Dynamic Assortment Optimization with a Multinomial Logit Choice Model and Capacity Constraint Paat Rusmevichientong * Zuo-Jun Max Shen † David B. Shmoys ‡ Cornell University UC Berkeley Cornell University April 7, 2008 Abstract The paper considers a stylized model of a dynamic assortment optimization problem, where given a limited capacity constraint, we must decide the assortment of products to offer to customers to maximize the profit. Our model is motivated by the problem faced by retailers of stocking products on a shelf with limited capacities and by the problem of placing a limited number of ads on a web page. We assume that each customer chooses to purchase the product (or to click on the ad) that maximizes her utility. We use the multinomial logit choice model to represent demand. However, we do not know the demand for each product. We can learn the demand distribution by offering different product assortments, observing resulting selections, and inferring the demand distribution from past selections and assortment decisions. We present an adaptive policy for joint parameter estimation and assortment optimization. To evaluate our proposed policy, we define a benchmark profit as the maximum expected profit that we can earn if we know the underlying demand distribution in advance. We show that the running average expected profit generated by our policy converges to the benchmark profit and establish its convergence rate. Numerical experiments based on sales data from an online retailer indicate that our policy performs well, generating over 90% of the optimal profit after less than two days of sales. 1. Motivation and Problem Formulation Companies have realized the importance of offering products that are tailored to the demand of customers in each region. For instance, Wal-mart stocks specific lines of clothes targeted exclu- sively to certain groups of customers (Zimmerman (2006)). Car manufacturers are well-known for * School of Operations Research and Information Engineering, Cornell University, Ithaca, NY 14853, USA. E-mail: [email protected]† Department of Industrial Engineering and Operations Research, University of California–Berkeley, 4129 Etchev- erry Hall, Berkeley, CA 94720, USA. E-mail: [email protected]‡ School of Operations Research and Information Engineering and Department of Computer Science, Cornell University, Ithaca, NY 14853, USA. E-mail: [email protected]1
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Dynamic Assortment Optimization with a Multinomial Logit
Choice Model and Capacity Constraint
Paat Rusmevichientong∗ Zuo-Jun Max Shen† David B. Shmoys‡
Cornell University UC Berkeley Cornell University
April 7, 2008
Abstract
The paper considers a stylized model of a dynamic assortment optimization problem, where
given a limited capacity constraint, we must decide the assortment of products to offer to
customers to maximize the profit. Our model is motivated by the problem faced by retailers
of stocking products on a shelf with limited capacities and by the problem of placing a limited
number of ads on a web page. We assume that each customer chooses to purchase the product
(or to click on the ad) that maximizes her utility. We use the multinomial logit choice model
to represent demand. However, we do not know the demand for each product. We can learn
the demand distribution by offering different product assortments, observing resulting selections,
and inferring the demand distribution from past selections and assortment decisions. We present
an adaptive policy for joint parameter estimation and assortment optimization. To evaluate our
proposed policy, we define a benchmark profit as the maximum expected profit that we can
earn if we know the underlying demand distribution in advance. We show that the running
average expected profit generated by our policy converges to the benchmark profit and establish
its convergence rate. Numerical experiments based on sales data from an online retailer indicate
that our policy performs well, generating over 90% of the optimal profit after less than two days
of sales.
1. Motivation and Problem Formulation
Companies have realized the importance of offering products that are tailored to the demand of
customers in each region. For instance, Wal-mart stocks specific lines of clothes targeted exclu-
sively to certain groups of customers (Zimmerman (2006)). Car manufacturers are well-known for∗School of Operations Research and Information Engineering, Cornell University, Ithaca, NY 14853, USA. E-mail:
[email protected]†Department of Industrial Engineering and Operations Research, University of California–Berkeley, 4129 Etchev-
erry Hall, Berkeley, CA 94720, USA. E-mail: [email protected]‡School of Operations Research and Information Engineering and Department of Computer Science, Cornell
Figure 6: Running average expected profit under the Adaptive Assortment algorithm.
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A. Proof of Proposition 3
The proof of Proposition 3 makes use of the following four lemmas which give us additional insight into theproperties of the sets generated in each iteration of the StaticMNL(v, w,C) algorithm. The first lemma(whose proof appears in Appendix A.1) provides a simple characterization of the ordering among intersectionpoints among any three lines.
