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Let [MATH] . A strategy function [MATH] is said to be a threshold strategy if it satisfies [EQUATION] Lemma 1 At equilibrium, all customers follow a threshold strategy. |
Proof. Consider any strategy function [MATH] . Since the expected waiting time is non-increasing with the priority, there is either a single potential priority, or an interval of potential priorities, or no potential priority such that |
[EQUATION] Note that the left hand side of Eq. ( ) is the expected total cost if making AR, while the right hand side is the expected total cost if not making AR. If Eq. ( ) holds for a single value [MATH] , then a customer with potential priority greater (respectively, smaller) than [MATH] is better off making (respec... |
If Eq. ( ) holds for an interval of values [MATH] , then all customers with potential priority [MATH] do not make AR (otherwise, [MATH] would not be a constant over that interval). Therefore, [MATH] is an equilibrium strategy only if it is a threshold strategy, with threshold [MATH] |
Finally, suppose that Eq. ( ) does not hold for any [MATH] . If [MATH] for all [MATH] , then all customers are better off not making AR. Therefore, [MATH] is an equilibrium strategy only if it is a threshold strategy, with threshold [MATH] |
Note that a situation where [MATH] for all [MATH] does not exist, since a customer with potential priority zero has the same expected waiting time if making or avoiding AR. |
Next, we define two types of equilibria. Definition 2 An equilibrium strategy with threshold [MATH] is called a some-make-AR equilibrium if [MATH] |
Definition 3 An equilibrium strategy with threshold [MATH] is called a none-make-AR equilibrium if [MATH] Since the structure of the equilibrium depends on the reservation cost, we aim to determine the equilibrium to which a given reservation cost leads. Given that all customers follow a threshold strategy, we define a... |
Given a strategy with threshold [MATH] , a threshold customer that makes AR observes three priority classes: 1. A lower priority class which contains all customers with potential priority smaller than the threshold customer (none of them makes AR). The arrival rate of customers belonging to this class is [MATH] |
2. A priority class which contains only the threshold customer (since the potential priority is a continuous random variable, the probability that two customers will have the same potential priority is zero). Thus, the arrival rate of customers belonging to this class is [MATH] |
3. A higher priority class which contains all customers with greater potential priority (they all made AR before the threshold customer). The arrival rate of customers belonging to this class is [MATH] |
A threshold customer that does not make AR only observes two classes: 1. A priority class which contains the threshold customer and all customers with smaller potential priority. The arrival rate of customers belonging to this class is [MATH] |
2. A higher priority class which contains all customers with greater potential priority. The arrival rate of customers belonging to this class is [MATH] |
Based on the priority classes defined above, we find the expected waiting of the threshold customer if making or not making AR. We apply the known formula of the waiting time in an [MATH] queue with preemptive-resume priorities (Conway et al. 2012 , p.175) and obtain the following: |
1. The expected waiting time of the threshold customer if making AR is [EQUATION] 2. The expected waiting time of the threshold customer if not making AR is |
[EQUATION] The condition for threshold [MATH] to be a some-make-AR equilibrium is [EQUATION] That is, a customer with potential priority equals to the threshold is indifferent between the two actions. The condition for threshold [MATH] to be a none-make-AR equilibrium is |
[EQUATION] That is, a customer with potential priority [MATH] (and hence, all customers) are better off not making AR. By isolating [MATH] in Eq. ( ), we define [MATH] to be a function that maps a threshold to the reservation cost that leads to that threshold |
[EQUATION] We conclude that given reservation cost [MATH] , the threshold [MATH] represents a some-make-AR equilibrium if and only if [MATH] . The threshold [MATH] represents a none-make-AR equilibrium if and only if [MATH] . In order to find the equilibrium structure, we next find the properties of [MATH] |
Lemma 2 If [MATH] , then [MATH] is a monotonically increasing function. If [MATH] , then [MATH] is a unimodal function with a global maximum. |
Proof. First, we compute the derivative of [MATH] [EQUATION] Since the denominator is negative for any [MATH] , the sign of the derivative is determined by the sign of [MATH] . If [MATH] , then this expression is negative for any [MATH] and the derivative of [MATH] is positive for any [MATH] . If [MATH] , then the deri... |
