Why Increment Variances Decide \(L^2\) Convergence
A martingale is built from successive increments whose conditional means are zero. This centering has an important consequence: increments from different times are orthogonal in \(L^2\). Therefore, the squared distance between two terms of a martingale is exactly the sum of the variances of the increments between them. An infinite martingale is Cauchy in \(L^2\) precisely when the tails of this variance sum tend to zero.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of sub-\(\sigma\)-fields of \(\mathcal{F}\). A random variable is square-integrable if \(\mathbb{E}[X^2]<\infty\). A sequence \((D_n)_{n\ge1}\) is a square-integrable martingale difference sequence if \(D_n\) is \(\mathcal{F}_n\)-measurable, \(\mathbb{E}[D_n^2]<\infty\), and \(\mathbb{E}[D_n\mid\mathcal{F}_{n-1}]=0\) almost surely for every \(n\).
The condition on the conditional means makes the increments orthogonal even when they are dependent. This is the key fact behind the test.
Proof. First take indices \(i<j\). Since \(D_i\) is \(\mathcal{F}_i\)-measurable, it is also \(\mathcal{F}_{j-1}\)-measurable. The conditional-expectation identity and the martingale-difference condition give $$ \mathbb{E}[D_iD_j] =\mathbb{E}\!\left[D_i\,\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]\right] =0. $$ The product is integrable by the Cauchy–Schwarz inequality, since both variables are square-integrable. Thus distinct increments are orthogonal. For \(m>n\), expand the square of the finite sum: $$ \mathbb{E}[(M_m-M_n)^2] =\mathbb{E}\!\left[\left(\sum_{k=n+1}^mD_k\right)^2\right] =\sum_{k=n+1}^m\mathbb{E}[D_k^2] +2\sum_{n<i<j\le m}\mathbb{E}[D_iD_j] =\sum_{k=n+1}^m\mathbb{E}[D_k^2]. $$ If the infinite variance sum is finite, its tails tend to zero, so this identity makes \((M_n)\) Cauchy in \(L^2\). Conversely, if \((M_n)\) is Cauchy in \(L^2\), then for every \(\varepsilon>0\) there is an \(N\) such that every finite sum \(\sum_{k=n+1}^m\mathbb{E}[D_k^2]\) is less than \(\varepsilon^2\) whenever \(m>n\ge N\). The partial sums of the nonnegative series are therefore bounded: the terms through \(N\) are finite, and every sum from \(N+1\) to \(m\) is bounded by \(\varepsilon^2\). Hence the series is finite. Finally, the standard completeness theorem for \(L^2\), viewed as a Hilbert space, says that every \(L^2\)-Cauchy sequence has an \(L^2\) limit. This completeness theorem is a standard result of functional analysis; its proof is not needed for the variance test. \(\square\)
The theorem tests convergence of the martingale sequence through the expectations of the squared increments. A conditional variance can help compute each term: when \(D_k\) is square-integrable and has conditional mean zero, \(\mathbb{E}[D_k^2]=\mathbb{E}[\mathbb{E}[D_k^2\mid\mathcal{F}_{k-1}]]\). But the test concerns the sum of these expectations, not merely whether a sum of random conditional variances is finite on individual sample paths.
Applying the Variance Test
Worked Example: Dyadically Shrinking Increments
Let \((\varepsilon_n)_{n\ge1}\) be independent random variables with \(\mathbb{P}(\varepsilon_n=1)=\mathbb{P}(\varepsilon_n=-1)=\tfrac12\). Take \(\mathcal{F}_n=\sigma(\varepsilon_1,\ldots,\varepsilon_n)\), and set \(D_n=2^{-n}\varepsilon_n\) and \(M_n=\sum_{k=1}^nD_k\), with \(M_0=0\). Determine whether \(M_n\) converges in \(L^2\).
Each \(\varepsilon_n\) is independent of \(\mathcal{F}_{n-1}\) and has mean zero, so \(\mathbb{E}[D_n\mid\mathcal{F}_{n-1}]=2^{-n}\mathbb{E}[\varepsilon_n]=0\). Also, \(\mathbb{E}[D_n^2]=4^{-n}\mathbb{E}[\varepsilon_n^2]=4^{-n}\). Every increment is square-integrable, and the variance sum is the geometric series $$ \sum_{k=1}^{\infty}\mathbb{E}[D_k^2] =\sum_{k=1}^{\infty}4^{-k} =\frac{1/4}{1-1/4} =\frac13. $$ The test gives \(L^2\) convergence. More precisely, for \(m>n\), $$ \mathbb{E}[(M_m-M_n)^2] =\sum_{k=n+1}^m4^{-k} =\frac{4^{-n}}{3}\bigl(1-4^{-(m-n)}\bigr) \le\frac{4^{-n}}{3}. $$ The upper bound tends to zero as \(n\) increases. This also verifies the Cauchy property directly.
Worked Example: A Harmonic Coefficient Still Gives \(L^2\) Convergence
Use the same fair, independent \(\varepsilon_n\) and filtration, but now set \(D_n=\varepsilon_n/n\). The absolute values have a divergent sum, but that is not the criterion for \(L^2\) convergence.
Independence and centering again give \(\mathbb{E}[D_n\mid\mathcal{F}_{n-1}]=0\). Since \(\varepsilon_n^2=1\), the variance of the \(n\)th increment is \(\mathbb{E}[D_n^2]=1/n^2\). To check that the sum is finite, for \(n\ge1\) use the decreasing function \(x\mapsto x^{-2}\): $$ \sum_{k=1}^{\infty}\mathbb{E}[D_k^2] =1+\sum_{k=2}^{\infty}\frac{1}{k^2} \le 1+\int_1^\infty\frac{1}{x^2}\,dx =2. $$ Thus the martingale partial sums converge in \(L^2\). For \(m>n\), the squared distance is \(\sum_{k=n+1}^m k^{-2}\), which is at most \(\int_n^\infty x^{-2}\,dx=1/n\). Substitution into the test shows that the squared distance tends to zero uniformly over \(m>n\) as \(n\) increases.