Lemma A.1. Under Assumption 1, for all 0 ≤ i < j < k ≤ N , one of the following conditions must hold:either I(i, j) < I(i, k) < I(j, k) or I(j, k) < I(i, k) < I(i, j).
38
The result of Lemma A.1 allows us to establish the following characterization of the sequence of permuta-tions generated by the StaticMNL(v, w,C) algorithm. We note that given a permutation τ : {1, . . . , N} →{1, . . . , N}, x is adjacent to y under τ if
∣∣τ−1(x)− τ−1(y)∣∣ = 1. The proof of this lemma appears in Appendix
A.2.
Lemma A.2. Let⟨σ0, . . . , σK
⟩denote the sequence of permutations generated by the StaticMNL(v, w,C)
algorithm. Then, under Assumption 1, for each t = 1, 2, . . . ,K, either σt = σt−1 or σt is obtained bytransposing two adjacent items under σt−1.
The next lemma shows that if a set Gt (corresponding to the first C elements in the permutation σt) isdistinct from all the previous sets, then one of its elements must be strictly smaller. The proof of this resultappears in Appendix A.3.
Lemma A.3. Let⟨G0, . . . , GK
⟩denote the sequence of sets generated by the StaticMNL(v, w,C) algo-
rithm. Under Assumption 1, for each t = 1, 2, . . . ,K, if Gt 6= Gt−1, then Gt is obtained from Gt−1 bytransposition between jt = σt−1
C and it = σt−1C+1 with 0 < it < jt, that is, Gt =
(Gt−1 \ {jt}
)∪ {it} and∑
`∈Gt ` <∑`∈Gt−1 `.
The next lemma shows that the size of the assortments generated by the StaticMNL algorithm is non-increasing and establishes the number of distinct assortments with sizes less than C. The proof appears inAppendix A.4.
Lemma A.4. Let⟨A0, . . . , AK
⟩denote the sequence of assortments generated by the StaticMNL(v, w,C)
algorithm. Then, under Assumption 1, for each s < C, there is exactly one distinct assortment of size s.
We are now ready to give a proof of Proposition 3.
Proof. Let⟨A0, . . . , AK
⟩denote the sequence of assortments generated by the StaticMNL(v, w,C) algo-
rithm. By Lemma A.4, we know that there are exactly C−1 distinct non-empty assortments of size C−1 orless. Thus, it suffices to count the number of distinct assortments of size C. Note that if At is an assortmentof size C, it must be the case that At = Gt. Therefore, the number of distinct assortments of size C isbounded above by the number of distinct subsets among G0, G1, . . . , GK .
Under Assumption 1, we know that, among h1(·), . . . , hN (·), each of the(N2
)pairs of lines will intersect
each other. Moreover, by Assumption 1, we have that σ0 = (N,N − 1, . . . , 2, 1), corresponding to the orderof hi(λ)’s at λ = −∞, and σK = (1, 2, . . . , N − 1, N) for λ = +∞. Note that∑
= NC − 2(1 + 2 + · · ·+ C − 1)− C = NC − C(C − 1)− C = C(N − C).
By Lemma A.3, whenever Gt is distinct from Gt−1, the total value∑`∈Gt ` is strictly less than
∑`∈Gt−1 `.
Thus, in addition to G0, there can be at most C(N − C) distinct subsets of Gt’s. Therefore, the number ofdistinct subsets among G0, G1, . . . , GK is at most C(N − C) + 1. Thus, the maximum number of distinctnon-empty assortments is at most C(N −C) + 1 + (C − 1) = C(N −C + 1), which is the desired result.
A.1 Proof of Lemma A.1
39
I(i, k)
hj(·)−−Option2hj(·)−−Option1
Line : hk(·)
Line : hi(·)
I(i, j) I(i, j)
I(j, k)
I(j, k)
Figure 7: A geometric proof of Lemma A.1.
Proof. As shown in Figure 7, the proof of this lemma follows from a simple geometric intuition. By As-sumption 1, we have that vi < vj < vk, which implies that the line hi(·) will intersect with hk(·). Since vj isbetween vi and vk, the are two possible options for line hj(·) as shown in the two dash lines in Figure 7. Inboth cases, we observe that I(i, k) is always between I(i, j) and I(j, k), giving the desired results. A morealgebraic proof is also straightforward. We omit the details due to space constraints.