Next, we define: [EQUATION] and [EQUATION] Note that if [MATH] , then [MATH] is the maximum value of [MATH] and if [MATH] , then [MATH] is the maximum value of [MATH] . We can now state the main result of this section: |
Theorem 1 The game has the following equilibrium structure. When [MATH] If [MATH] , then there is a unique some-make-AR equilibrium. |
If [MATH] , then there is a unique none-make-AR equilibrium. When [MATH] If [MATH] , then there is a unique some-make-AR equilibrium. |
If [MATH] , then there are two some-make-AR equilibria and a none-make-AR equilibrium. If [MATH] , then there is a unique none-make-AR equilibrium. |
Proof. We begin with [MATH] . If [MATH] , then there is a single value of [MATH] such that [MATH] has a solution. Hence, there is one some-make-AR equilibrium. A none-make-AR equilibrium does not exist since [MATH] . If [MATH] , then there is no value of [MATH] such that [MATH] has a solution. Hence, a some-make-AR equ... |
Next, consider [MATH] . In the range [MATH] , the function [MATH] is monotonically increasing. Thus, if [MATH] , then there is a single value of [MATH] such that [MATH] has a solution, and hence there is one some-make-AR equilibrium. In the range [MATH] , the function [MATH] is unimodal. Thus, if [MATH] , then there ex... |
Revenue Maximization In this section, we assume that the reservation cost is a fee determined by the service provider. We show that the fee that maximizes the revenue leads to a unique equilibrium if the utilization is smaller than [MATH] and to multiple equilibria if the utilization is greater than [MATH] . We also sh... |
The revenue per time unit, at equilibrium with threshold [MATH] , is the number of customers making AR multiplied by the AR fee that leads to that equilibrium. The expected revenue is |
[EQUATION] With some manipulation, we get that the revenue function does not depend on the values of [MATH] and [MATH] but only on the utilization [MATH] |
[EQUATION] At first glance, this result seems surprising since it implies that the revenue does not increase when scaling the system (i.e., increasing both arrival and service rates). However, in an [MATH] queue, the waiting time decreases as the system gets larger, and hence customers are less motivated to make AR. Th... |
By solving the equation [MATH] , we find that the optimal threshold is [MATH] . By substituting [MATH] into Eq. ( ), we get that the optimal fee is |
[EQUATION] Similarly, by substituting [MATH] into Eq. ( 12 ), we get that the maximum possible revenue is [EQUATION] Next we find the number of equilibria when [MATH] |
Theorem 2 The revenue maximizing fee [MATH] leads to a unique some-make-AR equilibrium if [MATH] and to multiple equilibria, including a none-make-AR equilibrium, otherwise. |
Proof. The optimal reservation cost [MATH] leads to multiple equilibria only if [MATH] and [MATH] (see Theorem ). Using Eq. ( ) and Eq. ( 13 ), we deduce that if [MATH] , then |
[EQUATION] One can show that the inequality above holds only if [MATH] Figure illustrates the game outcome when [MATH] 5.1 Price of Conservatism |
Assuming that [MATH] , the provider can either be risk-averse and charge a fee that leads to a unique equilibrium with guaranteed revenue, or it can be risk-taking and charge a higher fee that may lead to greater revenue but also to zero revenue. To compare between the two options, we use the Price of Conservatism (PoC... |
[EQUATION] Since [MATH] has exactly one extreme point (which is [MATH] ), it is increasing in the range [MATH] . Therefore, the maximum guaranteed revenue is achieved when choosing the largest [MATH] for which [MATH] . In other words, [MATH] should be slightly smaller than [MATH] . By solving [MATH] , we get two soluti... |
[EQUATION] By substituting [MATH] into Eq. ( 12 ), we get [EQUATION] and by dividing [MATH] by [MATH] , we get [EQUATION] We conclude with the following theorem. |
Theorem 3 If [MATH] , then [MATH] . Else, [MATH] Figure shows the maximum possible revenue and the maximum guaranteed revenue in a system with parameters [MATH] and [MATH] |
By computing the derivative of PoC with respect to [MATH] , we get that for any [MATH] [EQUATION] Thus, we obtain the following corollary: |
Corollary 1 The price of conservatism increases with the utilization. That is, as the utilization increases, the ratio between the potential revenue when the provider is risk-taking and the revenue when the provider is risk-averse increases and tends to [MATH] as [MATH] |
Dynamic Games 6.1 Learning Models In this section, we study dynamic versions of the game. In dynamic games (also known as learning models, since players learn over time the behavior of other players), it is assumed that the game repeats many times and that initially players do not necessarily follow an equilibrium stra... |
