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Proposition 2 Let [MATH] be a finite improvement path, and let [MATH] for an arbitrary improvement step [MATH] . It holds that [MATH] |
We now bound the utility the improving author obtains in the corresponding improvement step, when her document’s quality does not exceed the highest quality (on that particular topic) in the preceding improvement step. |
Proposition 3 Let [MATH] be a finite improvement path, and let [MATH] for an arbitrary improvement step [MATH] . If [MATH] , then |
[EQUATION] Next, we characterize a property that must hold in improvement cycles, under the false assumption that such exist. We prove that if an improvement cycle exists, the quality of the first-ranked document is constant throughout the improvement cycle; this must hold for every topic. |
Lemma 1 If [MATH] is an improvement cycle, then for every improvement step [MATH] and every topic [MATH] it holds that [MATH] Proof sketch. |
We give here a high-level overview of the proof and refer the reader to the appendix for the formal proof. Under the false assumption that an improvement cycle exists, assume that the claim does not hold. Namely, assume that [MATH] is an improvement cycle (w.l.o.g. [MATH] is a simple improvement cycle), and that there ... |
Recall that [MATH] , i.e., the topics are sorted according to the query distribution mass in a non-increasing order. We prove by induction on the topic index [MATH] that [MATH] holds for every [MATH] [MATH] . Clearly, if this holds for every improvement step [MATH] then |
[EQUATION] thus, all inequalities hold in equality and [MATH] cannot occur. Base, [MATH] Assume the assertion does not hold for [MATH] ; hence, there exists [MATH] [MATH] , such that [MATH] . This means that there exists an author who writes with the highest quality on topic 1 in the step [MATH] , and then she deviates... |
Step: Assume the assertion holds for [MATH] , i.e., [MATH] for every step [MATH] We show that [MATH] for a step [MATH] implies that the improving author in improvement step [MATH] deviates to a topic with a lower index. Using the bound obtained in Proposition and the induction hypothesis, we show that there must be an ... |
Lemma implies that the only element that varies throughout an improvement cycle, if such exists, is the number of authors who write on each topic. In particular, the highest quality on each topic remains constant. It also suggests that any improving author is not the only author writing the highest quality document on ... |
Consider an arbitrary improvement step, and denote by [MATH] the topic that the improving author writes on in the improvement step. The improving author joins a (non-empty) set of authors which are already writing documents with the highest quality on topic [MATH] . Since we deal with a cycle, at some point an author a... |
Lemma 2 If [MATH] is an improvement cycle, then for every improvement step [MATH] and topic [MATH] such that [MATH] there exist [MATH] such that [MATH] and |
[EQUATION] Proof sketch. Let [MATH] be such that [MATH] . By definition of improvement step [MATH] . From Lemma we know that [MATH] ; thus, [MATH] and [MATH] . Afterwards, we prove another claim which guarantees that there exists [MATH] such that |
[EQUATION] holds. In addition, [MATH] is the improving author, and so [EQUATION] Clearly, [MATH] . Lemma indicates that [MATH] ; hence, [MATH] . Having showed that the condition of Proposition holds, we invoke it for [MATH] and conclude that |
[EQUATION] Combining this fact with Equation ( ), we get [EQUATION] In Theorem below we leverage Lemma to show that improvement cycles cannot exist. |
Theorem 1 [MATH] is [MATH] -learnable. Proof of Theorem To show that [MATH] is [MATH] -learnable it suffices to show that every improvement path is finite. Moreover, every improvement path cannot contain more than a finite number of different strategy profiles, as [MATH] and [MATH] are finite; therefore, if [MATH] is i... |
Assume by contradiction that [MATH] contains an improvement cycle [MATH] . Let [MATH] be an arbitrary improvement step and denote by [MATH] the topic such that [MATH] From Lemma we know that there exist [MATH] such that [MATH] and |
[EQUATION] Since [MATH] , we can now use Lemma again in order to find [MATH] such that [MATH] and [EQUATION] This process can be extended to achieve additional [MATH] such that |
[EQUATION] Since there are only [MATH] topics and that the inequality above contains [MATH] elements, there are at least two elements which are identical; thus we obtain a contradiction. We deduce that an improvement cycle cannot exist. |
