A prophet inequality for \(L^p\)bounded dependent random variables
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Abstract
Keywords
Prophet inequality Optimal stopping Bellman function Best constantsMathematics Subject Classification (2010)
Primary 60G40 60E15 Secondary 62L151 Introduction
The purpose of the paper is to establish a sharp estimate between the expected supremum of a sequence \(X=(X_n)_{n\ge 1}\) of \(L^p\)bounded random variables and the optimal expected return (i.e., optimal stopping value) of \(X\). Such comparisons are called “prophet inequalities” in the literature and play a distinguished role in the theory of optimal stopping, as evidenced in the papers of Allaart [1], Allaart and Monticino [2], Assaf et al. [4, 5], Boshuizen [3, 6], Hill [8, 9], Hill and Kertz [10, 11, 12], Kennedy [13, 14], Kertz [15, 16], Krengel, Sucheston and Garling [17, 18, 19], Tanaka [24, 25] and many others.
We start with the necessary background and notation. Assume that \(X=(X_n)_{n\ge 1}\) is a sequence of (possibly dependent) random variables defined on the probability space \((\Omega ,\mathcal {F},\mathbb {P})\). With no loss of generality, we may assume that this probability space is the interval \([0,1]\) equipped with its Borel subsets and Lebesgue measure. Let \((\mathcal {F}_n)_{n\ge 1}=(\sigma (X_1,X_2,\ldots ,X_n))_{n\ge 1}\) be the natural filtration of \(X\). The problem can be generally stated as follows: under some boundedness condition on \(X\), find universal inequalities which compare \(M=\mathbb {E}\sup _nX_n\), the expected supremum of the sequence, with \(V=\sup _\tau \mathbb {E}X_\tau \), the optimal stopping value of the sequence; here \(\tau \) runs over the class \(\mathcal {T}\) of all finite stopping times adapted to \((\mathcal {F}_n)_{n\ge 1}\). The term “prophet inequality” arises from the optimalstopping interpretation of \(M\), which is the optimal expected return of a player endowed with complete foresight; this player observes the sequence \(X\) and may stop whenever he wants, incurring a reward equal to the variable at the time of stopping. With complete foresight, such a player obviously stops always when the largest value is observed, and on the average, his reward is equal to \(M\). On the other hand, the quantity \(V\) corresponds to the optimal return of the nonprophet player.
The main result of the paper is the following.
Theorem 1.1
Note that this statement generalizes the inequality (1.1) of Hill and Kertz: it suffices to take \(t=1\) and let \(p\) go to \(\infty \) to recover the bound. On the other hand, the expression on the right of (1.3) explodes as \(p\downarrow 1\), which indicates that there is no prophet inequality in the limit case \(p=1\).
A few words about the proof. Our approach is based on the following twostep procedure: first we show that it suffices to establish (1.3) under the additional assumption that \(X\) is a nonnegative supermartingale; second, we prove that in the supermartingale setting, the validity of (1.3) is equivalent to the existence of a certain special function which enjoys appropriate majorization and convexity properties. In the literature this equivalence is often referred to as Burkholder’s method or Bellman function method, and it has turned out to be extremely efficient in numerous problems in probability and analysis: consult e.g. [7, 20, 21, 26] and references therein.
We have organized the paper as follows. In the next section we reduce the problem to the supermartingale setting. Section 3 contains the description of Burkholder’s method (or rather its variant which is needed in the study of (1.3) for supermartingales). In Sect. 4 we apply the method and provide the proof of Theorem 1.1. In the final part of the paper we show that there are no interesting prophet inequalities in the case when the variables \(X_1\), \(X_2\), \(\ldots \) are only assumed to be bounded in \(L^p\).
2 A reduction
Lemma 2.1
Note that the additional assumption \(X_1\equiv 0\) is not restrictive at all: we can always replace the initial sequence \(X_1\), \(X_2\), \(\ldots \) with \(0\), \(X_1\), \(X_2\), \(\ldots \), and the prophet inequality remains the same. In the proof of the above lemma we will need the notion of essential supremum, a wellknown object in the optimal stopping theory. Let us briefly recall its definition, for details and properties we refer the reader to the monographs of Peskir and Shiryaev [22] and Shiryaev [23].
Definition 2.1
 (i)