At each step (game), a new set of customers participate (or the same set of participants but with new realizations of request times). At the first step, all customers have an initial belief about the strategy that is followed by all customers. Next, we assume: |
Assumption 1 Customers that are indifferent between actions [MATH] and [MATH] choose action AR’. Based on this assumption, and using the proof of Lemma , one can show that the best response of all customers to any initial belief is a threshold strategy. In order to simplify the analysis and since a threshold strategy i... |
Assumption 2 The initial belief is a threshold strategy. We denote by [MATH] the threshold of the strategy followed at step [MATH] . We denote by [MATH] the estimation of this strategy and we distinguish between two types of learning: |
1. Strategy learning . In this type of learning, the analysis assumes that at each step [MATH] [MATH] . That is, customers observe past strategies. |
2. Action learning . In this type of learning, customers observe previous actions and use the proportion of customers that chose [MATH] at the previous step as an estimation of the strategy that was followed at that step. Namely, if the demand and the number of reservations at step [MATH] are [MATH] and [MATH] respecti... |
[EQUATION] Since the best response of all customers to any belief is a threshold strategy, we can define a joint best response function [MATH] . The input is a belief about the threshold strategy that will be followed by all customers. The output is the best response threshold to that belief. Thus, we can describe the ... |
[EQUATION] where [MATH] represents the initial belief. Note that under strategy-learning this process is deterministic, while under action-learning this process is a Markov process (Gardiner et al. 1985 , Chapter 3) . In the following sections we analyze this dynamic process. |
Next, we focus on the behavior of customers at a given step. Thus, we remove the subscript [MATH] . We begin the analysis with the following observations: |
(i) Given a belief [MATH] (i.e., assuming that all other customers follow the threshold [MATH] ), if a tagged customer with potential priority [MATH] chooses [MATH] , then all customers with greater potential priority have higher priority and all customers with smaller potential priority have lower priority. Therefore,... |
[EQUATION] where [MATH] is defined in Eq. ( ). (ii) Given a belief [MATH] , if a tagged customer with potential priority [MATH] chooses [MATH] , then his/her (believed) expected waiting time is the same as the expected waiting time of the threshold customer (recall that each customer believes that he/she is the only on... |
[EQUATION] (iii) The expected waiting time of all customers that choose [MATH] are equal. Hence, [EQUATION] where [MATH] is defined in Eq. ( ). |
Those properties will be used later to prove our main results. Next, we separately explore the case of a unique some-make-AR equilibriun and the case of multiple equilibria. |
6.2 Learning with Unique Some-make-AR Equilibrium Consider a some-make-AR equilibrium with equilibrium threshold [MATH] . By computing the derivative of [MATH] and [MATH] , one can verify that both functions are decreasing with [MATH] . This property will be used in the proof of the following lemma. |
Lemma 3 Under a unique some-make-AR equilibrium: 1. If a belief [MATH] , then [MATH] 2. If a belief [MATH] , then [MATH] Proof. From Eq. ( ) and Eq. ( ), we deduce that [MATH] . Since, under unique some-make-AR equilibrium, [MATH] and [MATH] intersect once, we conclude that |
[EQUATION] and [EQUATION] For proving part 1, assume that [MATH] . Since [MATH] , and since [MATH] is a decreasing function we deduce that |
[EQUATION] From Eq. ( 27 ) and Eq. ( 29 ) we deduce that there exists a [MATH] such that [EQUATION] Given the equation above and using Eq. ( 24 ) and Eq. ( 25 ), one can see that all customers with potential priority [MATH] choose [MATH] , while all customers with potential priority [MATH] choose [MATH] . This complete... |
Now, let assume that [MATH] . Based on Eq. ( 24 ) and Eq. ( 25 ) and since [MATH] is a decreasing function, we deduce that [EQUATION] |
From Eq. ( 28 ) and Eq. ( 39 ), we deduce that [EQUATION] That is, all customers are better off choosing [MATH] and the best response to [MATH] is [MATH] |
Next, we study the long-term outcome of the dynamic game and establish the following result. Theorem 4 A game with unique some-make-AR equilibrium converges to equilibrium under strategy-learning and cycles under action-learning. |
Proof. We begin with the first part of the theorem. Let assume that the initial belief [MATH] . From Lemma , we deduce that [MATH] . Hence, [MATH] . By induction, we deduce that, for any [MATH] |