Theorem concludes the analysis of the exposure-targeted utility function. 3.2 Action-Targeted Utility After analyzing games with exposure-targeted utility, we proceed to action-targeted utility. The main result of this subsection is that [MATH] is [MATH] -learnable, which is analogous to the main result of the previous... |
Definition 6 Given a topic [MATH] and an improvement path [MATH] [EQUATION] In Proposition we bound the utility of an improving author in an improvement step. |
Proposition 4 Let [MATH] be a finite improvement path, and let [MATH] for an arbitrary improvement step [MATH] . If [MATH] , then |
[EQUATION] Notice that [MATH] for every [MATH] and every [MATH] ; thus, the bound given in Proposition trivially holds for [MATH] . However, proving this tighter bound becomes essential for refuting the existence of improvement cycles under [MATH] . By proving additional supporting lemmas (which are further elaborated ... |
Theorem 2 [MATH] is [MATH] -learnable. Non-Learnability under Other Mediators In the previous section we showed a powerful result: |
[MATH] is both [MATH] -learnable and [MATH] -learnable. In other words, when using [MATH] , any better-response dynamics converges; this is true for both utility schemes. In fact, [MATH] is not the only mediator under which such convergence occurs. For instance, Let [MATH] be the random mediator, such that for any auth... |
[EQUATION] By showing that under [MATH] any game with [MATH] can be reduced to a game with [MATH] , we conclude that Proposition 5 |
[MATH] is [MATH] -learnable. Proof sketch. We prove the claim by showing that under [MATH] any game with [MATH] can be reduced to a game with [MATH] , such that the two games are strategically equivalent. This is done by taking any game [MATH] with [MATH] as the mediator and a quality matrix [MATH] , and reduce it to a... |
Since both [MATH] consists of the exposure-targeted utility function, we omit the super-script [MATH] and use the super-script [MATH] to specify the utility of author [MATH] under the strategy profile [MATH] in [MATH] , i.e., [MATH] , and equivalently [MATH] for [MATH] . By definition of exposure-targeted utility and [... |
[EQUATION] Since [MATH] possesses [MATH] as the mediator, Theorem guarantees that [MATH] has the FIP property. Since we showed [MATH] and [MATH] are strategically equivalent, [MATH] also has the FIP property, and in particular does not contain improvement cycles. |
Notice that [MATH] treats every document the same, regardless of its quality. However, in many (and perhaps even most) scenarios mediators seek to promote high-quality content. Therefore, the reader may wonder whether other plausible mediators are [MATH] -learnable or [MATH] -learnable. We now focus on a wide and intui... |
Definition 7 Let [MATH] be a mediator. We say that [MATH] is a scoring mediator if there exists a non-decreasing function [MATH] such that for every [MATH] and author index [MATH] it holds that |
[EQUATION] It this case, we denote [MATH] for the corresponding [MATH] Under a scoring mediator every author receives a probability according to the proportion of her score over the sum of the scores of all author writing on that topic. Notice that if [MATH] is a scoring mediator such that the corresponding [MATH] is c... |
4.1 Exposure-Targeted Utility In this subsection we prove that, under mild assumptions, scoring mediators are not [MATH] -learnable (as opposed to [MATH] ). We restrict ourselves to mediators for which the corresponding function [MATH] is continuous, and exhibits the following property: the ratio between the score of t... |
Theorem 3 Let [MATH] be a scoring mediator. If [MATH] is a continuous function such that [MATH] , then [MATH] is not [MATH] -learnable. |
Proof sketch. It is sufficient to show that for every [MATH] that satisfies the theorem’s conditions, we can construct a game instance with an improvement cycle. We exploit the properties of [MATH] to construct a game with four authors and three topics, and show that an improvement cycle exists. Let [MATH] be a scoring... |
[EQUATION] For brevity, denote [MATH] and [MATH] , and observe that [MATH] . Consider a game with [MATH] authors, [MATH] topics and a quality matrix [MATH] such that |
[EQUATION] The only missing ingredient is the distribution [MATH] over the topics. The selection of such [MATH] is crucial: we shall select [MATH] to allow improvement cycles. Denote |
[EQUATION] for some [MATH] . It can be verified that [MATH] is a valid distribution over the set of topics. Consider the strategy profiles |
