A Maximal Inequality for Supermartingales

TL;DR

This paper establishes a precise upper bound on the probability that the maximum of a certain class of supermartingales exceeds a threshold, using semimartingale calculus and dynamic programming techniques.

Contribution

It provides a new, tight inequality for the distribution of supermartingale maxima, extending the understanding of their probabilistic behavior.

Findings

The probability that the maximum exceeds a threshold is bounded by 1/(1+a).

The result applies to semimartingales with specific supermartingale properties.

The proof employs semimartingale calculus and dynamic programming methods.

Abstract

A tight upper bound is given on the distribution of the maximum of a supermartingale. Specifically, it is shown that if $Y$ is a semimartingale with initial value zero and quadratic variation process $[Y,Y]$ such that $Y + [Y,Y]$ is a supermartingale, then the probability the maximum of $Y$ is greater than or equal to a positive constant $a$ is less than or equal to $1/(1+a).$ The proof makes use of the semimartingale calculus and is inspired by dynamic programming.

Equations

$$[Y,Y{]}_t=\underset{n\to \infty }{\lim }\sum _{i=0}^{k_n-1}(Y_{t_{i+1}^n}-Y_{t_i^n}{)}^2$$

$$P\{Y^{\ast }\ge a\}\le \frac{1}{1+\gamma a}.$$

$$P\{Y^{\ast }\ge a\}\ge \frac{1}{5(1+a\gamma )}$$

$$Y_{t}=\left\{\begin{array}[]{cl}-y(t)&t<T\\ h(y(T))-\mu(t-T)&t\geq T\end{array}\right.$$

$$\dot{y}=\mu +\frac{{\sigma }^2}{y+h(y)},y(0)=0$$

$$P\{T\ge t\}=\exp \left(-{\int }_0^t\kappa (y_s)ds\right)\text{mbox}with\kappa (y)=\frac{{\sigma }^2}{(y+h(y){)}^2}.$$

$$E[(Y_{t+\eta }-Y_t{)}^2|T>t]$$

$$E[Y_{t+\eta }-Y_t|T>t]$$

$$P\{Y^{\ast }\ge h(c)\}$$

$$I(c)$$

$$P\{Y^{\ast }\ge a\}$$

$$P\{Y^{\ast }\ge b+c\}=\exp (-I(c))-\exp (-I(\infty ))$$

$$I(c)$$

$$P\{Y^{*}\geq a\}=\left\{\begin{array}[]{cl}\left(\frac{\mu b}{\mu b+\sigma^{2}}\right)^{\frac{1}{2}}\left\{\left(1+\frac{\sigma^{2}}{\mu(2a-b)}\right)^{\frac{1}{2}}-1\right\}&a\geq b\\ 1-\left(\frac{\mu b}{\mu b+\sigma^{2}}\right)^{\frac{1}{2}}&0<a\leq b\end{array}\right.$$

$$P\{Y^{\ast }\ge a\}$$

$$P\{T<\infty \}\le \frac{1}{1+\gamma a}.$$

$$(\tilde{X}-\gamma[\tilde{X},\tilde{X}])_{t}-(X-\gamma[X,X])_{t\wedge T}=\left\{\begin{array}[]{cl}0&0\leq t<T\\ D+\gamma(2DX_{T-}+D^{2})\geq 0&t\geq T\end{array}\right.$$

$$\begin{array}{rl}p(X_t)& =p(X_s)+{\int }_s^t{p}^{\prime }(X_{u-})d{X}_u\\ & +\sum _{s<u\le t}\left(p(X_u)-p(X_{u-})-p^{\prime }(X_{u-})△X_u\right)+\frac{1}{2}{\int }_s^t{p}^{\prime \prime }(X_u)d[X,X{]}_u^c.\end{array}$$

