{F}_{n-1}\), check that the next increment is independent of it (or directly calculate its conditional mean), verify integrability, and then evaluate \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]\). A zero unconditional mean by itself is not enough.