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Theorem 5 For every instance of the box problem with exogenous ordering, there is a [MATH] -thresholding policy whose expected net value is at least half that of Weitzman’s optimal search procedure (which endogenously selects the ordering of the boxes). |
Proof. Lemma reduces the analysis of [MATH] -thresholding policies to a question about prophet inequalities. In particular, the expected net value of running a |
[MATH] -thresholding policy is equal to the expected covered call value of the random element selected from the sequence [MATH] by a particular threshold stopping rule. Since Samuel-Cahn, ’s 1984 ) prophet inequality Theorem |
above) implies that threshold stopping rules can always attain at least half the expectation of the maximum random variable in the sequence, it follows that there is a [MATH] -thresholding policy whose expected net value is at least half the expectation of the maximum covered call value. |
ensures that the latter is an upper bound on the expected net value of Weitzman’s optimal search procedure. 4.3 An approximately optimal mechanism |
Recall that for a half-infinite interval [MATH] or [MATH] the mechanism [MATH] is defined to be the single proposal mechanism whose eligible set consists of solutions [MATH] that are feasible and satisfy [MATH] |
Theorem 6 There exists a choice of [MATH] such that the expected net value of mechanism [MATH] — i.e., the principal’s value for adopting the agent’s proposal, if adopted, minus combined cost of all the alternatives explored — is at least half of the expected net value the principal could achieve by performing the opti... |
Proof. Convert the delegated search problem into a box problem with exogenous order, where the order is defined by sorting the solutions [MATH] in non-increasing order of the agent’s priority value [MATH] and the value [MATH] inside box [MATH] is defined to be |
[MATH] if [MATH] turns out to be feasible, 0 otherwise. According to Theorem there exists a choice of [MATH] such that the [MATH] -thresholding policy with target set [MATH] |
attains at least half the expected net value of the optimal search procedure. This thresholding policy goes through boxes in the given order, i.e. descending [MATH] , and opens only those with [MATH] , selecting the first one such that [MATH] . Note that among the boxes which the policy opens, the first one with [MATH] |
is also the first one corresponding to a feasible [MATH] This is because an infeasible [MATH] has [MATH] hence [MATH] , whereas a feasible [MATH] has [MATH] hence [MATH] |
Recall from Section 4.1 that the agent’s best response to mechanism [MATH] is to go through the elements of [MATH] in decreasing order of [MATH] , stopping and proposing the first one that is discovered to be feasible. This is exactly the behavior of the [MATH] -thresholding policy with target set [MATH] as derived in ... |
[MATH] coupled with the agent’s best response behavior emulates the [MATH] -thresholding policy which attains at least half the expected net value of the optimal search procedure. |
4.4 Limiting the number of samples In some cases the number of distinct potential solutions, [MATH] , may be prohibitively large, and the agent may only have the power to explore the feasibility of a limited number of them, [MATH] . In this case, if the principal were to conduct the search autonomously without delegati... |
The key observation is the following lemma, which provides a useful upper bound on the value of running the optimal search procedure. |
Lemma 4 In the box problem with [MATH] boxes, if the searcher is limited to open at most [MATH] boxes before claiming a prize, then the expected net value of any search procedure is bounded above by [MATH] where |
[MATH] is the random set of boxes that the procedure opens. Proof. Sum up the inequality ( ) over all boxes and note that [MATH] for [MATH] to derive |
[EQUATION] The lemma follows by noting that [MATH] because [MATH] Lemma 5 There exists a (non-random) set [MATH] of cardinality [MATH] such that [MATH] where [MATH] is the random set of solutions explored by the optimal search procedure subject to a contraint of exploring at most [MATH] solutions. |
Proof. The problem of adaptively exploring a random set [MATH] of at most [MATH] solutions to maximize [MATH] is a special case of the stochastic monotone submodular function maximization problem studied by Asadpour and Nazerzadeh, ( 2016 in which the role of the monotone submodular function |
[MATH] is played by the function [MATH] and role of the matroid constraint is played by the cardinality constraint that at most [MATH] elements may be probed. Theorem 1 of Asadpour and Nazerzadeh, ( 2016 which asserts that the adaptivity gap of stochastic monotone submodular maximization is [MATH] , specializes in the ... |
Theorem 7 Consider delegated search in the binary model with a constraint that no more than [MATH] solutions can be examined for feasibility. There exists a mechanism that attains at least |