$$p^{\prime }(X_{u-})=-\gamma p^2(X_{u-})$$

$$p(X_u)-p(X_{u-})-p^{\prime }(X_{u-})△X_u$$

$$p^{\prime \prime }(X_{u-})$$

$$p(X_t)\le p(X_s)-\gamma {\int }_s^{t}p(X_{u-}{)}^{2}d{G}_u$$

$$E\left[\int_{s}^{t}p(X_{u-})^{2}dG_{u}\bigg{|}{\cal F}_{s}\right]\geq 0$$

$$P\{S^{\ast }\ge a\}\le \frac{1}{1+\gamma a}$$

$$E[U_k^2|{\mathcal{F}}_{k-1}]$$

$$E[U_k-\gamma U_k^2|{\mathcal{F}}_{k-1}]\ge \tilde{\mu }-\gamma ({\tilde{\mu }}^2+{\tilde{\sigma }}^2)=0$$

$$P[S^{\ast }\ge a]\ge P[Y^{\ast }\ge \tilde{a}]=\frac{1}{1+\tilde{\gamma }\tilde{a}}=\frac{1}{1+\left[\frac{\gamma }{1+\tilde{\mu }\gamma }\right][\tilde{\mu }+a]}$$

$$E[Y^{\ast }]\le \frac{1}{2\gamma },\text{mbox}(underSII)$$

$$P\{Y^{\ast }\ge a\}\le \frac{1}{2a\gamma }.\text{mbox}(underSII)$$

$$P\{Y^{\ast }\ge a\}\le \exp (-2\gamma a)\text{mbox}(underSII,skip-freepositive)$$

Full text

\SHORTTITLE

A Maximal Inequality for Supermartingales \TITLEA Maximal Inequality for Supermartingales ††thanks: Supported by the National Science Foundation under Grant ECCS 10-28464.

\AUTHORSBruce Hajek111University of Illinois at Urbana-Champaign, USA. \EMAILb-hajek@illinois.edu \KEYWORDSmaximal inequality; semimartingale calculus; drift; martingale \AMSSUBJ60G40;93E20;60G42;60G44 \SUBMITTEDJanuary 2, 2014 \ACCEPTEDJuly 30, 2014 \VOLUME0 \YEAR2014 \PAPERNUM0 \DOIvVOL-PID \ABSTRACTA tight upper bound is given on the distribution of the maximum of a supermartingale. Specifically, it is shown that if $Y$ is a semimartingale with initial value zero and quadratic variation process $[Y,Y]$ such that $Y+[Y,Y]$ is a supermartingale, then the probability the maximum of $Y$ is greater than or equal to a positive constant $a$ is less than or equal to $1/(1+a).$ The proof makes use of the semimartingale calculus and is inspired by dynamic programming.

1 Introduction

The basic framework of the classical martingale calculus will be used, as described, for example, in [6, 7, 9, 10, 11]. All random processes are assumed to be defined on a complete probability space $(\Omega,{\cal F},P)$ with a filtration of $\sigma$-algebras $({\cal F}_{t}:t\geq 0)$ that is assumed to satisfy the usual conditions of right-continuity and inclusion of all sets of probability zero. Semimartingales are processes that are càdlàg (right continuous with finite left limits) and can be represented as the sum of a càdlàg local martingale and a càdlàg, adapted process of locally finite variation. The quadratic variation process of a semimartingale $Y$, denoted by $[Y,Y],$ is the nondecreasing process defined by

[TABLE]

for any choice of $0=t_{0}^{n}<t_{1}^{n}<\cdots<t_{k_{n}}^{n}=t$ such that $\max_{i}|t^{n}_{i+1}-t^{n}_{i}|\rightarrow 0$ as $n\rightarrow\infty.$ The process $[Y,Y]$ can be decomposed as $[Y,Y]_{t}=\sum_{s\leq t}(\triangle Y_{s})^{2}+[Y,Y]^{c}_{t},$ where $\triangle Y_{s}$ is the jump of $Y$ at $s$, and $[Y,Y]^{c}$ is the continuous component of $[Y,Y].$

Recall that a process $Y$ is a supermartingale if $E[Y_{t}-Y_{s}|{\cal F}_{s}]\leq 0$ for $0\leq s\leq t.$ It means that the process has a downward drift. The following condition is stronger than the supermartingale condition, requiring that the downward drift be at least as strong as a constant $\gamma$ times the rate of variation of the process, as measured by the quadratic variation process.