Worked Example: Constant-Variance Increments Fail the Test
Let \((\varepsilon_n)\) again be independent fair signs, and define \(D_n=\varepsilon_n\), \(M_n=\sum_{k=1}^nD_k\), with \(\mathcal{F}_n=\sigma(\varepsilon_1,\ldots,\varepsilon_n)\). This is a square-integrable martingale, since each new sign is independent of the past and has mean zero. Its increment variance is \(\mathbb{E}[D_k^2]=1\).
For any \(m>n\), the exact distance formula gives $$ \mathbb{E}[(M_m-M_n)^2]=\sum_{k=n+1}^m1=m-n. $$ In particular, taking \(m=n+1\) gives squared distance \(1\), no matter how large \(n\) is. Choosing \(\varepsilon=\tfrac12\) in the definition of an \(L^2\)-Cauchy sequence shows that the Cauchy condition fails. Equivalently, \(\sum_{k=1}^{\infty}\mathbb{E}[D_k^2]=\sum_{k=1}^{\infty}1=\infty\), so the test rules out \(L^2\) convergence.
Worked Example: Finite Accumulated Variance on Each Path Is Not Enough
Let \(N\) be a positive-integer-valued random variable with \(\mathbb{P}(N=r)=1/(r(r+1))\) for \(r\ge1\), independent of fair signs \((\varepsilon_n)\). These probabilities sum to one because $$ \sum_{r=1}^{R}\frac{1}{r(r+1)} =\sum_{r=1}^{R}\left(\frac1r-\frac{1}{r+1}\right) =1-\frac{1}{R+1}\longrightarrow1. $$ Moreover, \(\mathbb{P}(N\ge n)=1/n\), by the same telescoping sum. Put \(N\) in \(\mathcal{F}_0\), set \(\mathcal{F}_n=\sigma(N,\varepsilon_1,\ldots,\varepsilon_n)\), and define \(D_n=\mathbf{1}_{\{N\ge n\}}\varepsilon_n\).
The indicator is known at time \(n-1\), and \(\varepsilon_n\) is independent of \(\mathcal{F}_{n-1}\) with mean zero. Therefore \(\mathbb{E}[D_n\mid\mathcal{F}_{n-1}]=0\). Each increment is square-integrable, with $$ \mathbb{E}[D_n^2]=\mathbb{P}(N\ge n)=\frac1n. $$ The variance sum diverges: for each positive integer \(r\), the block from \(r+1\) to \(2r\) has \(r\) terms, each at least \(1/(2r)\), so its sum is at least \(1/2\). Consequently the partial sums of \(\sum_n1/n\) are unbounded, and the martingale is not Cauchy in \(L^2\).
Nevertheless, \(N\) is finite almost surely, since \(\mathbb{P}(N\ge n)=1/n\) tends to zero. On each such path, \(D_n=0\) for every \(n>N\), so the series has only finitely many nonzero terms and \(M_n\) is eventually constant. This pathwise observation does not contradict the \(L^2\) test: the expected sum of the squared increments is infinite.
What the Test Does—and Does Not—Say
The useful quantity is \(\mathbb{E}[D_k^2]\), not \(\mathbb{E}[|D_k|]\), and not just a pathwise sum of conditional variances. Orthogonality is what makes the squared \(L^2\) distance an additive sum of variances. Without the martingale-difference condition, cross terms need not vanish, and this formula cannot be used without further justification.
Common mistake. For the fair-sign example with \(D_k=\varepsilon_k/k\), an incorrect calculation might treat the divergent series \(\sum_k\mathbb{E}[|D_k|]=\sum_k1/k\) as evidence that the martingale partial sums fail to converge in \(L^2\). That uses the wrong quantity. The correct calculation is \(\mathbb{E}[D_k^2]=1/k^2\), whose sum is finite; the variance test therefore gives \(L^2\) convergence. Conversely, finite accumulated squared increments on almost every path, as in the random-\(N\) example, does not establish \(L^2\) convergence when the expected variance sum is infinite.
The test is an exact Cauchy criterion, not a claim that every form of convergence is equivalent. If the variance sum diverges, the theorem says the martingale is not Cauchy in \(L^2\); by itself, it does not rule out convergence in some other sense. Keeping the mode of convergence explicit prevents a variance calculation from being used to answer a different question.
Check Your Understanding
For each question, identify the filtration and check the martingale-difference condition before applying the variance test.
- If \(D_n=3^{-n}\varepsilon_n\) for independent fair signs, compute \(\sum_{n=1}^{\infty}\mathbb{E}[D_n^2]\) and decide whether the partial sums converge in \(L^2\).
- For \(D_n=\varepsilon_n/\sqrt{n}\), compute the squared \(L^2\) distance between partial sums \(M_m\) and \(M_n\), where \(m>n\). Does the Cauchy test hold?
- Why does \(\mathbb{E}[D_iD_j]=0\) for \(i<j\) hold for square-integrable martingale differences?
- In the random-\(N\) example, what is \(\mathbb{P}(N\ge n)\), and why can the sum of squared increments be finite on each path while its expectation is infinite?
- State the precise condition on \(\sum_n\mathbb{E}[D_n^2]\) that is necessary and sufficient for the partial sums to converge in \(L^2\).