[MATH] fraction of the expected net value of the optimal search procedure subject to the same limitation of examining at most [MATH] solutions. |
Proof. According to Lemmas and , there is an [MATH] -element set [MATH] such that the optimal search procedure that is limited to explore only solutions in [MATH] is able to attain at least [MATH] fraction of the expected net value of the optimal search procedure that is limited to examine at most [MATH] solutions but ... |
during its search. When the set of solutions is restricted to [MATH] , the constraint that at most [MATH] solutions can be examined becomes irrelevant since [MATH] only has [MATH] elements. Thus, |
Theorem guarantees the existence of a delegated search mechanism that is at least half as good as the optimal search procedure limited to [MATH] , and is consequently at least |
[MATH] times as good as the optimal search procedure limited to examine at most [MATH] solutions. Moreover, by applying the algorithm in |
Asadpour and Nazerzadeh, ( 2016 used to prove Lemma , we can implement this policy in polynomial time with a loss of a further additive [MATH] |
in the approximation ratio, thus obtaining a bound of [MATH] efficiently. Acknowledgements. This work was supported in part by NSF grants CCF-1512964, CCF-1740822, and SES-1741441, a grant from the MacArthur Foundation, and a Simons Investigator Award. The authors would like to thank Brendan Lucier, Jens Ludwig, Sendhi... |
Appendix A Proofs of Prophet Inequalities In this appendix, we provide proofs of the three main prophet inequalities used in this work, |
Theorems and A.1 Threshold stopping rules for independent distributions In this section we provide a proof of the following theorem from Section |
3.1 . The theorem (stated in a different form) was originally proven by Samuel-Cahn, ( 1984 ; we provide a proof here for the sake of making the paper self-contained. |
Theorem 8 There is a prophet inequality with factor [MATH] for [MATH] Proof. Consider any distribution [MATH] in [MATH] and let [MATH] denote independent random variables representing the [MATH] -coordinates of random samples from [MATH] , respectively. Note that the subscripts on the variables [MATH] |
represent the distributions from which they were sampled, not necessarily the order in which they arrive, since their corresponding time coordinates [MATH] may not be in ascending order. However, this issue will be immaterial in the proof because our argument is insensitive to the arrival order of [MATH] |
Let [MATH] and choose a threshold [MATH] defining two semi-infinite intervals [EQUATION] such that [MATH] and [MATH] Let [MATH] be the solution to the equation [MATH] and let [MATH] denote the random variable defined by the following sampling process: first choose the interval |
[MATH] with probability [MATH] and [MATH] with probability [MATH] Then apply the threshold stopping rule with eligible set [MATH] to select an element [MATH] and let [MATH] denote the [MATH] -coordinate of that element. We will prove that [MATH] Since [MATH] is a convex combination of the expected value of applying the... |
To compare [MATH] with [MATH] we reason as follows. First, we have the following easy upper bound on [MATH] [EQUATION] To put a lower bound on [MATH] , let [MATH] denote an indicator random variable for the event that [MATH] Note that this event, when it happens, implies that |
[MATH] [EQUATION] where the last line follows because [MATH] again by our construction of the random set [MATH] . Finally, the theorem follows by combining ( ) with ( ). |
A.2 Threshold stopping rules for i.i.d. atomless distributions This section provides a proof of the following lemma, which is equivalent to Theorem |
from Section 3.1 Lemma 6 If [MATH] is a sequence of i.i.d. random variables, each sampled from an atomless distributions, and [MATH] |
is a threshold stopping rule such that [MATH] then [EQUATION] Proof. Let [MATH] . We have [EQUATION] To compare the integrands at any specified [MATH] , we consider the cases [MATH] and [MATH] separately. (The case [MATH] is omitted because it contributes zero to both integrals.) When [MATH] the inequality [MATH] |
is satisfied whenever [MATH] . Since we are assuming [MATH] are identically distributed, and the threshold stopping rule has the same behavior at every point in time, we have |
[EQUATION] Hence the probability that the stopping rule does not stop at any time [MATH] is [EQUATION] and therefore for [MATH] [EQUATION] |
Meanwhile, for [MATH] , the probability that the stopping rule stops at time [MATH] and selects an element of value greater than [MATH] is |
[MATH] . Summing over [MATH] we have [EQUATION] Using the fact that [MATH] , we find that [EQUATION] Since we have proven that [MATH] |
for all [MATH] , we may integrate over [MATH] and combine with ( )-( to obtain ( ). A.3 Oblivious stopping rules applied to i.i.d. atomless product distributions |
