Tutorials › Stochastic Calculus › Worked Applications of Doob’s Maximal Inequalities

Continuous-Time Martingales · Tutorial 30 of 600

Worked Applications of Doob's Maximal Inequalities

Learn to turn maximal inequalities into explicit probability and moment estimates, and to assess how sharp those estimates are in finite-state martingale examples.

Advanced 12 min read

What You'll Learn

  • Apply the weak-type inequality to bound the chance a nonnegative submartingale crosses a threshold.
  • Convert Doob’s \(L^p\) maximal inequality into an explicit tail bound.
  • Derive lower-order running-maximum moment bounds from a terminal \(p\)-moment.
  • Compute exact running maxima for finite-state martingales and compare them with inequality bounds.
  • Check threshold conventions and identify when a maximal bound is uninformative.

From Maximal Inequalities to Useful Bounds

Doob’s maximal inequalities turn information about a process at a fixed terminal time into control of its entire path up to that time. This is useful when the quantity of interest is a crossing event, such as whether a martingale ever exceeds a prescribed level, or a moment of its running maximum. The resulting estimates need not be sharp for a particular process: comparing them with an exact calculation helps show what the inequalities do and do not tell us.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). The arguments below do not use Brownian motion or stochastic integrals. In particular, no integrand class is needed. The example filtrations may contain information independent of \(B_t\); adjoining the Brownian filtration then leaves the conditional expectations in the examples unchanged.

Definition: For a real-valued process \(M\) on \([0,T]\), its running absolute maximum is \(M_T^*=\sup_{0\le t\le T}|M_t|\). For a nonnegative process \(X\), write \(X_T^*=\sup_{0\le t\le T}X_t\). We use right-continuous paths, so these suprema are measurable: they equal the suprema over rational times in \([0,T]\), together with the endpoint \(T\).

Two inequalities will be used. The weak-type maximal inequality says that if \(X\) is a nonnegative, right-continuous submartingale and \(\mathbb{E}[X_T]<\infty\), then for every \(\lambda>0\), \(\mathbb{P}(X_T^*\ge\lambda)\le\mathbb{E}[X_T]/\lambda\). The \(L^p\) maximal inequality says that if \(M\) is a right-continuous martingale and \(\mathbb{E}[|M_T|^p]<\infty\) for some \(p>1\), then \(\mathbb{E}[(M_T^*)^p]\le (p/(p-1))^p\mathbb{E}[|M_T|^p]\). These are the standard Doob maximal inequalities established in the earlier material on continuous-time martingales. In particular, \(|M_t|\) is a nonnegative submartingale, by conditional Jensen’s inequality, so the weak-type bound applies to it.

Tail Bounds from Terminal-Time Information

Proposition: If \(M\) is a right-continuous martingale on \([0,T]\), \(\mathbb{E}[|M_T|^p]<\infty\) for \(p>1\), and \(K>0\), then \(\mathbb{P}(M_T^*\ge K)\le (p/(p-1))^p\mathbb{E}[|M_T|^p]/K^p\). The bound may be replaced by its minimum with \(1\).

Proof. The \(L^p\) maximal inequality gives \(\mathbb{E}[(M_T^*)^p]\le (p/(p-1))^p\mathbb{E}[|M_T|^p]\), so \(M_T^*\) has finite \(p\)-th moment. On the event \(\{M_T^*\ge K\}\), the nonnegative random variable \((M_T^*)^p\) is at least \(K^p\). Consequently \(\mathbb{E}[(M_T^*)^p]\ge K^p\mathbb{P}(M_T^*\ge K)\). Combining these inequalities and dividing by \(K^p>0\) proves the claim. A probability is at most \(1\), which gives the stated improvement. \(\square\)

The weak-type inequality often gives a more direct crossing estimate when a first moment is available. For a martingale \(M\) with \(\mathbb{E}[|M_T|]<\infty\), applying it to \(|M|\) yields \(\mathbb{P}(M_T^*\ge K)\le\mathbb{E}[|M_T|]/K\). The threshold event here is inclusive: it counts a path that reaches \(K\) exactly. The bounds also hold for the strict event \(\{M_T^*>K\}\), since that event is contained in \(\{M_T^*\ge K\}\).

