in Mobile In-App Advertising

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1 Online Appendices to Not Just a Fad: Optimal Sequencing in Mobile In-App Advertising Appendix A: Proofs of echnical Results Proof of heorem : For ease of exposition, we will use a concept of dummy ads in this proof. Let D a, k : a A, k, 2,..., K a }. Each element d D is a dummy ad. hat is, each exposure of an ad a A is treated as a distinct dummy ad. o connect the concept of dummy ads with the ads discussed in the main paper, let ad and kd represent the first and the second components of dummy ad d, respectively. hen, the revenue-per-click of dummy ad d is α d α ad, and the probability of a click on dummy ad d is p d p ad kd. In other words, dummy ad d a, k has the same revenue-per-click and the probability of click as those of the k th exposure of ad a A. Using the notion of dummy ads and the assumption that p a k is decreasing in k, our proposed } policy can be rewritten as follows: In time slot t, display ad d t arg max αd p d, where D λ+ λp D d D d t and D t+ D t \d t }. Let denote the ad sequence resulting from this policy. It is easy to verify that, in the sequence, we have α d t p dt λ+ λp dt α d t+ p dt+ λ+ λp dt+ for all t, 2,..., a K a }. Note that our policy respects the implicit ordering constraint that dummy ad a, k must be allocated before a, k +. o see this, consider two dummy ads m a, k and n a, k +. We want to show that, in our policy, α mp m λ+ λp m dummy ad n. Since α m α n α a, we need we have p m p n. herefore, αnpn λ+ λp n, which implies that dummy ad m will be displayed before p m λ+ λp m pn λ+ λp n. Since p m p a k and p n p a k +, λp m λp n λp m + λp m p n λp n + λp m p n p m λ + λp n p n λ + λp m p m λ + λp m p n λ + λp n.

2 Consider an optimal ad sequence that is different from. We will show that the expected revenue generated from the sequence is no less than that of the sequence, thus establishing the optimality of. Let u denote the earliest time slot in which the two sequences differ from each other, and let ζ d u denote the ad that is placed in slot u by the sequence. Since u is the earliest time slot in which the two sequences differ, ad ζ appears in a time slot that is later than u in sequence ; let s > u denote this time slot in sequence. Next, we will show that by switching the two ads placed in time slots s and s of the sequence, the expected revenue of the new sequence is the same as that of the sequence. For the sequence, we know that α ζ p ζ λ + λp ζ α d t p dt λ + λp dt for all t > u. Let µ denote the ad that is placed in slot s of the sequence. hus, we have α ζ p ζ λ + λp ζ α µp µ λ + λp µ. Let denote the sequence obtained after switching ads µ and ζ in the sequence, and let b i denote the ad that is placed in slot i of the sequence. Note that µ b s and ζ b s. he difference in the expected revenues of and i.e., the expected revenue of minus that of is R R α µ p µ λ p bi + α ζ p λs ζ p bi p µ i i [ ] α ζ p λ ζ p bi + α µ p µ λs p bi p ζ i i λ p bi [ α µ p µ + α ζ p ζ λ pµ α ζ p ζ α µ p µ λ pζ ] i λ p bi [ ] α µ p µ λ + λpζ αζ p ζ λ + λpµ i λ p bi [ ] α µ p µ λ + λpζ αζ p ζ λ + λpµ 0, i λ p bi [ ] α ζ p ζ λ + λpµ αζ p ζ λ + λpµ i where the inequality follows from. hus, since is an optimal sequence, so is the sequence. If the sequence is the same as the sequence, we are done. Otherwise, we repeat the same argument to eventually obtain after a finite number of such adjacent-ad switches. Since the 2