A.2 Proof of Lemma A.2
Proof. We will prove the lemma by induction on t. Consider the case when t = 1. By Assumption 1, weknow that σ0 = (N,N − 1, . . . , 2, 1). We want to show that either σ1 = σ0 or σ1 is obtained by transposingtwo adjacent products. Suppose that σ1 6= σ0. It then follows from the definition that σ1 is obtained fromσ0 by a transposition of i1 and j1, where I (i1, j1) corresponds to the smallest intersection point. Since thisis the first intersection point, it follows from Lemma A.1 that we must have j1 = i1 + 1, which is the desiredresult.
Now, suppose that the result holds for ` = 1, 2, . . . , t; that is, for each s = 1, . . . , t, σs is either equal toσs−1 or is obtained from σs−1 by transposing two adjacent products. We will now establish the result forσt+1 via proof by contradiction.
Suppose on the contrary that σt+1 6= σt and σt+1 is obtained by σt by transposing it and jt, which areNOT adjacent under σt. This implies there is another product m that is between it and jt under σt, thatis, it = σtu, m = σtv, and jt = σtw where either 1 ≤ u < v < w ≤ N or 1 ≤ w < v < u ≤ N .
We first claim that we cannot have 1 ≤ u < v < w ≤ N . The follows because we know from Assumption1(a) that σ0 = (N,N − 1, . . . , 2, 1). If 1 ≤ u < v < w ≤ N holds, this implies that we have σt =(. . . , it, . . . ,m, . . . , jt, . . .) . Since it < jt and the transpositions in all of the previous iterations involveadjacent items by induction, it must be the case that it and jt have switched places once before, that is, wemust have already encountered the intersection point I(it, jt) in the earlier iterations. Contradiction!
So, we have 1 ≤ w < v < u ≤ N , which implies that σt = (. . . , jt, . . . ,m, . . . , it, . . .) . By construction,we know that it < jt. Thus, there are three cases to consider for the value of m.
Case 1: m < it < jt. In this case, m is smaller than it but appears earlier in the ordering σt. Sinceσ0 = (N,N − 1, . . . , 2, 1) and all previous transpositions involve adjacent items by induction, it must be thecase that it and m interchange their positions in the earlier iteration, implying that I(m, it) < I(it, jt). It
40
thus follows from Lemma A.1 that
I(m, it) < I(m, jt) < I(it, jt),
implying that we must have already encountered the intersection point I(m, jt) before this iteration. Byinduction, this means that m and jt should have interchanged their places and we must have m before jt(note that m ≥ 1). This contradicts our definition of σt!
Case 2: it < m < jt. In this case, it follows from Lemma A.1 that either
We will show that either one of the above condition will lead to a contradiction. Suppose that I(it,m) <I(it, jt) < I(m, jt). This implies that we must have encountered the intersection point I(it,m) before thecurrent iteration. Therefore, by induction, m and it should have switched places and we must have m afterit in σt. This again contradicts our definition of σt! Now, suppose that I(m, jt) < I(it, jt) < I(it,m).Then, we must have encountered the intersection point I(m, jt) before the current iteration. Therefore, byinduction, m and jt should have switched places and we must have m before jt in σt. This again contradictsour definition of σt!
Case 3: it < jt < m. The proof for this case is similar to Case 1 and we omit the details.
Thus, all three cases lead to contradictions. Therefore, it must be the case that σt+1 is obtained fromσt by transposing two adjacent products. This completes the induction.
A.3 Proof of Lemma A.3
Proof. By definition, if Gt 6= Gt−1, it must be the case that we encounter an intersection point I (it, jt)with 0 < it < jt. In this case, σt is obtained from σt−1 by transposition of it and jt. Moreover, we knowfrom Lemma A.2 that it is adjacent to jt. Note that Gt and Gt−1 correspond to the first C elements for theordering σt and σt−1, respectively. Thus, in order for Gt to be different from Gt−1, it must be the case thateither 1) σt−1
C = jt and σt−1C+1 = it, or 2) σt−1
C = it and σt−1C+1 = jt.
We will first show that option 2) is not feasible. Recall that under Assumption 1, we have σ0 =(N,N − 1, . . . , 2, 1). Since it < jt and every transposition only occur between adjacent items (by LemmaA.2), if σt−1
C = it and σt−1C+1 = jt, then it must be the case that it and jt have interchanged places before; that
is, we have already encountered the intersection point I (it, jt) in the earlier iterations. This is a contradiction!Therefore, we can only have σt−1
C = jt and σt−1C+1 = it. In this case, we have Gt =
(Gt−1 \ {jt}
)∪{it}, which
implies that∑`∈Gt `−
∑`∈Gt−1 ` = it − jt < 0, which is the desired result.