[EQUATION] The set [MATH] is a monotonically increasing sequence bounded by [MATH] . Thus, it has a limit, denoted by [MATH] . since [MATH] , and sense [MATH] , we conclude that the limit [MATH] is a fixed point of [MATH] , and hence it must be the equilibrium point [MATH] |
Next, we assume that [MATH] . In this case, based on Lemma [MATH] and the game converges to equilibrium as in the case of [MATH] |
From Eq. ( 21 ), we deduce that, under action-learning , at any step [MATH] , if [MATH] , then [MATH] (i.e., if the strategy followed at step [MATH] is greater than zero then there is a positive probability that the fraction of customers not making AR will be greater than [MATH] ). Once [MATH] , then [MATH] . Thus, cus... |
Next, we determine, for a given reservation cost, whether a service provider who wishes to maximize the number of reservations is better off under strategy-learning or under action-learning |
Theorem 5 In a dynamic game with a unique some-make-AR equilibrium, the average number of customers making AR under action-learning is greater than under strategy-learning. |
Proof. Denote the unique equilibrium by [MATH] . Consider an arbitrary step [MATH] and assume that action-learning is applied. If at step [MATH] the strategy [MATH] , then all customers will choose [MATH] at the next step. If [MATH] , then [MATH] . Thus, in any realization, the strategy followed by all customers in all... |
Next, we present a simulated example that compares between the revenue under action-learning and under strategy-learning . The pseudo-code of the simulation is given in Algorithm 1. The inputs of the procedure are the arrival rate [MATH] , the initial belief [MATH] , the reservation cost [MATH] and the number of steps ... |
Algorithm 1 Learning Simulation ( [MATH] [MATH] for [MATH] to [MATH] {iterating over all steps} do [MATH] {variable counting the number of reservations} [MATH] |
generate Poisson random variable {the number of customers} for [MATH] to [MATH] {iterating over all customers} do [MATH] generate random variable from U(0,1) {the potential priority} |
if [MATH] {check if the potential priority is greater than the current belief} then if [MATH] {check if the customer is better off making AR} then |
[MATH] {increase the number of reservations by one} end if else if [MATH] {check if the customer is better off making AR} then [MATH] {increase the number of reservations by one} |
end if end if end for if strategy-learning then [EQUATION] {compute the current strategy} end if if action-learning then [MATH] {estimate the current strategy} |
end if end for Example 1 Consider a queue with parameters [MATH] and [MATH] . Let the reservation cost be [MATH] . The unique equilibrium (computed using Eq. ( ) and Eq. ( )) is [MATH] (i.e., on a static game, on average, [MATH] of the customers make AR). We run a simulation of [MATH] steps. Each step lasts for one tim... |
We conclude that if the provider interest is that as many customers as possible will make reservations, then it is better off if customers gain information about previous actions rather than strategies. |
6.3 Learning with Multiple Equilibria Lemma 4 In a game with multiple equilibria with thresholds [MATH] [MATH] and [MATH] 1. If a belief [MATH] , then [MATH] |
2. If a belief [MATH] , then [MATH] 3. If a belief [MATH] , then [MATH] Proof. From Eq. ( ) and Eq. ( ), we deduce that [MATH] . Since, under unique some-make-AR equilibrium, [MATH] and [MATH] intersect twice at [MATH] and [MATH] , we conclude that |
[EQUATION] From Eq. ( 35 ) and Eq. ( 37 ) we deduce that there exists a [MATH] such that [EQUATION] Given the equation above and using Eq. ( 24 ) and Eq. ( 25 ), one can see that all customers with potential priority [MATH] choose [MATH] , while all customers with potential priority [MATH] choose [MATH] . This complete... |
Now, assume that [MATH] . Based on Eq. ( 24 ) and Eq. ( 25 ) and since [MATH] is a decreasing function, we deduce that [EQUATION] |
From Eq. ( 36 ) and Eq. ( 39 ), we deduce that [EQUATION] That is, all customers are better off choosing [MATH] and the best response to [MATH] is [MATH] The third part of the lemma can be proved using the same arguments as in the proof of the first part of the lemma. |
Next, we study the long-term outcome of a dynamic game with multiple equilibria. Theorem 6 A game with multiple equilibria converges to some-make-AR or none-make-AR equilibrium (depend on the initial belief) under strategy-learning and to none-make-AR equilibrium under action-learning. |
Proof. Using the same arguments as in the proof of Theorem and based on Lemma , one can show the following. Under strategy-learning , a game with initial belief [MATH] converges to [MATH] , while a game with initial belief [MATH] converges to [MATH] |