3 expected revenue does not reduce throughout this process, the optimality of follows. Proof of Lemma : Our lower bound on the expected revenue generated from an arbitrary sequence is LB i where p i δ i γk i i for i, 2,...}. i αi p i β λ i j p j, Consider any positive integer. Let LB denote the value generated by the first time slots of LB. hat is, LB i αi p ψ i i i j p j, where ψ β λ. Observe that this expression is of the same form as 3, with ψ taking the place of λ. hus, by an application of heorem, we can conclude that the policy which displays ad } α a t arg max ap ak at+ in every period t, 2,..., } maximizes LB. Notice that a β λ+β λp ak at+ the definition of the sequence of ads to display under this policy does not depend on. In other words, the rule for picking an ad in a period can be used to construct an infinite ad sequence, which we denote by. Since ψ 0,, it is easy to verify from the expressions for LB and LB that max LB max LB + ψ + ψ LB ψ, since maximizes LB. Now, for any given ɛ > 0, we can pick ɛ large enough such that ψ ɛ ɛ. hus, we have LB LB ɛ ɛ max LB. Since the choice of ɛ was arbitrary and weak inequalities are preserved in the limit, we have LB max LB, which implies that LB max LB, as required. Proof of Lemma 2: Our upper bound on the expected revenue generated from an arbitrary sequence is UB αi δ i γk i i i β i λi δ min β j γ j i j i where ϕ j λβ δ min β j γ j for j, 2,...}. i αi δ i i γk i j ϕ j, 3

4 We will show that the sequence resulting from the proposed policy maximizes the upper bound. We again use the notion of dummy ads. Let D a, k : a A, k, 2,...}. Each element d D is a dummy ad. hat is, each exposure of an ad a A is treated as a distinct dummy ad. Let ad and kd represent the first and the second components of dummy ad d, respectively. Let α d α ad and p d δ ad γ kd. hen, the revenue-per-click of dummy ad d is α d, and the probability of a click on dummy ad d if it is shown in time slot t is p d β t. Let w d d j ϕ j. he optimization problem to find a sequence that maximizes the upper bound is } max UB } max α d p d w d, 2 where α d and p d d are ad-specific parameters of the dummy ad shown in time slot d by the sequence. Notice that w d only depends on the index d of the time slot and not on the dummy ad placed in that time slot. Using the notion of dummy ads, the proposed policy can be rewritten as follows: In time slot t, display dummy ad d t arg maxα d p d }, where D D and D t+ D t \d t }. Since w d decreases d D t with d, it is immediate that the sequence generated by this policy solves 2. Proof of heorem 2: Let and denote the sequences that optimize the lower and upper bounds, respectively. Let LB and UB denote the values from the first time slots of the expressions for LB and UB, respectively. hat is, LB UB αi δ i γk i i i β λ i i j αi δ i γk i i i β i λi i j δ j γk j j and δ min β j γ j. It is easy to verify that LB LB 0 and UB UB 0 for any arbitrary 0. Since δ max max a δ a }, we have LB UB LB since LB LB UB i αi δ i γk i i β λ i i j δj γk j j i αi δ i γk i i β i λ i i j δ minβ j γ j i α i δ i γk i i β i λi i j i αi δ i γk i i βi λ i δ j γk j j 4

5 i α i δ i γk i i β i λi i αi δ + i2 [ i j i γk i i βi λ i δ i + δj γk j j i2 j δ max i since γ 0, ] i δ max δ max, δ max δ max ] j γk j j where inequality follows from Chebyshev s sum inequality. Let E Rev and OP denote the expected revenue generated by sequence for the first time slots and the highest expected revenue that can be generated from the first time slots in problem 6, respectively. Using the relationships LB < E Rev < OP < UB, we have E Rev OP LB UB LB UB δ max. δ max Let OP denote the optimum objective value of problem 6. Next, we will establish an upper bound on OP. Since p a k, t δ a β t γ k decreases with k and t, the expected revenue generated from time slot i + to time slot i +, for i, 2,...}, is no more than OP. herefore, an upper bound on OP is U OP + λ + λ 2 + OP. Since U OP, we have λ herefore, OP OP λ. hus, we have E Rev OP OP OP. 3 λ E Rev OP Using E Rev E Rev, we have E Rev OP herefore, ERev OP max,2,3,...} δ max δ max δ max δ max OP OP δ max δ max λ. λ for any, 2, 3,...}. 4 λ }. Proof of Corollary : Consider 6, λ 0.5 and δ max 0.0 in 4. We have, E Rev OP