A.4 Proof of Lemma A.4
Proof. By definition,∣∣A0∣∣ = C. Let θ denote the smallest index such that
∣∣Aθ∣∣ < C. Note that by definition,|As| = C for all 0 ≤ s < θ, which implies that As = Gs. Thus, by Lemma A.3, if As 6= As−1, we must haveAs =
(As−1 \ {jt}
)∪{it}. To complete the proof of Lemma A.4, it suffices to show the following two results:
(a) Aθ = Aθ−1 \ {jθ}(b) For any 1 ≤ t ≤ K, if
∣∣At−1∣∣ < C, then either At = At−1 or At = At−1 \ {jt}
We will first prove part (a). The set Aθ is created when we encounter the θth intersection point I (iθ, jθ).We claim that we must have iθ = 0. Suppose on the contrary that we have 0 < iθ < jθ. In this case,
41
Bθ = Bθ−1. Thus, for∣∣Aθ∣∣ < C =
∣∣Aθ−1∣∣, it must be the case that Gθ 6= Gθ−1. It follows from Lemma
A.3 that Gθ is obtained from Gθ−1 by transposition of jθ = σθ−1C and iθ = σθ−1
C+1 with 0 < iθ < jθ. Since∣∣Aθ∣∣ < C, it follows that Bθ 3 iθ, which implies that we must have already encountered the intersectionI (0, iθ) earlier. By Lemma A.1,
I (0, iθ) < I (0, jθ) < I (iθ, jθ) ,
which implies that the intersection point I(0, jθ) appeared before the current iteration, and thus, Bθ−1 3 jθ,which implies that
∣∣Aθ−1∣∣ < C. Contradiction!
Therefore, iθ = 0. Then, we know that σθ = σθ−1, Gθ = Gθ−1, and Bθ = Bθ−1 ∪ {jθ}. Since∣∣Aθ−1∣∣ = C >
∣∣Aθ∣∣, Gθ−1 ∩ Bθ−1 = ∅ and Gθ ∩ Bθ 6= ∅. Since only jθ is added to the set Bθ−1, we haveGθ ∩Bθ = {jθ}, and therefore, Aθ = Gθ \Bθ = Gθ \ {jθ} = Gθ−1 \ {jθ} = Aθ−1 \ {jθ}, which is the desiredresult.
To prove part (b), consider At−1 such that∣∣At−1
∣∣ = r < C for some r. Note that by the def-inition, σt−1 =
(σt−1
1 , . . . , σt−1N
)represent the ordering of the lines h1(·), . . . , hN (·) during the interval(
I (it−1, jj−1) , I (it, jt))
with
hσt−11
(I (it, jt)) ≥ hσt−12
(I (it, jt)) ≥ · · · ≥ hσt−1N
(I (it, jt)) (9)
Since Gt−1 ={σt−1
1 , . . . , σt−1C
}and
∣∣At−1∣∣ = r < C, σt−1
r+1 ∈ Bt−1, which implies that
hσt−1r
(I (it, jt)) ≥ 0 > hσt−1r+1
(I (it, jt)) ≥ · · · ≥ hσt−1N
(I (it, jt)) .
Since the functions hi(·)’s are strictly decreasing, we have Bt−1 ={σt−1r+1, σ
t−1r+2, . . . , σ
t−1N
}.
We will now show that either At = At−1 or At = At−1\{jt}. Consider the tth intersection point I (it, jt).There are two cases to consider: it = 0 and it ≥ 1. Suppose that it = 0. Then, we claim that jt = σt−1
r . SinceBt−1 =
{σt−1r+1, σ
t−1r+2, . . . , σ
t−1N
}, we know that jt ∈
{σt−1
1 , . . . , σt−1r
}. If, on the contrary, jt = σt−1
k wherek < r, this means that I
(0, σt−1
k
)< I
(0, σt−1
r
); that is, the value associated with the line hσt−1
rremains non-
negative. However, by the ordering in (9), it must be the case that 0 > hσt−1k
(I (it, jt)) ≥ hσt−1r
(I (it, jt)).Contradiction! Therefore, jt = σt−1
r , which implies that At = At−1 \{σt−1r
}, which is the desired result.