Condition 1.1**.**

The process $Y$ is a semimartingale with $Y_{0}=0$, and $\gamma\geq 0$ such that $Y+\gamma[Y,Y]$ is a supermartingale.

Condition 1.1 is a joint condition on $Y$ and $[Y,Y]$, and is preserved under continuous, adapted time changes. A stronger condition, involving separate constraints on $Y$ and $[Y,Y]$, is the following.

Condition 1.2**.**

$Y$* is a random process with $Y_{0}=0$, and $\gamma>0$, $\mu>0$ and $\sigma^{2}\geq 0$ are such that $\gamma=\frac{\mu}{\sigma^{2}}$ and $(Y_{t}+\mu t:t\geq 0)$ and $([Y,Y]-\sigma^{2}t:t\geq 0)$ are supermartingales.*

Condition 1.2 implies Condition 1.1 because, under Condition 1.2,

$Y_{t}+\gamma[Y,Y]_{t}=(Y_{t}+\mu t)+\gamma([Y,Y]_{t}-\sigma^{2}t).$ Let $Y^{*}=\sup\{Y_{t}:t\geq 0\}.$

Proposition 1.1**.**

*Suppose $Y$ and $\gamma>0$ satisfy Condition 1.1 (or Condition 1.2) and $a\geq 0.$

(a) The following holds:*

[TABLE]

(b) Equality holds in (1.1) if and only if the following is true, with $T=\inf\{t\geq 0:Y_{t}\geq a\}:$ $(Y_{t\wedge T}:t\geq 0)$ has no continuous martingale component, $Y$ is sample-continuous over $[0,T)$ with probability one, $P(Y_{T}=a|T<\infty)=1$, and $(Y+\gamma[Y,Y])_{t\wedge T}$ is a martingale.

Proposition 1.1 raises the question of whether there is a finite upper bound on $E[Y^{*}]$ depending only on $\gamma.$ Also, the necessary conditions in Proposition 1.1(b) cannot be satisfied for two distinct strictly positive values of $a.$ This raises the question as to how close to equality the bound (1.1) can be for all values of $a$, for a single choice of $Y$ not depending on $a$. The following proposition addresses these two questions.

Proposition 1.2**.**

Given $\gamma\geq 0$ there exists $Y$ satisfying Condition 1.2 such that

[TABLE]

for all $a\geq 0.$ In particular, $E[Y^{*}]=+\infty$ for this choice of $Y.$

The remainder of this paper is organized as follows. Section 2 describes a construction, showing there is a process meeting the conditions of Proposition 1.1(b), and providing a proof of Proposition 1.2. Proposition 1.1 is reformulated and proved in Section 3. A discrete time version of Proposition 1.1 is stated and proved in Section 4. Discussion about intuition and possible extensions is given in Section 5. Some well known related inequalities are discussed in Section 6.

2 The big jump construction

A construction is given for a process meeting the bound of Proposition 1.1 with equality, and for a process providing a proof of Proposition 1.2. These processes satisfy Condition 1.2.

Let $\mu>0$, $\sigma^{2}>0,$ and let $h:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ be such that $h$ is nondecreasing. Consider a random process $Y$ of the form:

[TABLE]

for a deterministic, continuous function $y=(y(t):t\geq 0)$ and random variable $T$ described below. Note that if $T<+\infty$ then $Y_{T-}=-y(T)$, $Y_{T}=h(y(T)),$ and $\triangle Y_{T}=y(T)+h(y(T)).$ Let $y$ be a solution to the differential equation

[TABLE]

and let $T$ be an extended nonnegative random variable such that for all $t\geq 0,$

[TABLE]

The function $\kappa(y(t))$ is the failure rate function of $T$: $P(T\leq t+\eta|T\geq t)=\kappa(y(t))\eta+o(\eta).$ The function $\kappa$ was chosen so that

[TABLE]

and the differential equation for $y$ was chosen so that

[TABLE]

Therefore, $Y$ satisfies Condition 1.2.