6 δ Since the guarantee offered by the policy in heorem 2 is max max,2,3,...} δ max λ }, the result follows. Proof of heorem 3: Using an argument similar to that in Section 6., we have the following lower and upper bounds on the expected revenue of an arbitrary sequence : LB αi δ i γk i i i β i λ j δj j γk j, and UB i i j αi δ i γk i i i β i j λ j δ min β j γ j. Let 0 denote the sequence generated by our heuristic policy. Consider a sequence, say 0, that maximizes LB. Every ad occurs infinitely often in, for otherwise it is easy to obtain a contradiction to the optimality of. Let u denote the earliest time slot in which the sequence 0 differs from the sequence and let ς denote the ad that is placed in slot u by the sequence 0. Let s > u denote the earliest time slot in which the sequence displays ad ς. Next, we will prove that by switching the two ads placed in time slots s and s of the sequence, we obtain a sequence, say 2, such that LB 2 LB. hus, since maximizes LB, so does 2. o proceed with the proof, let µ denote the ad that is placed in slot s of the sequence. Since ad ς is placed no later than ad µ in the sequence 0, we have α ς δ ς γ k 0 ς u βλ u + βλ u δ ς γ k 0 ς u α µ δ µ γ k 0 µ u βλ u + βλ u δ µ γ. 5 k 0 µ u As defined earlier, 2 is the sequence obtained after switching ads µ and ς in sequence. Let b i denote the ad that is placed in slot i of sequence. Note that b s µ and b s ς. Let k µ k µ s k 2 µ s and k ς k ς s k 2 ς s. hus, LB LB 2 α µδ µγ kµ β λ i δ bi γ k b i i + α ςδ ςγ kς β s λ i δ bi γ k b i i λ s δ µγ kµ i [ ] α ςδ ςγ kς β λ i δ bi γ k b i i + α µδ µγ kµ β s λ i δ bi γ k b i i λ s δ ςγ kς i β λ i δ bi γ k b i i [ α µδ µγ kµ + α ςδ ςγ kς β λ s δ µγ kµ α ςδ ςγ kς α µδ µγ kµ β λ s δ ςγ kς ] i β λ i δ bi γ k b i i [ α µδ µγ kµ β λ s δ ςγ kς α ςδ ςγ kς β λ s δ µγ kµ ] i β λ i δ bi γ k b i i [ α µδ µγ kµ β λ s + β λ s δ ςγ kς α ςδ ςγ kς β λ s + β λ kµ] s δ µγ i i i 6

7 0, β λ i δ bi γ k b i i [ α ςδ ςγ kς β λ s + β λ kµ s δ µγ α ςδ ςγ kς β λ s + β λ kµ] s δ µγ i where the inequality follows from assumption * and 5. hus, LB 2 LB. By repeating this adjacent switch argument a finite number of times, we eventually obtain a sequence s that maximizes LB with the property that our original sequence 0 and the sequence s are identical in the first s time slots. In this manner, for any finite integer > 0, we can obtain a sequence that maximizes LB and is identical to 0 in the first time slots. Consequently, for any given ɛ > 0, we can pick ɛ large enough and obtain a sequence ɛ that maximizes LB and is identical to 0 in the first ɛ time slots and satisfies LB ɛ ɛ LB ɛ LB 0 ɛ LB ɛ ɛ. hus, we have LB ɛ ɛ LB ɛ ɛ. Since the choice of ɛ was arbitrary and weak inequalities are preserved in the limit, we have LB 0 max LB, which implies that the sequence 0 maximizes the lower bound LB. Following the same argument in the proof of Lemma 2, we can show that the following sequence maximizes the upper bound: In time slot t, display ad a t arg maxα a δ a γ kat }. Let denote a the sequence that maximizes the upper bound. hus, we have UB max UB }. Let LB and UB denote the values from the first time slots of the expressions for LB and UB, respectively. hus, we have LB 0 UB LB UB i αi δ i γk i i β i i λ i j j j δ j γ k j j i αi δ i γk i i β i i λ i j j j δ minβ j γ j i αi δ i γk i i β i i λ i j j j δ j γ k j j i α i δ i αi δ i γk i i β i i i i β i i λ j j i γk λ j j ] δ j γ k j j + [ i i2 j i αi δ i γk i i β i i λ j j 7