On the other hand, suppose that it ≥ 1. In this case, we have Bt = Bt−1. We claim that we musteither have {it, jt} ⊂ Bt or {it, jt} ⊂ (Bt)c; that is, either both elements are in Bt or both or not. Thiswill show that At = At−1, which is the desired result. To prove this, note that since 0 < it < jt, itfollows from Lemma A.1 that either I(0, it) < I(0, jt) < I(it, jt) or I(it, jt) < I(0, jt) < I(0, it). IfI(0, it) < I(0, jt) < I(it, jt) holds, then the intersection points I(0, it) and I(0, jt) appear in the earlieriterations, implying that {it, jt} ⊂ Bt. On the other hand, if I(it, jt) < I(0, jt) < I(0, it), then it /∈ Bt andjt /∈ Bt.
B. Proof of Proposition 4
Let us introduce the following notation that will be used throughout the rest of this section. Let the functiong : R→ R be defined by:
g(λ) = max
{∑`∈X
v` (w` − λ) : X ⊆ {1, 2, . . . , N} and |X| ≤ C
}.
The following lemma establishes important properties of the function g and its relationship with the optimalprofit Z∗.
42
Lemma B.1. Under Assumption 1, we have the following properties.
• The function g is piecewise linear, non-increasing, convex, and continuous.
• For any λ ∈ R, if λ ∈[I (it, jt) , I (it+1, jt+1)
)for some t = 0, . . . ,K, then g(λ) =
∑`∈At v` (w` − λ)
• Z∗ = max {λ : g(λ) ≥ λ}
Proof. Observe that g(λ) can be expressed as a solution of the following linear program:
g(λ) = max
{N∑`=1
z`v` (w` − λ) :N∑i=1
zi ≤ C and 0 ≤ zi ≤ 1 for all i
}
The first part of Lemma B.1 follows immediately from standard results in linear programming.
To establish the second part of the lemma, note that to compute g(λ), we would order the products in adescending order of hi(λ) = vi (wi − λ) and choose the first C product with non-negative values. By defini-tion, we know that within the interval [I (it, jt) , I (it+1, jt+1)), none of the N + 1 lines h0(·), h1(·), . . . , hN (·)intersects each other. Thus, for λ ∈ [I (it, jt) , I (it+1, jt+1)), the ordering of hi(λ) is given by σt and thefirst C items with non-negative value is simply At. This is exactly what we need to show.
The final part of Lemma B.1 follows from the following argument. By definition of the profit functionf , there exists X ⊂ {1, . . . , N} with |X| ≤ C such that f(X) ≥ λ if and only if
∑i∈X vi (wi − λ) ≥ λ,
which occurs if and only if maxA:|A|≤C∑i∈A vi (wi − λ) ≥ λ, which is equivalent to g(λ) ≥ λ. Therefore,
Z∗ = max {λ : g(λ) ≥ λ}
The next lemma provides an expression for the change in the profit for the sequence of assortmentsgenerated by the StaticMNL algorithm. For x ∈ R, we let
sign(x) =
+1, if x > 0,0, if x = 0,−1, if x < 0
Lemma B.2. Let⟨A0, . . . , AK
⟩denote the sequence of assortments generated by the StaticMNL(v, w,C)
algorithm. Then, under Assumption 1, for each t = 1, 2, . . . ,K, if At 6= At−1,
sign(f(At)− f
(At−1
))= sign (g (I (it, jt))− I (it, jt)) .
