From a Martingale to Its Absolute Value
A martingale has no predictable drift in conditional expectation, but its absolute value can increase on average: fluctuations away from zero count as positive magnitude whether they are upward or downward. This makes \(|M_n|\) a natural process for asking how large a martingale gets. The key step is to show that taking absolute values turns a martingale into a submartingale, after which Doob’s \(L^p\) maximal inequality controls the running maximum.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_k)_{k\ge0}\), meaning an increasing sequence of sigma-fields. A real-valued process \((M_k)_{k\ge0}\) is a martingale with respect to this filtration if each \(M_k\) is \(\mathcal{F}_k\)-measurable, \(\mathbb{E}|M_k|<\infty\), and \(\mathbb{E}[M_{k+1}\mid\mathcal{F}_k]=M_k\) almost surely for every \(k\). A submartingale satisfies the same measurability and integrability conditions, with \(\mathbb{E}[X_{k+1}\mid\mathcal{F}_k]\ge X_k\) almost surely.
Proof. The process \(|M_k|\) is \(\mathcal{F}_k\)-measurable and integrable because \(\mathbb{E}|M_k|<\infty\). For any integrable real random variable \(Y\), conditional expectation preserves order, so \(\mathbb{E}[|Y|\mid\mathcal{F}_k]\ge\mathbb{E}[Y\mid\mathcal{F}_k]\) and \(\mathbb{E}[|Y|\mid\mathcal{F}_k]\ge-\mathbb{E}[Y\mid\mathcal{F}_k]\), almost surely. Together these inequalities give \(\mathbb{E}[|Y|\mid\mathcal{F}_k]\ge|\mathbb{E}[Y\mid\mathcal{F}_k]|\). Set \(Y=M_{k+1}\) and use the martingale identity to obtain \(\mathbb{E}[|M_{k+1}|\mid\mathcal{F}_k]\ge|M_k|\). This is precisely the submartingale condition. \(\square\)
The conditional inequality used in the proof is conditional Jensen’s inequality for the convex function \(x\mapsto|x|\). It is an inequality, not generally an equality: averaging values of opposite signs can reduce their absolute value. That distinction is why \((|M_k|)\) is generally a submartingale rather than a martingale.
The Running-Maximum Bound
For a fixed integer \(n\ge0\), define the running maximum \(M_n^*=\max_{0\le k\le n}|M_k|\). Doob’s \(L^p\) maximal inequality says that if \((X_k)_{0\le k\le n}\) is a nonnegative submartingale and \(p>1\), then, provided \(\mathbb{E}[X_n^p]<\infty\),
This is the standard Doob \(L^p\) maximal inequality established in the earlier material on that inequality. Its hypotheses matter: it concerns a submartingale over a finite index range, requires \(p>1\), and controls the maximum using the \(p\)th moment at the final time. Apply it with \(X_k=|M_k|\). The proposition supplies the submartingale condition. Thus a martingale with \(\mathbb{E}|M_n|^p<\infty\) satisfies the following bound.
Proof. By the proposition, \(X_k=|M_k|\) is a nonnegative submartingale. The terminal \(p\)th moment of \(X\) is \(\mathbb{E}[X_n^p]=\mathbb{E}|M_n|^p<\infty\), so Doob’s \(L^p\) maximal inequality applies. Its maximum is \(\max_{0\le k\le n}X_k=M_n^*\). Substitution gives the stated inequality. The moment condition also ensures integrability at earlier times: the martingale property iterated through conditional expectations gives \(M_k=\mathbb{E}[M_n\mid\mathcal{F}_k]\), and conditional Jensen for \(x\mapsto|x|^p\) gives \(\mathbb{E}|M_k|^p\le\mathbb{E}|M_n|^p\) for each \(k\le n\). \(\square\)
The result is a bound, not an assertion that the maximum’s moment equals the right-hand side. The constant depends on \(p\), and the inequality does not provide a strong \(L^1\) bound: the factor \(p/(p-1)\) is not finite at \(p=1\). The following examples illustrate how the argument works and where its conditions enter.
Worked Example: A Two-Step Symmetric Random Walk
Let \(\Omega=\{-1,1\}^2\), with each outcome having probability \(1/4\). Let \(\varepsilon_1,\varepsilon_2\) be the two coordinates, set \(S=\{-2,0,2\}\), and define \(M_0=0\), \(M_1=\varepsilon_1\), and \(M_2=\varepsilon_1+\varepsilon_2\). Use the filtration \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_1=\sigma(\varepsilon_1)\), and \(\mathcal{F}_2=\sigma(\varepsilon_1,\varepsilon_2)\). Each increment has conditional mean zero given the preceding information, so \((M_k)\) is a martingale.
For \(p=2\), the four equally likely coordinate pairs give the following paths and maxima:
Therefore \(\mathbb{E}[(M_2^*)^2]=\tfrac14(4+1+1+4)=\tfrac52\). At the terminal time, \(M_2\) equals \(2\) and \(-2\) once each and \(0\) twice, so \(\mathbb{E}|M_2|^2=\tfrac14(4+0+0+4)=2\). Doob’s bound is \(\tfrac52\le(2/(2-1))^2\cdot2=8\). The computed maximum moment is below the bound, as required.
Check: the terminal second moment is \(2\), while the running-maximum second moment is \(5/2\). The maximum includes the intermediate value \(M_1\), which is why it need not have the same moment as the terminal value.