If the function $h$ is strictly increasing, then for any $c\geq 0$, a change of variable of integration from $t$ to $y$ yields:

[TABLE]

where

[TABLE]

Example 1: Meeting Proposition 1.1 with equality.

Take $h(y)\equiv a$ for some $a>0.$ We don’t use (2.1) because $h$ is not strictly increasing, but similar reasoning yields:

[TABLE]

Thus, the process $Y$ satisfies the bound of Proposition 1.1 with equality for $\gamma=\frac{\mu}{\sigma^{2}}.$

Example 2: Proof of Proposition 1.2.

Take $h(y)=b+y$ for some $b>0.$ Equation (2.1) yields that for $b\geq 0$,

[TABLE]

where

[TABLE]

Using this and (2.1), and setting $c=a-b,$ yields

[TABLE]

Let $b=\frac{16\sigma^{2}}{9\mu}.$ Then $P\{Y^{*}\geq a\}=\frac{1}{5}\geq\frac{1}{5(1+\frac{\mu a}{\sigma^{2}})}$ for $0\leq a\leq b.$ By checking derivatives, it is easy to verify that $(1+\frac{\alpha}{2})^{\frac{1}{2}}-1\geq\frac{\alpha}{4(1+\alpha)}$ for any $\alpha>0$. Therefore, for this choice of $b$, and $a\geq b$,

[TABLE]

This bound for the process $Y$ proves Proposition 1.2.

3 Reformulation and proof of Proposition 1.1

Proposition 1.1 concerns the probability that a process with downward drift, starting from [math], reaches level $a.$ It is more convenient for the proof, to consider the equivalent problem, of the probability a process with upward drift, starting from $a$, reaches zero. The correspondence between the two formulations is obtained by setting $X=a-Y,$ so the proof of Proposition 3.1 below also establishes Proposition 1.1.

Condition 3.1**.**

$X$* is a semimartingale and $\gamma\geq 0,$ such that $X-\gamma[X,X]$ is a submartingale.*

Proposition 3.1**.**

*(Equivalent to Proposition 1.1.) Suppose $X$ and $\gamma$ satisfy Condition 3.1 and $X_{0}=a$ for some $a\geq 0.$ Let $T=\inf\{t:X_{t}\leq 0\}$ (so $T=\infty$ if $X_{t}>0$ for all $T$).

(a) The following holds:*

[TABLE]

(b) Equality holds in (3.1) if and only if $(X_{t\wedge T}:t\geq 0)$ has no continuous martingale component, $X$ is sample-continuous over $[0,T)$ with probability one, $P\{X_{T}=0|T<\infty\}=1$, and $\left((X-\gamma[X,X])_{t\wedge T}:t\geq 0\right)$ is a martingale.

Proof 3.2**.**

(Proof of (a))* Suppose $X$ and $\gamma$ satisfy Condition 3.1 and $X_{0}=a$ for some $a\geq 0.$ Let $\tilde{X}_{t}=\max\{X_{t\wedge T},0\}.$ Let $D=-X_{T}$ on the event $T<\infty$, and let $D=0$ otherwise. Note that $X_{t}=\tilde{X}_{t}$ and $[\tilde{X},\tilde{X}]_{t}=[X,X]_{t}$ for $t\in[0,T),$ while, if $\{T<\infty\}$, $0=\tilde{X}_{T}\geq X_{T}=-D$ and $\triangle[\tilde{X},\tilde{X}]_{T}=(X_{T-})^{2}\leq(D+X_{T-})^{2}=\triangle[X,X]_{T}.$ Thus,*

[TABLE]