Worked Example: A Nonnegative Martingale Crossing a Level

Let \(A\) be an event with \(\mathbb{P}(A)=1/4\), and set \(X=4\mathbf{1}_A\). Use the filtration that reveals \(A\) at time \(1/2\) and reveals \(X\) at time \(1\). Before time \(1/2\), the conditional expectation \(M_t=\mathbb{E}[X\mid\mathcal{F}_t]\) is \(1\); from time \(1/2\) up to time \(1\), it is \(4\) on \(A\) and \(0\) on \(A^c\); at time \(1\), \(M_1=X\). This is a right-continuous martingale. Its terminal mean is \(\mathbb{E}[M_1]=4\mathbb{P}(A)=4\cdot(1/4)=1\).

If \(A\) occurs, \(M_1^*=4\). If \(A^c\) occurs, the process has value \(1\) before the reveal and then value \(0\), so \(M_1^*=1\). Thus \(\mathbb{P}(M_1^*\ge3)=\mathbb{P}(A)=1/4\). The weak-type bound gives \(\mathbb{P}(M_1^*\ge3)\le\mathbb{E}[M_1]/3=1/3\), which is valid but not exact. Check: the two running-maximum values give \(\mathbb{E}[M_1^*]=(1/4)\cdot4+(3/4)\cdot1=7/4\); this is at least the terminal mean \(1\), as expected.

Moment Bounds and Lower-Order Moments

Corollary: Suppose \(M\) is a right-continuous martingale with \(\mathbb{E}[|M_T|^p]<\infty\) for \(p>1\). For every \(q\) with \(0<q<p\), \(\mathbb{E}[(M_T^*)^q]\le (p/(p-1))^q(\mathbb{E}[|M_T|^p])^{q/p}\).

Proof. The \(L^p\) maximal inequality first gives \(\mathbb{E}[(M_T^*)^p]\le (p/(p-1))^p\mathbb{E}[|M_T|^p]<\infty\). Apply the monotonicity of probability-space \(L^r\) norms, which follows from Hölder’s inequality: for a nonnegative random variable \(Y\) with finite \(p\)-th moment and \(q<p\), \(\mathbb{E}[Y^q]\le(\mathbb{E}[Y^p])^{q/p}\). Taking \(Y=M_T^*\), then substituting the \(L^p\) maximal bound, proves the displayed inequality. This argument also covers \(M_T^*=0\); no division by that random variable is used. \(\square\)

The \(L^p\) inequality also yields a tail bound by the proposition. This can be useful when the terminal \(p\)-th moment is known but the first absolute moment is not the most convenient quantity. The dependence on \(K^{-p}\) shows how a higher terminal moment controls the probability of very large excursions. It does not imply that the bound is close to the true probability for moderate \(K\).

Worked Example: The Running Maximum of a Two-Step Martingale

Let \(\xi_1,\xi_2\) be independent random variables, each equal to \(1\) or \(-1\) with probability \(1/2\). Reveal \(\xi_1\) at time \(1\) and \(\xi_2\) at time \(2\); let \(M_t=0\) for \(t<1\), \(M_t=\xi_1\) for \(1\le t<2\), and \(M_2=\xi_1+\xi_2\). This is a martingale: the first reveal has conditional mean zero, and the second increment has conditional mean \(\mathbb{E}[\xi_2\mid\mathcal{F}_1]=0\).

The four equally likely paths are \((1,1),(1,-1),(-1,1),(-1,-1)\). Their terminal values are \(2,0,0,-2\), and their running absolute maxima are \(2,1,1,2\), respectively. Therefore \(\mathbb{E}[(M_2^*)^2]=(4+1+1+4)/4=10/4=5/2\). The terminal second moment is \(\mathbb{E}[|M_2|^2]=(4+0+0+4)/4=8/4=2\). For \(p=2\), Doob’s \(L^p\) inequality gives \(\mathbb{E}[(M_2^*)^2]\le(2/(2-1))^2\cdot2=4\cdot2=8\). The exact value \(5/2\) is below this bound.