[EQUATION] In the rest of the proof we show that [MATH] can be selected such that the cycle [MATH] is an improvement cycle of the game we constructed. More precisely, we prove that for every [MATH] [MATH] [MATH] . This suggests that [MATH] is not [MATH] -learnable. |
While all it takes to prove Theorem is to show a single game instance with an improvement cycle, we can actually construct infinitely many games which do not possess FIP. Moreover, our construction can be viewed as a sub-game in a much broader game, i.e., with more authors and topics. |
4.2 Action-Targeted Utility When analyzing scoring mediators, an additional difference between the two utility schemes emerges. In the improvement cycle constructed in the proof of Theorem , there exists an improvement step in which the improving author decreases the quality of her document but still increases her util... |
Theorem 4 Let [MATH] be a scoring mediator. If [MATH] is a continuous function such that [MATH] for some [MATH] , then [MATH] is not [MATH] -learnable. |
Notice the resemblance between the condition of Theorem to that of Theorem . Due to space limitations, additional results on the non-learnability of other scoring mediators under [MATH] are omitted and further elaborated in the appendix. |
Discussion We introduced the study of learning dynamics in the context of information retrieval games. Our results address learning in the framework introduced by ), where authors are action-targeted as well as for a complementary type of information retrieval game in which the authors’ aim is to maximize their exposur... |
One prominent question is the time required for the authors to converge, namely, finding the worst-case length of an improvement path. It turns out that there is a class of games where the length of the best-response paths is easy to analyze. |
Consider the exposure target utility, and assume that [MATH] is strictly decreasing, the number of authors equals the number of topics, and that the matrix [MATH] is generic, i.e., has [MATH] distinct values. The induced game exhibits a unique equilibrium: topic 1 is assigned to the author with the highest quality w.r.... |
Our model, as any other novel model that pretends to explain theoretical aspects of real-world systems, has its limitations. To name a few, we assume the set of authors and topics are fixed, while in reality they are often dynamic; we assume that the quality of documents is perfectly observed by the mediator, which onl... |
An interesting future direction is to expand the information retrieval setting to a setup where each author’s document may include several topics. This issue is treated in a preliminary manner in ( ) and it may be of interest to see whether our results can be extended to that context as well. It may be also interesting... |
Appendix A Omitted Proofs from Section Proof of Proposition The proof makes use of the following theorems by , which are stated slightly different for ease of presentation. |
Theorem 5 (Theorem 2.8,( )) Let [MATH] be a two-by-two bimatrix game such that [EQUATION] Then [MATH] is an exact potential game if and only if |
[EQUATION] Further, Theorem 6 (Corollary 2.9,( )) A game [MATH] is an exact potential game if and only if every two-by-two subgame of [MATH] is an exact potential game. |
To show that the class of games induced by the PRP mediator and [MATH] (equivalently, [MATH] ) does not have an exact potential, it is sufficient to show a subgame of a larger game which is not an exact potential game. |
Consider a game [MATH] with [MATH] authors, [MATH] topics, [MATH] and a quality matrix [EQUATION] We first focus on [MATH] . Observe that the bimatrix game describing the utilities of authors 1 (rows) and 2 (columns) induced by setting [MATH] is |
[EQUATION] The above bimatrix game does not satisfy the condition given in Equality ( ), since [EQUATION] hence, Theorem implies that [MATH] does not have an exact potential. |
Next, consider a game [MATH] with the same [MATH] , and let [MATH] be the utility function. Let [MATH] , and observe that the utility bimatrix of authors 1 and 2 is |
[EQUATION] Here again, [EQUATION] thus, Theorem implies that [MATH] does not have an exact potential. Appendix B Omitted Proofs from Subsection 3.1 |
Proof of Proposition Since author [MATH] improves her utility, [MATH] . By definition of [MATH] , if [MATH] then [MATH] , which results in a contradiction. |
Proof of Proposition Combined with Proposition , we know that [EQUATION] Notice that [MATH] and [MATH] ; hence, together with Equation ( ) we obtain |