Since $X-\gamma[X,X]$ is a submartingale, so is $(X-\gamma[X,X])_{t\wedge T},$ and therefore so is $\tilde{X}-\gamma[\tilde{X},\tilde{X}].$ Therefore, $\tilde{X}$ and $\gamma$ also satisfy the conditions of the proposition, and $T=\inf\{t\geq 0:\tilde{X}\leq 0\}.$ Therefore, to prove (a), it suffices to prove (a) with $X$ replaced by $\tilde{X}.$ Equivalently, in proving (a), we can, and do, make the assumption, without loss of generality, that $X_{t}\equiv 0$ for $t\geq T.$

Let $p(x)=\frac{1}{1+\gamma x}$ for $x\geq 0.$ Let $0\leq s<t.$ By the Doléan-Dade Meyer change of variables formula for semimartingales,

[TABLE]

Observe that

[TABLE]

[TABLE]

and

[TABLE]

Combining (3.3) - (3.6) and the fact $[X,X]_{u}=[X,X]^{c}_{u}+\sum_{v\leq u}(\triangle X_{v})^{2}$ yields

[TABLE]

where $G=X-\gamma[X,X].$ By assumption, $G$ is a submartingale, so

[TABLE]

Therefore, $p(X)$ is a supermartingale, so that $E[p(X_{t})]\leq p(X_{0})=p(a).$ For $t\geq 0,$ $\{T\leq t\}=\{p(X_{t})=1\}.$ Therefore, $P\{T\leq t\}\leq E[p(X_{t})]\leq p(a)$ for all $t$. Thus, $P\{T<\infty\}=\lim_{t\rightarrow\infty}P\{T\leq t\}\leq p(a)$, completing the proof of part (a).

(Proof of (b))* To begin, assume that $X_{t}\equiv 0$ for $t\geq T,$ with probability one. Note that there are three places in the proof of part (a) where the bounds might not hold with equality. These are associated with inequalities (3.5), (3.6), and (3.8). First, if $\triangle X_{s}\neq 0$, then (3.5) holds with equality if and only if $X_{s}\neq 0$. Therefore, in order for (3.1) to be tight, with probability one, any jumps of $X$ happening in $[0,T]$ must happen at time $T.$ Second, the inequality (3.6) for $p^{\prime\prime}(X_{u-})$ is strict for any $u<T$. This inequality is applied to the last term in (3.3). Therefore, the resulting upper bound on the last term in (3.3) holds with equality if and only if $[X,X]^{c}\equiv 0$ over the interval $(s,t].$ Finally, (3.8) holds with equality if and only if $G$ is a martingale (not just a supermartingale) over $(s,t].$ These observations prove (b) under the additional assumption that $X_{t}\equiv 0$ for $t\geq T.$*

To prove (b) in general, suppose $X$ satisfies the conditions of the proposition, and let $\tilde{X}$ be defined in part (a). The conditions in part (b) involve $X$ only up to time $T.$ The process $X$ and $\tilde{X}$ are equal for $t<T$, so the only difference between them over $[0,T]$ could possibly be at time $T$. If (3.1) holds with equality, then by the special case of part (b) already proved, $\tilde{X}$ must satisfy the conditions in (b). In particular, $(X-\gamma[X,X])_{t\wedge T}$ is a martingale. In view of (3.2), $(X-\gamma[X,X])_{t\wedge T}$ must also be a martingale and $P\{D=0\}=1$, or equivalently, $P\{X_{T}=0|T<\infty\}=1.$ Therefore, $X$ also satisfies the conditions in (b), and the proof of the proposition is complete.

The above proof is inspired by dynamic programming. The problem of finding $X$ to maximize $P\{T<\infty\}$ subject to the given constraints can be viewed as a stochastic control problem. Intuitively speaking, given $X_{s}=x$ for some time $s$ and $x>0$, the conditional distribution of the increment $X_{s+\eta}-X_{s}$ must be selected subject to first and second moment constraints. The function $p$ plays the role of the value function in dynamic programming.

4 Discrete time processes

Suppose $(\Omega,{\cal F},P)$ is a complete probability space with a filtration of $\sigma$-algebras $({\cal F}_{k}:k\in\mathbb{Z}_{+})$.