For the crossing threshold \(K=3/2\), exactly the paths \((1,1)\) and \((-1,-1)\) satisfy \(M_2^*\ge3/2\), so the probability is \(2/4=1/2\). The weak-type bound gives \(\mathbb{P}(M_2^*\ge3/2)\le\mathbb{E}[|M_2|]/(3/2)\). Here \(\mathbb{E}[|M_2|]=(2+0+0+2)/4=1\), so the bound is \(1/(3/2)=2/3\). Check: the exact crossing probability \(1/2\) is no larger than \(2/3\). The \(L^2\) moment bound followed by Markov’s inequality is weaker here: it gives \(8/(3/2)^2=32/9>1\), so the probability bound reduces to the trivial bound \(1\).

Worked Example: Bounding a Lower-Order Maximum Moment

Let \(A\) have probability \(1/4\), let \(X=\mathbf{1}_A\), and take \(M_t=\mathbb{E}[X\mid\mathcal{F}_t]\), where the filtration reveals \(A\) at time \(1\). Then \(M_t=1/4\) before time \(1\), and \(M_1=X\). Hence \(M_1^*=1\) on \(A\) and \(M_1^*=1/4\) on \(A^c\). Its first moment is \(\mathbb{E}[M_1^*]=(1/4)\cdot1+(3/4)\cdot(1/4)=1/4+3/16=7/16\).

The terminal second moment is \(\mathbb{E}[|M_1|^2]=\mathbb{E}[X^2]=1/4\), since \(X^2=X\). Apply the corollary with \(p=2\) and \(q=1\): \(\mathbb{E}[M_1^*]\le(2/(2-1))(1/4)^{1/2}=2\cdot(1/2)=1\). The direct value \(7/16\) satisfies this estimate. Check: the same process has \(\mathbb{E}[M_1]=1/4\), matching \(\mathbb{E}[X]\); the larger running-maximum mean \(7/16\) reflects the possible early value \(1/4\) on \(A^c\), even though the terminal value there is zero.

Choosing and Interpreting a Bound

For a nonnegative submartingale, the weak-type estimate uses the terminal first moment and is often the natural choice for a single crossing event. For a martingale with a known \(p\)-th terminal moment, the \(L^p\) inequality controls a moment of the whole path, and the proposition converts that moment into a tail estimate. If the resulting numerical bound exceeds \(1\), it remains mathematically correct but gives no information beyond the fact that a probability is at most \(1\).

A common mistake is to apply the terminal-time estimate directly to the running maximum. For example, in the two-step example the calculation \(\mathbb{P}(M_2^*\ge3/2)\le\mathbb{E}[|M_2|]/(3/2)\) is justified by the weak-type inequality applied to the submartingale \(|M_t|\), not by treating \(M_2^*\) as though it were \(M_2\). A different incorrect calculation would replace \(\mathbb{E}[|M_2|]\) by \(\mathbb{E}[M_2]=0\), giving a zero upper bound. The correction is essential: the weak-type inequality here concerns the nonnegative process \(|M_t|\), whose terminal expectation in the example is \(1\), not the signed mean \(\mathbb{E}[M_2]=0\).

These inequalities concern true martingales or submartingales with the stated integrability; no local-martingale version is being asserted here. Checking the hypotheses first, then comparing the bound with an exact distribution when one is available, keeps the estimate both valid and informative.

Check Your Understanding

For each question, specify whether the event uses an inclusive or strict threshold, and state the moment or integrability assumption needed.

  1. For a nonnegative right-continuous submartingale \(X\), what weak-type bound controls \(\mathbb{P}(X_T^*\ge\lambda)\)?
  2. If \(\mathbb{E}[|M_T|^3]<\infty\), write the \(L^3\) bound for \(\mathbb{E}[(M_T^*)^3]\).
  3. In the two-step example, compute \(\mathbb{P}(M_2^*>1)\) and compare it with the weak-type bound at threshold \(1\).
  4. Why does a bound on \(\mathbb{E}[M_T]\) alone not give the weak-type crossing estimate for a signed martingale?
  5. State the lower-order moment bound when \(p=4\) and \(q=2\).