\(\mathbb {P}(Z_\alpha \le \overline{Z})=1\) for each \(\alpha \in I\),
 (ii)
if \(\tilde{Z}\) is another random variable satisfying (i) in the place of \(\overline{Z}\), then \(\mathbb {P}(\overline{Z}\le \tilde{Z})=1\).
Therefore, it suffices to establish the inequality (1.3) under the additional assumption that the process \(X\) is a finite supermartingale and the variable \(X_1\) is constant almost surely. By some standard approximation arguments, we may further restrict ourselves to the class of simple supermartingales; recall that the sequence \(X=(X_n)_{n\ge 1}\) is called simple, if for each \(n\) the random variable \(X_n\) takes only a finite number of values. We are ready to apply Burkholder’s method, which is introduced in the next section.
3 Burkholder’s method
Theorem 3.1
The function \(\mathbb {B}\) is the least element of the class \(\mathcal {C}\).
Proof
It is convenient to split the reasoning into two parts.
1. If \(\mathbb {P}(\tau =1)>0\), then the set \(\{\tau \le 1\}\) is nonempty; combining this with the facts that \(X_1\) is constant and \(\tau \) is a stopping time of \(X\), we see that \(\{\tau \le 1\}=\Omega \), or \(\tau \equiv 1\). Then \(\mathbb {E}X_\tau ^p=x^p\le t\) by the definition of \(D\).
4 Proof of Theorem 1.1
4.1 Proof of \(\mathbb {B}\le B\)
4.2 Proof of \(\mathbb {B}\ge B\)
For the sake of clarity, we have split the reasoning into a few parts.
4.3 On the construction of extremal examples
The arguments presented in Steps 1–5 can be easily translated into the construction of extremal supermartingales \(X\) (“extremizers”) corresponding to \(\mathbb {B}(x,y,t)\), i.e., those for which the supremum defining \(\mathbb {B}(x,y,t)\) is almost attained. The purpose of this section is to explain how to extract this construction from the above calculations. The reasoning will be a little informal, as our aim is to present the idea of the connection.
 (i)
Any point of the form \((\lambda ,\lambda ,\lambda ^p)\) (with some \(\lambda >0\)) leads to \((0,\lambda ,0)\) or to \((\lambda (1+\delta ),\lambda (1+\delta ),\lambda ^p(1+\delta )^p)\) with probabilities \(1(1+\delta )^{p}\) and \((1+\delta )^{p}\), respectively.
 (ii)
The states of the form \((0,\lambda ,0)\) are absorbing.
 (i’)
Any point of the form \((\lambda ,\lambda ,\lambda ^pt)\) (with some \(\lambda >0\) and \(t>1\)) leads to \((0,\lambda ,0)\) or to \((\lambda (1+\delta ),\lambda (1+\delta ),\lambda ^pt(1+\delta ))\) with probabilities \(\delta /(1+\delta )\) and \(1/(1+\delta )\), respectively.
Now, suppose that \(\delta \) is chosen appropriately as in Step 2: \(\delta =t^{1/(n(p1))}1\) for some large positive integer \(n\). Then, after \(n\) steps, the process \((X,Y,T)\) gets to the point \(((1+\delta )^n,(1+\delta )^n,(1+\delta )^{np})\) with positive probability. Now the procedure described in (i’) does not apply since the point is not of appropriate form. In Step 2 we encountered a similar phenomenon: the number \(\mathbb {B}(1,1,1)\) came into play and the arguments of Step 2 did not apply. To overcome this difficulty, we exploited Step 1. Here we do the same and apply the procedure (i) to the point \(((1+\delta )^n,(1+\delta )^n,(1+\delta )^{np})\). In other words, the Markov process \((X,Y,T)\) is given by the starting position \((1,1,t)\) and the conditions (i), (ii) and (i’). It remains to stop the process \(X\) after a large number of steps to obtain the extremizer corresponding to \(\mathbb {B}(1,1,t)\).
The remaining extremal processes, corresponding to the values of \(\mathbb {B}\) at remaining points, are found similarly. We leave the details to the reader.
5 Lack of prophet inequalities for \(L^p\) bounded variables

If \(\omega \in [0,1/L)\), then \(X_n(\omega )=Y_n(L\omega )\) for \(n=1,\,2,\,\ldots ,\,N\) and \(X_n(\omega )=0\) for other \(n\).

If \(\omega \in [1/L,2/L)\), then \(X_n(\omega )=Y_{nN}(L\omega 1)\) for \(n=N+1,\,N+2,\,\ldots ,\,2N\) and \(X_n(\omega )=0\) for other \(n\).

If \(\omega \in [2/L,3/L)\), then \(X_n(\omega )=Y_{n2N}(L\omega 2)\) for \(n=2N+1,\,2N+2,\,\ldots ,\,3N\) and \(X_n(\omega )=0\) for other \(n\).

...

If \(\omega \in [11/L,1)\), then \(X_n(\omega )=Y_{n(L1)N}(L\omega (L1))\) for \(n=(L1)N+1,\,(L1)N+2,\,\ldots ,\,LN\) and \(X_n(\omega )=0\) for other \(n\).
Notes
Acknowledgments
The author would like to thank an anonymous Referee for the careful reading of the first version of the paper and several helpful suggestions.
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