Condition 4.1**.**

$S=(S_{k}:k\in\mathbb{Z}_{+})$* is an adapted random process with $S_{0}=0$ and, with $U_{j}=S_{j}-S_{j-1}$ for $j\geq 1$, $(S_{k}+\gamma\sum_{1\leq j\leq k}U_{j}^{2}:k\geq 0)$ is a supermartingale.*

Let $S^{*}=\sup\{S_{k}:k\in\mathbb{Z}_{+}\}.$

Proposition 4.1**.**

*Suppose $S$ and $\gamma\geq 0$ satisfy Condition 4.1 and $a\geq 0.$

(a) The following holds:*

[TABLE]

(b) For any $\gamma\geq 0,a\geq 0$ and $\epsilon>0$, there is a process $S$ satisfying Condition 4.1 such that $P\{S^{*}\geq a\}\geq\frac{1}{1+\gamma a}-\epsilon.$

Proof 4.2**.**

(Proof of (a))* The filtration $({\cal F}_{k}:k\in\mathbb{Z}_{+})$ can be extended to a filtration $({\cal F}_{t}:t\in\mathbb{R}_{+})$ by letting ${\cal F}_{t}={\cal F}_{\lfloor t\rfloor}$ for $t\in\mathbb{R}_{+}$, and the process $S$ can be extended to a piecewise constant process $(Y_{t}:t\in\mathbb{R}_{+})$ by letting $Y_{t}=S_{\lfloor t\rfloor}$ for $t\in\mathbb{R}_{+}.$ Then $S^{*}=Y^{*}$ and $Y$ satisfies Condition 1.1. Thus, by Proposition 1.1, $P\{S^{*}\leq a\}=P\{Y^{*}\leq a\}\leq\frac{1}{1+\gamma a}.$ This establishes (a).*

(Proof of (b))* If $\gamma=0$, the process $S$ can be a mean zero random walk with finite variance jumps, and then $S^{*}=\infty$ with probability one, so that equality holds in (4.1) for such choices of $S.$ So assume for the remainder of the proof that $\gamma>0.$ Given $\tilde{\mu}>0,$ let $\tilde{\sigma}^{2}=\frac{\tilde{\mu}}{\gamma}-\tilde{\mu}^{2}.$ Assume that $\tilde{\mu}$ is so small that $\tilde{\sigma}^{2}>0.$ Let $\tilde{a}=a+\tilde{\mu}.$ Let $Y$ be a continuous time martingale as constructed in Example 1, for the parameters $\tilde{\mu},\tilde{\sigma}^{2},$ and $\tilde{a}.$ Thus, $Y$ follows a deterministic trajectory up to some random time $T$, which could be infinite. If $T<\infty$, then $Y$ jumps up to $\tilde{a}$ at time $T$. Furthermore, modify $Y$ on the event $\{T<\infty\}$ by letting $Y_{t}=\tilde{a}-\tilde{\mu}(t-T)$ for $t\geq T.$ Then $Y_{t}-\tilde{\mu}t$ and $[Y,Y]-\tilde{\sigma}^{2}(t\wedge T)$ are martingales, and $P\{Y^{*}\geq\tilde{a}\}=\frac{1}{1+\tilde{\gamma}\tilde{a}},$ where $\tilde{\gamma}=\frac{\tilde{\mu}}{\tilde{\sigma}^{2}}.$ Let $M$ denote the martingale defined by $M_{t}=Y_{t}-\tilde{\mu}t.$ Since $M$ differs from $Y$ by a continuous function with finite variation, $[M,M]\equiv[Y,Y]$ with probability one. Since $M$ is a martingale with finite second moments, $M^{2}-[M.M]$ is also a martingale.*