8 + i δ j γ k j j i2 j δ max i i δ max δ max, δ max δ max where inequality follows from Chebyshev s sum inequality. Let E Rev 0 and OP denote the expected revenue generated by sequence 0 for the first slots and the highest expected revenue that can be generated by the first slots in problem 8, respectively. Using the relationships LB 0 < E Rev 0 < OP < UB, we have ERev 0 OP LB 0 UB δ max δ max. Let OP denote the optimum objective function value of problem 8. We will establish an upper bound for OP. Since p a k, t δ a β t γ k decreases with k as well as t, and λ t decreases with t, the revenue generated from slot i + to slot i +, for i, 2,...}, is no more than OP. herefore, an upper bound on OP is 2 U OP + λ i + λ i + OP. i λi Since U OP, we have OP OP λ. herefore, i i i we have ERev 0 OP ERev 0 OP E Rev 0, we have E Rev 0 OP OP OP δ max δ max i δmax δ max i OP OP i λi. hus, i λi. Using E Rev 0 λ i for any, 2, 3,...}. 6 herefore, ERev 0 δ OP max max,2,3,...} δ max } i λi. Proof of Corollary 2: Consider 6, λ i 0.5 and δ max 0.0 in 6. We have, E Rev 0 OP Since the guarantee offered by the policy in heorem 3 is the result follows δ max max,2,3,...} δ max } i λi, Proof of heorem 4: Following the logic in the proof of heorem 3, it is easy to verify that the sequence generated by our heuristic policy maximizes the lower bound. Similarly, following 8

9 the logic in the proof of Lemma 2, we can verify that the following sequence maximizes the upper bound: In time slot t, display ad a t arg maxα a δ a + ρ a γ kat }. a Let and denote the sequences which maximize the lower and upper bounds, respectively. he remainder of the argument in this proof is similar to that in the proof of heorem 2. Let LB and UB denote the values from the first time slots of the expressions for LB and UB, respectively. hat is, and hus, we have LB UB i i α i δ i + ρ i γ k i i β λ i i α i δ i + ρ i γ k i i β i λi j i j δ j γk j j, δ min β j γ j. LB UB LB UB i α i δ i + ρ i i α i δ i + ρ i i [αi δ i + ρ i i α [ i αi δ i + ρ i + + i δ i2 j γ k i i β λ i i j γ k i i β i λ i i γ k i i β λ i ] i i δ i + ρ i j δ j j γk j j δ minβ j γ j δ j j γ k i i βi λ i γ k i i β λ i ] + i α j j γk j i δ max i2 j δ max i i δ max i δ i + ρ i j γk i2 [ i γ k i i βi λ i δ max δ max, δ max j δ j] j γk j where inequality follows from Chebyshev s sum inequality. Let E Rev and OP denote the expected revenue generated by sequence in the first slots and the highest expected revenue can be generated in the first slots in problem 9, respectively. Using the relationship LB < E Rev < OP < UB, we have 9