[EQUATION] Observe that Equation ( ) suggests that [EQUATION] which concludes the proof of this proposition. Proposition 6 If [MATH] is an improvement cycle and [MATH] is a topic such that the following properties hold |
1. there exists an improvement step [MATH] satisfying [MATH] , and 2. for every improvement step [MATH] [MATH] then there exist an index [MATH] such that [MATH] and |
[EQUATION] Proof of Proposition From Property we know that there exists an improvement step [MATH] such that [MATH] Assume w.l.o.g. that [MATH] ; hence |
[EQUATION] By the definition of [MATH] we know that there exists an improvement step [MATH] such that [EQUATION] From Property we get that for every improvement step [MATH] [MATH] , which implies that |
[EQUATION] Combining Equations ( ),( ) and ( ) with the fact that [MATH] is an improvement cycle leads to the fact that there must exist an improvement step [MATH] such that [MATH] [MATH] and [MATH] . This implies that [MATH] and [MATH] ; therefore, |
[EQUATION] Proof of Lemma Assume w.l.o.g. that [MATH] is a simple improvement cycle. First, we prove by induction on the topic index [MATH] that [MATH] holds for every [MATH] [MATH] . Later, we leverage this result to prove the statement of the lemma. |
Base: Assume the assertion does not hold for [MATH] ; hence, there exists [MATH] [MATH] , such that [MATH] . As a result, it holds for the improving author [MATH] in step [MATH] that [MATH] and [MATH] . In words, the quality of [MATH] ’s document exceeds all other qualities under [MATH] on topic 1; thus, |
[EQUATION] In addition, [MATH] is an improvement step for author [MATH] , and so [MATH] . Combined with Equation ( ), [EQUATION] |
On the other hand, [MATH] holds; thus, Equation ( ) implies that [MATH] , which is clearly a contradiction since [MATH] Step: Suppose the assertion holds for every [MATH] where [MATH] , but does not hold for [MATH] . Similarly to the base case, there exists [MATH] [MATH] , such that [MATH] . As a result, [MATH] and [MA... |
[EQUATION] In addition, [MATH] holds since [MATH] is the improving author; hence, with Equation ( 10 ) we get [EQUATION] Let [MATH] denote the topic that author [MATH] is writing on under [MATH] , i.e., [MATH] . By definition of [MATH] we obtain |
[EQUATION] Recall that [MATH] ; hence, Equations ( 11 ) and ( 12 ) suggest that [MATH] holds, and therefore we are guaranteed that [MATH] |
Since [MATH] , the induction hypothesis hints that [MATH] ; therefore, [MATH] holds and by Proposition we get that [MATH] . Notice that [MATH] is a finite improvement path, and that the condition of Proposition holds; thus, by invoking it for [MATH] , we get |
[EQUATION] Together with Equation ( 11 ), we conclude that [EQUATION] Next, we wish to find an improvement step such that the improving author’s utility strictly bounds the right-hand-side of Equation ( 13 ). Since [MATH] and [MATH] we get that [MATH] . In addition, from the induction hypothesis, we get that for every ... |
[EQUATION] Since [MATH] is the improving author [MATH] holds, which together with Equation ( 14 ) implies [EQUATION] Let [MATH] . By definition of [MATH] , we know that |
[EQUATION] Observe that [MATH] must hold. To see this, assume otherwise that [MATH] , and [MATH] follows. Incorporating this assumption with Equations ( 13 ),( 15 ) and ( 16 ) we obtain |
[EQUATION] which is a contradiction; hence, [MATH] . The induction hypothesis hints that [MATH] , implying [MATH] Here again, the condition of Proposition holds; thus, by invoking it for [MATH] we conclude that |
[EQUATION] Together with Equation ( 15 ), we conclude that [EQUATION] We have therefore bound the right-hand-side of Equation ( 13 ) as desired. |
This process can be extended to obtain additional [MATH] , such that for all [MATH] [MATH] and [EQUATION] While the inequality above contains [MATH] elements, there are only [MATH] topics with index lower than [MATH] ; hence, at least two of them must be identical, and we obtain a contradiction. We deduce that [MATH] f... |
This concludes the proof of the induction. Ultimately, to end the proof of this lemma, fix a topic [MATH] . Due to the induction above, [MATH] holds for every [MATH] , i.e., |
[EQUATION] The left-hand-side and the right-hand-side of the inequality above are identical; thus, they must all hold in equality. This concludes the proof of this lemma. |
Proof of Lemma Let [MATH] such that [MATH] . From Lemma we know that for every improvement step [MATH] [MATH] ; thus, [MATH] which by Proposition leads to |