Let $S=(S_{k}:k\in\mathbb{Z}_{+})$ be obtained by sampling $Y$ at nonnegative integer times, namely, $S_{k}=Y_{k}$ for $k\in\mathbb{Z}_{+}.$ We first check that $S$ satisfies Condition 4.1. For $k\geq 1,$ set $U_{k}=S_{k}-S_{k-1}=Y_{k}-Y_{k-1}.$ Then $E[U_{k}|{\cal F}_{k-1}]=E[Y_{k}-Y_{k-1}|{\cal F}_{k-1}]=\tilde{\mu},$ and

[TABLE]

Therefore, by the choice of $\tilde{\sigma}^{2},$

[TABLE]

Thus, $S$ satisfies Condition 4.1 as required. On the event $\{T<\infty\}$, $Y_{T}=\tilde{a}=a+\mu$ and $Y$ decreases with constant slope $-\tilde{\mu}$ after time $T$, implying that $S_{\lceil T\rceil}=Y_{\lceil T\rceil}\geq a$, so that $S^{*}\geq a$. Therefore

[TABLE]

For $\tilde{\mu}$ sufficiently small, (4.2) implies the inequality of part (b), and the proof is complete.

5 Discussion

There is a large literature on bounds on processes implied by drift conditions and concentration inequalities. Typically the constraints placed on the sizes of the increments of the processes are stronger than those imposed here. For example, in the discrete-time case, Condition 4.1 constrains the second moment of the increments in terms of the drift, but stronger constraints such as bounded increments, or uniformly exponentially dominated increments [4], yield stronger bounds. The motivation of this paper is to explore the implication of a relatively mild condition on the variation of a process with negative drift. A second motivation of this paper is that the bound represents the solution of an interesting optimal stochastic control problem that may arise in some applications such as finance.

The proof of Proposition 1.1 might leave the reader wanting some additional insight into why the big jump processes are the ones meeting the bound with equality. As noted in Section 3, it is useful to view the bound $p(a)=\frac{1}{1+\gamma a}$ as a value function in the sense of dynamic programming. Given its explicit form, it is trivial to verify that it is decreasing, convex, and the third derivative is negative. However, one might expect these properties even without knowing the function explicitly. The fact it is decreasing means that the chances of $X$ ever reaching zero decrease with the initial state $a$. Due to Jensen’s inequality, the fact the function is convex means that the chances of ever reaching zero from an initial state $a$ would increase if, before imposing the constraints on $X$, the process could follow a martingale trajectory. Finally, the fact that the third derivative is negative means that the second derivative is decreasing. Therefore, it is better to use the variations of $X$ at smaller values. In this connection, it is useful to remember that martingales, even discontinuous ones, can be obtained from time changes of brownian motion. That explains why the process, for initial state $a$, has only two possible values at future times, and why there are no upward jumps in the process.

We close by speculating about two possible extensions. Proposition 1.1 shows that, for continuous-time processes, even though Condition 1.2 is more restrictive than Condition 1.1, the same maximum value of $P\{Y^{*}\geq a\}$ can be achieved. However, the situation is different in discrete-time. The discrete-time version of Condition 1.2 would constrain the process $Y$ to take downward jumps that are not vanishingly small, and the maximum possible value of $P\{Y^{*}\geq a\}$ would be strictly smaller than the bound in Proposition 4.1. The intuition in the first paragraph above would indicate that in such a case, the optimal process would still be one following a deterministic path, with the possible exception of a single big jump, reaching the target value in one step.

Proposition 1.2 shows that under Condition 1.1 it is possible that $E[Y^{*}]=+\infty.$ However, it might be interesting to consider maximizing the expected value of sublinear functions of $Y$, such as $E[(Y^{*})^{\alpha}]$ for a fixed $\alpha$ with $0<\alpha<1.$ The process of Proposition 1.2 achieves at least one-fifth of the maximum possible value. Perhaps the big jump construction of Section 2 still yields the extremal processes, for a suitable choice of the function $h.$

6 Related inequalities

6.1 Kingman’s bound

If $Y$ has stationary, independent increments (SII) and $Y$ and $\gamma$ satisfy Condition 1.2, the moment upper bound of Kingman [8] for the waiting time in a $GI/G/1$ queue implies222Kingman’s moment bound is for discrete time SII processes, but it is easily extended to continuous time by sampling the continuous time processes at times of the form $2^{-n},$ and letting $n\rightarrow\infty.$