Under action-learning , if at some step [MATH] [MATH] and none-make-AR is an equilibrium, then at all future steps all customers will keep not making AR. Given any threshold strategy [MATH] followed by all customers, there is a positive probability that the potential priority of all customers will be smaller than [MATH... |
Example 2 Consider a queue with parameters [MATH] and [MATH] . Let the reservation cost be [MATH] . Using Eq. ( ) and Eq. ( ) we compute the set of equilibria: [MATH] . We set three different initial strategies: [MATH] and [MATH] . We apply strategy-learning. As Figure shows, within a few steps, the system converges to... |
6.4 Profit Maximization in Dynamic Games In this section, we assume that the reservation cost is a fee collected by the provider. Our goal is to find the fee that maximizes the provider revenue in the dynamic games setting. Under action learning, any fee that leads to multiple equilibria will eventually lead to zero re... |
Under strategy-learning with multiple equilibria, the initial belief determines to which equilibrium the game will converge. To execute the analysis, we assume that the initial belief [MATH] is a continues random variable that takes values between zero and one. |
Consider a game with multiple equilibria [MATH] . From Lemma , Lemma and Theorem , we deduce that if the initial belief is in [MATH] , then the game converges to [MATH] , otherwise it converges to [MATH] (i.e., zero reservations). Thus, with probability [MATH] the strategy converges to [MATH] and with probability [MATH... |
[EQUATION] where [MATH] is defined in Eq. ( ). Since the expected revenue depends on both [MATH] and [MATH] , we next find the relation between those two thresholds. By manipulating the equation [MATH] (see Eq. ( ) for definition of [MATH] ), we get the following relation. |
[EQUATION] Given the distribution of [MATH] and using Eq. ( 41 ), Eq. ( 42 ) and Eq. ( 12 ), one can find the value of [MATH] that maximizes the revenue and, in turn, the optimal fee. For instance, let assume that [MATH] is uniformly distributed in [MATH] . In this case, the revenue as a function of [MATH] is |
[EQUATION] By computing the derivative of [MATH] with respect to [MATH] , one can show that it decreases with [MATH] . Thus, when considering multiple equilibria, the optimal value of [MATH] is [MATH] . From Eq. ( 17 ) we know that this value is [MATH] and is obtained when [MATH] |
Combining this result with Corollary leads to the following theorem: Theorem 7 Under strategy-learning, if the initial belief is uniformly distributed between [MATH] and [MATH] , then the optimal fee is [MATH] when [MATH] and [MATH] when [MATH] |
Conclusion and future work In this paper, we analyzed an M/D/1 queue that supports advance reservations. We associated the act of making reservation with a fixed reservation cost and studied the impact of this cost on the behavior of customers. First, we showed that if the utilization of the queue is greater than [MATH... |
In the second part of the paper, we studied a dynamic version of the game. We showed that if the customers observe previous strategies, then the game converges to an equilibrium. If the customers observe previous actions, then the game converge to a none-make-AR equilibrium, if such an equilibrium exists, and cycles ot... |
# Source: arxiv 1806.08444 # Title: What Makes An Asset Useful? # Sections: all # Downloaded: 2026-03-03T04:47:31.373076+00:00 WHAT MAKES AN ASSET USEFUL? |
Abstract Given a new candidate asset represented as a time series of returns, how should a quantitative investment manager be thinking about assessing its usefulness? This is a key qualitative question inherent to the investment process which we aim to make precise. We argue that the usefulness of an asset can only be ... |
We identify four features that the time series of returns of an asset should exhibit for the asset to be useful to an investment manager, two primary and two secondary. As primary criteria, we propose that the new asset should provide sufficient incremental diversification to the reference universe of assets/benchmarks... |
INTRODUCTION The idea of diversification is ancient. The related well-known phrase ‘don’t put all your eggs in one basket’ can be traced back to the classical novel Don Quixote by Miguel de Cervantes Saavedra as early as 1605 |
, where it is phrased as “ It is the part of a wise man to keep himself today for tomorrow, and not venture all his eggs in one basket. ” The idea of diversification itself is much older, and can be found for instance in the book of Ecclesiastes — “But divide your investments among many places, for you do not know what... |
, where it is illustrated that an investor can reduce portfolio risk simply by holding combinations of assets that have the same expected return and are not perfectly (positively) correlated. |
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