10 E Rev LB OP UB LB UB δ max δ max Let OP denote the optimal value of problem 9. Since p a k, t and e a k, t decrease with k and t, the expected revenue generated from time slot i + to slot i +, for i, 2,...}, is no more than OP. herefore, an upper bound on OP is U OP + λ + λ 2 + Using U OP, we have OP OP λ. herefore, hus, we have ERev OP ERev OP OP OP E Rev, we have E Rev OP herefore, ERev OP max,2,3,...} δ max δ max δ max δ max OP λ. OP δmax δ max OP λ.. λ. Using E Rev λ for any, 2, 3,...}. 7 λ }. Proof of Corollary 3: Consider 6, λ 0.5, and δ max 0.0 in 7. We have, E Rev OP Since the guarantee offered by the policy in heorem 4 is result follows δ max max,2,3,...} δ max λ }, the Proof of heorem 5: We first note the following lower and upper bounds on the expected revenue of an arbitrary ad display sequence for the unconstrained problem 6: LB αi δ i γk i i i β i λi δ max β j since δ max δ a a and γ <, i j UB αi δ i γk i i i β i λi δ min β j γ j since δ min δ a a and ki j j. i j Let denote the sequence generated in Lemma 2 for the unconstrained problem 6. Following the argument in the proof of Lemma 2, we can easily verify that UB max UB } and LB max LB }. Let LB and UB denote the values from the first time slots of the expressions for LB and UB, respectively. As shown in the proof of the heorem 2, we have E Rev, where E Rev and OP denote the expected OP LB UB δmax δ max revenue from the first time slots of sequence and the highest expected revenue from the first 0

11 time slots in problem 6, respectively. Since the sequence generated by the heuristic policy for the constrained problem 0 matches with the first N slots of the sequence, E Rev N is equal to the expected revenue generated from the heuristic policy for the constrained problem. hus, we have E Rev N LB N. Let OP resp., OP c denote the highest expected revenue OP N UB N δmaxn δ maxn for the unconstrained problem 6 resp., the constrained problem 0. Clearly, OP OP c. From inequality 3, we have OP c OP OP N. herefore, λ N we have E Rev N OP c E Rev N OP N OP N OP δmaxn δ maxn E Rev for all positive integer N, we have E Rev N OP c δ max δ max OP N OP c λ N. Since E λ for any, 2, 3,..., N}. λ N. hus, Rev N herefore, E Rev N δ OP c max max,2,...,n} δ max λ }.

12 Appendix B: Micro-Foundation of the Sojourn Decay Let p be the probability that a visitor belongs to the clicker type and p be the probability that a visitor belongs to the non-clicker type. Given that a clicker non-clicker has arrived at a certain time slot, let a resp. ε be the probability that she will click on an ad shown in that time slot. Clearly, a > ε. Let A he event of a click in time slot t, B he event of no click in time slots to t, C he visitor is of the clicker type, Ĉ he visitor is of the non-clicker type. herefore, P C p, P Ĉ p, P A BC a, P A BĈ ε. We next show that the probability that an ad shown in time slot t will be clicked, given that there is no click in the earlier t slots, decreases with t. In other words, we will show that P A B decreases with t. We have P A B P AB P B P ABC P B + P ABĈ P B P ABC P BC P BC P B + P ABĈ P BĈ P BĈ P B P A BC P BC P B a P BC P B + εp BĈ P B + P A BĈP BĈ P B P BC/P C P C P BĈ/P Ĉ P Ĉ a + ε P B P B P B CP C P B ĈP Ĉ a + ε P B P B a P B Cp P B P B Ĉ p + ε P B app B C + ε pp B Ĉ P B app B C + ε pp B Ĉ P BC + P BĈ 2

13 app B C + ε pp B Ĉ P BC/P C P C + P BĈ/P Ĉ P Ĉ app B C + ε pp B Ĉ P B CP C + P B ĈP Ĉ app B C + ε pp B Ĉ P B Cp + P B Ĉ p ap at + ε p ε t a t p + ε t p. Differentiate the above expression with respect to t, we have [ d ap a t + ε p ε t ] dt p a t + p ε t ap a t + ε p ε t [ a t p log a + ε t p log ε ] p a t + p ε t 2 + p a t + p ε t [ a t ap log a + ε t ε p log ε ] p a t + p ε t 2 ε p εt a t p log a + ap a t ε t p log ε p a t + p ε t 2 + p ε t a t ap log a + p a t ε t ε p log ε p a t + p ε t 2 ε t a t p p p a t + p ε t 2 [a ε log a + ε a log ε] ε t a t p p a p a t + p ε t 2 a ε log ε < 0, where the inequality follows from the assumption that a > ε. herefore, P A B decreases with t. In other words, there is sojourn decay. 3