Finally, in this subsection we furnish the proof of Theorem from Section 3.1 . The oblivious stopping rule that fulfills a prophet inequality with factor |
[MATH] is more complicated than the threshold stopping rules analyzed earlier. It is defined by first solving a differential equation |
[EQUATION] with initial condition [MATH] to define a function [MATH] mapping [MATH] to [MATH] . The constant [MATH] is chosen so that [MATH] . Since the differential equation ( ) implies |
[EQUATION] for all [MATH] such that [MATH] is defined, the boundary condition [MATH] requires the equation [EQUATION] to be satisfied, and we treat this equation as the definition of [MATH] |
Denoting the cumulative distribution functions of [MATH] and [MATH] by [MATH] and [MATH] , respectively, we will be analyzing the oblivious stopping rule which selects the earliest [MATH] |
satisfying [EQUATION] We will prove that this rule satisfies a prophet inequality with a specific factor [MATH] to be determined later. (Equation ( 26 ) below defines [MATH] .) Let [MATH] denote the |
[MATH] -coordinate of the element [MATH] selected by the oblivious stopping rule, and let [MATH] denote the random variable [MATH] . As in the proof of Lemma |
, we will make use of the equations [EQUATION] which reduces the task of proving a prophet inequality with factor [MATH] to the task of proving |
[EQUATION] At this point a couple of observations will slightly simplify the analysis. If we change coordinates to replace [MATH] with [MATH] this has no effect on the behavior of the stopping rule, and of course if has no effect on [MATH] , so we are free to adopt this reparameterization and assume henceforth that [MA... |
is uniformly distributed in [MATH] . In particular, this means the stopping rule simplifies to choosing the first [MATH] such that |
[MATH] The next simplification comes from introducing the variable [MATH] and writing the event [MATH] in the form [MATH] The probability of this event is |
[EQUATION] The function [MATH] will be appearing frequently throughout this calculation so we will assign it a name: [MATH] . (This name is inspired by the fact that [MATH] |
as [MATH] for any fixed [MATH] , so [MATH] is a degree- [MATH] polynomial approximation to the exponential function.) We have derived that for [MATH] |
[EQUATION] Now what about [MATH] We can calculate this probability by integrating, with respect to [MATH] the probability that the stopping rule selects a point in [MATH] In order for [MATH] to be this point, the following things must happen. |
1. [MATH] . This has probability [MATH] 2. [MATH] 3. [MATH] 4. None of the points [MATH] , for [MATH] satisfy the stopping condition with [MATH] |
The conjunction of the first three events has probability [MATH] Let [MATH] so that [MATH] is the probability that a specific point [MATH] satisfies [MATH] along with the stopping condition [MATH] The fourth event above has probability |
[MATH] Hence [EQUATION] We are left with proving that for a suitable choice of [MATH] , the inequality [EQUATION] holds for all [MATH] At this point a useful identity comes to our aid. |
Lemma 7 The functions [MATH] and [MATH] defined above satisfy the equation [EQUATION] for every [MATH] Proof. The proof is a brief but fairly opaque calculation using the differential equation ( ) that defines the function [MATH] . First, observe that [MATH] |
satisfies the equation [MATH] because if one differentiates both sides of ( one obtains [EQUATION] Second, observe that the function [MATH] is constant: its derivative satisfies |
[EQUATION] Since [MATH] , we may conclude that [MATH] for all [MATH] , which means [EQUATION] Integrating both sides of ( 13 ) we obtain |
[EQUATION] Let [MATH] . The function [MATH] is increasing in [MATH] so [MATH] equals [MATH] for [MATH] , and it equals [MATH] for [MATH] Hence, |
[EQUATION] which matches the equation claimed in the lemma statement. Compared to the equation ( 12 asserted by Lemma the inequality ( 11 ) that we need to prove differs in three important respects. |
1. The [MATH] function appearing on both sides of ( 11 ) has been replaced by the exponential function in ( 12 ). 2. The left side of ( 12 ) has an additional [MATH] term that does not correspond to any term on the left side of ( 11 ). |
3. The constant on the left side of ( 12 is [MATH] instead of [MATH] . In fact, we will be choosing [MATH] to be very close to [MATH] but slightly smaller. The difference between [MATH] |
and [MATH] will be necessary to compensate for error terms arising from substituting [MATH] for [MATH] and from the additional [MATH] term on the left side. |
The process of accounting for these differences between ( 11 ) and ( 12 begins with the following technical lemma that provides an upper bound on the value of |
[MATH] for [MATH] Lemma 8 If [MATH] then [MATH] for all [MATH] Proof. First, recall from equation ( 10 ) that [MATH] As a function of [MATH] , the value of |
[MATH] is monotonically decreasing, so the fact that [MATH] (as can be verified numerically) implies that [MATH] . Since [MATH] is increasing in [MATH] , this also implies |