Worked Example: A Three-Time Martingale with a Nonzero Intermediate Value
Let \(\Omega=\{a,b,c,d\}\), with probability \(1/4\) on each outcome. Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1\) distinguish \(\{a,b\}\) from \(\{c,d\}\), and let \(\mathcal{F}_2\) reveal the outcome. Define \(M_0=0\), \(M_1=2\) on \(\{a,b\}\) and \(M_1=-2\) on \(\{c,d\}\), and set \(M_2(a)=3\), \(M_2(b)=1\), \(M_2(c)=-1\), \(M_2(d)=-3\). The average of the terminal values is \(0=M_0\). Within the first \(\mathcal{F}_1\)-atom their average is \((3+1)/2=2\); within the second it is \((-1-3)/2=-2\). Thus \(M_1=\mathbb{E}[M_2\mid\mathcal{F}_1]\), and the process is a martingale.
Use \(p=3\). On outcomes \(a,b,c,d\), respectively, the running maxima \(M_2^*\) are \(3,2,2,3\). Hence
The terminal absolute third moment is
The maximal bound is \(\tfrac{35}{2}\le(\tfrac{3}{2})^3\cdot14=\tfrac{27}{8}\cdot14=\tfrac{189}{4}\). Since \(\tfrac{35}{2}=\tfrac{70}{4}\le\tfrac{189}{4}\), the calculated values satisfy the inequality. Here \(M_1\) matters on outcomes \(b\) and \(c\): its absolute value \(2\) exceeds the terminal absolute value \(1\) on each of those outcomes.
Why the Endpoint \(p=1\) Is Different
The restriction \(p>1\) is not just a technical way to avoid a problematic constant. There is no finite constant that bounds the expected running maximum of every finite-time martingale by a constant times its terminal first absolute moment. A dyadic conditional-expectation martingale shows this directly.
Worked Example: A Family Showing the Failure of a Strong \(L^1\) Bound
Take \(\Omega=[0,1)\) with the uniform probability measure. Fix an integer \(N\ge1\). For each \(k=0,\ldots,N\), let \(\mathcal{F}_k\) be generated by the partition into half-open dyadic intervals of length \(2^{-k}\). Define \(A=[0,2^{-N})\), \(Y=\mathbf{1}_A\), and \(M_k=\mathbb{E}[Y\mid\mathcal{F}_k]\). Conditional expectations form a martingale, and \(\mathbb{E}|M_N|=\mathbb{E}Y=2^{-N}\).
At level \(k\), the dyadic cell containing 0 is \([0,2^{-k})\). On that cell, the conditional probability of \(A\) is \(2^{-N}/2^{-k}=2^{k-N}\); on every other level-\(k\) cell it is zero. For \(k=0\), the cell is all of \([0,1)\) and \(M_0=2^{-N}\). For \(1\le k\le N\), the same formula gives \(M_k=2^{k-N}\) on \([0,2^{-k})\) and zero elsewhere.
On \(A\), the running maximum is \(1\). For each \(k=0,\ldots,N-1\), the half-open ring \(R_k=[2^{-(k+1)},2^{-k})\) has probability \(2^{-(k+1)}\). A point in \(R_k\) lies in \([0,2^{-k})\), so \(M_k=2^{k-N}\); at every later level the cell containing that point no longer intersects \(A\), and its conditional expectation is zero. Earlier levels have values no larger than \(2^{k-N}\). Thus \(M_N^*=2^{k-N}\) on \(R_k\). These disjoint rings and \(A\) cover \([0,1)\), apart from endpoints of probability zero, so
Dividing by the terminal first absolute moment gives \(\mathbb{E}M_N^*/\mathbb{E}|M_N|=1+N/2\). This ratio grows without bound as \(N\) increases. Consequently no single finite constant works in an inequality \(\mathbb{E}M_N^*\le C\mathbb{E}|M_N|\) for all finite-time martingales. This example explains why the \(p>1\) condition in the strong \(L^p\) maximal bound cannot simply be replaced by \(p=1\).
Common Mistake and Practical Use
Common mistake: replacing conditional Jensen’s inequality by an equality. For example, let \(Y\) be equally likely to be \(1\) or \(-1\), and let \(\mathcal{G}\) be the trivial sigma-field. Then \(\mathbb{E}[Y\mid\mathcal{G}]=0\), so \(|\mathbb{E}[Y\mid\mathcal{G}]|=0\), whereas \(\mathbb{E}[|Y|\mid\mathcal{G}]=1\). The correct relation is \(0\le1\), not equality. Accordingly, \(|M_k|\) is guaranteed to be a submartingale; it is not generally a martingale.
In applications, first verify that the process is a martingale for the filtration being used, and identify a finite terminal time \(n\). Next check \(p>1\) and \(\mathbb{E}|M_n|^p<\infty\). Conditional Jensen then gives the submartingale \(|M_k|\), and Doob’s inequality controls the largest absolute value over all times \(0,\ldots,n\), not just the terminal value. This is useful when a pathwise event depends on whether the martingale ever becomes large before time \(n\). The inequality supplies a moment bound for that running maximum; it does not by itself assert an infinite-time bound or a limiting distribution.
Check Your Understanding
For each question, identify the relevant hypothesis before applying the maximal inequality.
- Why does conditional Jensen’s inequality imply \(\mathbb{E}[|M_{k+1}|\mid\mathcal{F}_k]\ge |M_k|\)? State the martingale identity used.
- If \(p=2\) and \(\mathbb{E}|M_n|^2=5\), what upper bound does Doob’s inequality give for \(\mathbb{E}[(M_n^*)^2]\)?
- In the four-outcome example, which outcomes have a running maximum strictly larger than their terminal absolute value?
- For the dyadic example with \(N=4\), compute \(\mathbb{E}M_N^*\) and \(\mathbb{E}|M_N|\) from the displayed formulas, and compare their ratio with \(1+N/2\).
- Why does the dyadic family rule out a uniform strong \(L^1\) maximal bound, even though every individual martingale in the family has a finite expected maximum?