[TABLE]

which together with Markov’s inequality implies that

[TABLE]

If $Y$ is skip-free positive, a simple argument shows that $Y^{*}$ has an exponential distribution, so that

[TABLE]

If $Y$ is SII and both skip-free positive and skip-free negative, it is, of course, continuous. If in addition both $(Y_{t}+\mu t:t\geq 0)$ and $([Y,Y]-\sigma^{2}t:t\geq 0)$ are martingales, $Y$ can be written as $Y=\sigma B_{t}-\mu t,$ where $B$ is a standard Brownian motion, in which case equality holds in both (6.1) and (6.2).

6.2 Doob’s $L^{p}$ inequalities

Doob’s well known $L^{p}$ inequality for a nonnegative submartingale $X=(X_{t}:0\leq t\leq T)$, holding for $p>1$, is:

[TABLE]

Dubins and Gilat [3] gave a construction of a martingale showing that the constant in (6.3) is the best possible. That martingale can be expressed as a big jump process as follows. Let $h$ be a positive, nondecreasing function on the interval $[0,1],$ let $U$ be uniformly distributed on the interval $[0,1],$ let $0<c<1,$ and let $X=(X_{t}:0\leq t\leq 1)$ denote the process:

[TABLE]

In words, $X$ follows $h$ until time $U$, at which time it jumps downward to a fraction $c$ of itself, and sticks. We now determine a choice of $h$ for which $X$ is a martingale. The failure rate function of $U$ (or jump intensity, given the jump hasn’t yet happened) at time $t$ is $\frac{1}{1-t}.$ Therefore, the drift of $X$ at time $t$ is $h^{\prime}(t)-\frac{(1-c)h(t)}{1-t}.$ Setting the drift to zero determines $h$, up to a constant factor, to be $h(t)=\frac{1}{(1-t)^{1-c}}.$ Let $T=1.$ Note that $X^{*}=h(U)$ and $X_{T}=cX^{*}.$ Given $p>1$, $X_{T}$ is in $L^{p}$ if $(1-c)p>1.$ We thus have:

[TABLE]

which shows that the constant in (6.3) is the best possible.

Cox [2] described the dynamic programming approach for Doob-like inequalities, and showed that Doob’s $L^{2}$ martingale inequality can’t be satisfied with equality.

The analysis of Dubins and Gilat [3] is related to work of Blackwell and Dubins [1], which makes a connection to the Hardy-Littlewood maximal function [5], $h$, of a nondecreasing integrable function $g$ on $[0,1]$, defined as follows:

[TABLE]

In fact, $h$ is the unique function such that a process that follows $h$ up until time $U$, and then jumps and sticks to value $g(U),$ is a martingale. Blackwell and Dubins [1] used the inequality

[TABLE]

to show that for a specified distribution of the end variable $X_{T},$ the Hardy-Littlewood maximal function [5] provides the stochastically largest possible distribution for the supremum of a martingale. The idea is the following. Given the distribution of $X_{T}$, one can ask how large $P\{E\}$ can be for an event $E$ such that

[TABLE]

Without loss of generality assume that the underlying probability space is $[0,1]$ and that $X_{T}=g$, where $g$ is a nondecreasing function on $[0,1].$ Clearly $E$ should be an event of the form $[\gamma,1],$ and

[TABLE]

So if $h(t)=\frac{\int_{t}^{1}X(t)dt}{1-t}$, $P(E)\leq P\{h\geq\lambda\}$. Thus, $P\{X^{*}\geq\lambda\}\leq P\{h\geq\gamma\}.$ That is, $X^{*}$ is stochastically dominated by $h.$

In contrast, the inequalities given in Proposition 1.1 can be achieved with equality. Also, there is no maximal probability distribution for $Y^{*}$ subject to Condition 1.1, because equality cannot hold in (1.1) for two distinct positive values of $a.$