[MATH] for all [MATH] Now, as [MATH] is increasing in [MATH] , to prove the inequality [MATH] it suffices to show that [MATH] . Recalling equation ( 13 ), we have |
[EQUATION] where the final inequality used the fact that [MATH] Taking the logarithm of both sides, we find that [MATH] as desired. |
The next lemma bounds the multiplicative error in using [MATH] to approximate [MATH] Lemma 9 For [MATH] and [MATH] [EQUATION] Proof. |
Recalling the definition of the [MATH] function and using the inquality [MATH] , we have [MATH] which furnishes the first inequality in the lemma statement. To prove the second inequality, note that |
[EQUATION] where the first and last inequalities made use of the fact that [MATH] , and the second inequality depends on the fact that [MATH] , so that the exponent [MATH] is non-positive. |
The following simple inequality will be used twice in the sequel. Lemma 10 For all [MATH] [EQUATION] Proof. The function [MATH] attains its minimum value at [MATH] , where its value is strictly positive. Since this function is the derivative of [MATH] , we may conclude that the latter function is strictly increasing, a... |
Combining the steps above, we can obtain a useful lower bound on the right side of ( 11 ). Lemma 11 For [MATH] [EQUATION] Proof. |
Lemma justifies applying Lemma to the factor [MATH] in the integrand, which implies [EQUATION] where the second line is derived from Lemma |
Now, the function [MATH] is an increasing function of [MATH] in the range [MATH] , as can be verified by computing that the derivative is [MATH] |
and observing that all three factors are strictly positive. Hence, for [MATH] we have [EQUATION] where the second inequality makes use of Lemma |
10 Rearranging terms in ( 19 ) we obtain [EQUATION] and substituting this into the right side of ( 18 ) we obtain [EQUATION] The observation that [MATH] |
concludes the proof. Next we need a lemma accounting for the discrepancy between the function [MATH] appearing on the left side of ( 11 ) and the function [MATH] |
in ( 17 ). Lemma 12 For [MATH] [EQUATION] Proof. If [MATH] then [MATH] so both [MATH] and [MATH] are in the interval [MATH] , and the lemma follows easily in this case. Assume henceforth that [MATH] |
For [MATH] the inequality [MATH] implies that [MATH] Integrating from [MATH] to [MATH] , we derive [EQUATION] Multiplying both sides by [MATH] we obtain |
[EQUATION] Now exponentiate both sides of ( 24 ). [EQUATION] where the final inequality used the fact that [MATH] Lemma 10 implies that [MATH] so |
[EQUATION] and the lemma follows upon rearranging terms. Combining equation ( 17 ) with Lemma 12 we obtain the inequality [EQUATION] |
which proves ( 11 ) with [EQUATION] since [MATH] Finally, let us bound the difference between [MATH] and the constant [MATH] defined by the equation |
[EQUATION] Perform the substitution [MATH] to deduce that [EQUATION] Set [MATH] , which implies [MATH] Noting that [EQUATION] and that [MATH] is a monotonically decreasing function of [MATH] we deduce that [MATH] which implies |
[MATH] Hence [MATH] Appendix B Tightness of bounds In this section we prove that the bounds in each part of Theorem are tight under their respective assumptions. |
Proposition 1 For any [MATH] there exists a joint distribution of the agent’s and principal’s utilities such that the principal’s expected utility is always less than [MATH] |
regardless of the choice of mechanism. Proof. For a parameter [MATH] define the joint distribution of [MATH] to be a mixture of uniform distributions on two rectangles. The first is the rectangle |
[MATH] the second is the rectangle [MATH] The pair [MATH] is drawn from the uniform distribution on [MATH] with probability [MATH] and from the uniform distribution on [MATH] with probability [MATH] When we draw [MATH] i.i.d. samples [MATH] |
from this joint distribution, the random variable [MATH] is uniformly distributed in [MATH] with probability [MATH] , and it is uniformly distributed in [MATH] with probability |
[MATH] . Therefore, letting [MATH] [EQUATION] Now we bound the principal’s expected utility under any mechanism. By Lemma it is enough to consider single proposal mechanisms. Let [MATH] denote the eligible set of the mechanism, and let [MATH] denote the probability that a random sample from the distribution of [MATH] b... |
then the agent will propose an element of [MATH] . This event [MATH] happens with probability [MATH] and the principal’s expected utility conditional on [MATH] |
is bounded above by [MATH] . The other case in which the principal gets non-zero utility is if all [MATH] samples belong to the complement of [MATH] , but they don’t all belong to the complement of [MATH] . The probability of this event [MATH] is |
[MATH] and principal’s expected utility conditional on [MATH] is bounded above by [MATH] . Hence, denoting the principal’s utility by [MATH] |
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