[EQUATION] By definition of improvement step [MATH] ; hence together with Equation ( 17 ) we get that [MATH] Notice that [MATH] is a finite improvement path, and that the condition of Proposition holds; hence, by invoking it for [MATH] we conclude the existence of an index [MATH] such that [MATH] and |
[EQUATION] In addition, [MATH] is the improving author, and so [EQUATION] Clearly, [MATH] . Lemma indicates that [MATH] ; hence, [MATH] . Having showed the condition of Proposition holds, we invoke it for [MATH] and conclude that |
[EQUATION] Combining this fact with Equation ( 18 ), we get [EQUATION] Appendix C Omitted Proofs from Subsection 3.2 Proposition 7 |
Let [MATH] be a finite improvement path, and let [MATH] for an arbitrary improvement step [MATH] . It holds that [MATH] Proof of Proposition |
Similarly to the proof of Proposition , since author [MATH] improves her utility, [MATH] . By definition of [MATH] , if [MATH] then [MATH] , which results in a contradiction. |
Proof of Proposition Combined with Proposition , we know that [EQUATION] Notice that [MATH] and [MATH] ; hence, together with Equation ( 19 ) we obtain |
[EQUATION] Observe that Equation ( 20 ) suggests that [EQUATION] which concludes the proof of this proposition. Proposition 8 If [MATH] is an improvement cycle and [MATH] is a topic such that the following properties hold |
[EQUATION] Combining Equations ( 21 ),( 22 ) and ( 23 ) with the fact that [MATH] is an improvement cycle leads to the fact that there must exist an improvement step [MATH] such that [MATH] [MATH] , and [MATH] . This implies that [MATH] and [MATH] ; therefore, |
[EQUATION] C.1 Proof of Theorem To ease presentation of the proof, throughout this subsection we re-index the topics according to the following order |
[EQUATION] The proof of Theorem relies on several supporting lemmas, which are proven first. Lemma 3 If [MATH] is an improvement cycle, then for every improvement step [MATH] and every topic [MATH] it holds that [MATH] |
Proof of Lemma Assume w.l.o.g. that [MATH] is a simple improvement cycle. First, we prove by induction on the topic index [MATH] that [MATH] holds for every [MATH] [MATH] . Later, we leverage this result to prove the statement of the lemma. |
Base: By the definition of [MATH] we know that there exists an improvement step [MATH] [MATH] such that [MATH] . Now Assume that the assertion does not hold for [MATH] ; hence, there exists [MATH] [MATH] , such that [MATH] . Therefore, combining the above with the fact that [MATH] is an improvement cycle implies that t... |
[EQUATION] In addition, [MATH] is an improvement step for author [MATH] , and so [MATH] . Combined with Equation ( 24 ), [EQUATION] |
On the other hand, [MATH] holds; thus, Equation ( 25 implies that [MATH] , which is clearly a contradiction since [MATH] Step: Suppose the assertion holds for every [MATH] where [MATH] , but does not hold for [MATH] . Similarly to the base case, by the definition of [MATH] , there exists [MATH] [MATH] such that [MATH] ... |
As a result, it holds for the improving author [MATH] in step [MATH] that [MATH] and [MATH] . In words, the quality of [MATH] ’s document exceeds all other qualities under [MATH] on topic K; thus, |
[EQUATION] In addition, [MATH] holds since [MATH] is the improving author; hence, with Equation ( 26 ) we get [EQUATION] Let [MATH] denote the topic that author [MATH] is writing on under [MATH] , i.e [MATH] . By definition of [MATH] we obtain |
[EQUATION] Recall that [MATH] ; hence, Equations ( 27 ) and ( 28 ) suggest that [MATH] holds, and therefore we are guaranteed that [MATH] |
Since [MATH] , the induction hypothesis hints that [MATH] ; therefore, [MATH] holds and by Proposition we get that [MATH] .Notice that [MATH] is a finite improvement path, and that the condition of Proposition holds; thus, by invoking it for [MATH] , we get |
[EQUATION] Together with Equation ( 27 ), we conclude that [EQUATION] Next, we wish to find an improvement step such that the improving author’s utility strictly bounds the right-hand-side of Equation 29 ). Since [MATH] and [MATH] we get that [MATH] . In addition from the induction hypothesis, we get that for every imp... |
and [EQUATION] Since [MATH] is the improving author [MATH] holds, which together with Equation ( 30 ) implies [EQUATION] Let [MATH] . By definition of [MATH] , we know that |
[EQUATION] Observe that [MATH] must hold. To see this, assume otherwise that [MATH] , and [MATH] follows. Incorporating this assumption with Equations ( 29 ),( 31 ) and ( 32 ) we obtain |
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