43
Proof. There are two cases to consider: |At| = C and |At| < C. Suppose that |At| = C. It then follows fromLemma A.4 that At =
(At−1 \ {jt}
)∪ {it} with 1 ≤ it < jt. Let X = At−1 \ {it, jt}. Then, we have
f(At)− f
(At−1
)=
∑`∈X w`v` + witvit
1 +∑`∈X v` + vit
−∑`∈X w`v` + wjtvjt
1 +∑`∈X v` + vjt
=
(1 +
∑`∈X v`
)(witvit − wjtvjt)−
(∑`∈X w`v`
)(vit − vjt) + (wit − wjt) vitvjt(
1 +∑`∈X v` + vit
) (1 +
∑`∈X v` + vjt
)=
(vit − vjt) ·{(
1 +∑`∈X v`
) (witvit−wjtvjt)vit−vjt
−(∑
`∈X w`v`)
+ (wit−wjt)vitvjt
vit−vjt
}(1 +
∑`∈X v` + vit
) (1 +
∑`∈X v` + vjt
)=
(vit − vjt) ·{(
1 +∑`∈X v`
)I (it, jt)−
(∑`∈X w`v`
)− hit (I (it, jt))
}(1 +
∑`∈X v` + vit
) (1 +
∑`∈X v` + vjt
)=− (vit − vjt) ·
{−I(it, jt) + hit (I (it, jt)) +
∑`∈X v` (w` − I (it, jt))
}(1 +
∑`∈X v` + vit
) (1 +
∑`∈X v` + vjt
)=− (vit − vjt) ·
{−I(it, jt) +
∑`∈At h` (I(it, jt))
}(1 +
∑`∈X v` + vit
) (1 +
∑`∈X v` + vjt
) ,
where the fourth equality follows from the fact that
I (it, jt) =(witvit − wjtvjt)
vit − vjtand hit (I (it, jt)) =
− (wit − wjt) vitvjtvit − vjt
Note that the denominator is always positive. Also, since it < jt, it follows from Assumption 1 that− (vit − vjt) is always positive. Thus, the sign of f (At)−f
(At−1
)is determined by−I(it, jt)+
∑`∈At h` (I(it, jt)),
which is equal to −I(it, jt) + g (I (it, jt)) by Lemma B.1. This is the desired result. A similar argumentapplies to the case when |At| < C.
We are now ready to give a proof of Proposition 4.
Proof. Let m∗ ∈ {0, 1, . . . ,K} denote the largest index such that
I (i1, j1) ≤ · · · ≤ I (im∗ , jm∗) ≤ Z∗ < I (im∗+1, jm∗+1) < · · · < I (iK , jK) .
Consider any t = 1, . . . ,m∗, if At 6= At−1, then it follows from Lemma B.2 that
sign(f(At)− f
(At−1
))= sign (g (I(it, jt))− I(it, jt)) ≥ 0,
where the last inequality follows from the fact that I (it, jt) ≤ Z∗ and Z∗ = max {λ : g(λ) ≥ λ}. Thisimplies that g (I(it, jt)) ≥ I(it, jt). Thus, the sequence 〈f (At) : t = 1, 2, . . . ,K〉 is non-decreasing for t =0, 1, . . . ,m∗.
On the other hand, consider any t ≥ m∗ + 1. In this case, we have I (it, jt) > Z∗, which implies thatg (I(it, jt)) < I(it, jt). Using the same arguments as above, we can show that
sign(f(At)− f
(At−1
))≤ 0,
and thus, the sequence 〈f (At) : t = 1, 2, . . . ,K〉 is non-increasing for t ≥ m∗ + 1. So, we have that
f(A0)≤ f
(A1)≤ · · · ≤ f
(Am
∗)
and f(Am
∗)≥ f
(Am
∗+1)≥ · · · ≥ f
(AK),
which is the desired result.
44
C. Proof of Theorem 5
The proof of Theorem 5 relies on the following series of lemmas. Let us first introduce the following notation.For each E ∈ E , let fE : ({0} ∪ E)× RC → R be defined as follows: for each ` ∈ {0} ∪ E and xE ∈ RC ,
fE(`, xE) = ln
(1
ex`
/ (1 +
∑k∈E e
xk
)) = ln
1
ex`
/(∑k∈{0}∪E e
xk
) ,
where we define x0 = 0. Recall that the utility U` that each customer assigns to the option ` is given byU` = µ∗`+ζ`, where µ∗` = 0 and ζ`’s are i.i.d. Gumbel random variables with mean zero and scaling parameterone. Let the random variable C (E) denote the option that has the maximum utility given the assortmentE; that is, C(E) = arg max`∈E U`. For each xE ∈ RC , let g (xE) be defined by
gE (xE) := E [fE (C(E), xE)] =∑
`∈{0}∪E
P {C(E) = `} fE(`, xE)
=∑
`∈{0}∪E
eµ`∑k∈{0}∪E e
µkln
(1
ex`
/∑k∈{0}∪E e
xk
)
The following standard lemma establishes a relationship between the true mean utilities µi’s and the functiong. For the proof, the reader is referred to Cover and Thomas (2006).
Lemma C.1. For each E ∈ E, gE (µE) = minxE∈RC gE(xE), and for each xE ∈ RC ,
∑`∈{0}∪E
∣∣∣∣∣ eµ`∑k∈{0}∪E e
µk− ex`∑
k∈{0}∪E exk
∣∣∣∣∣ ≤ 2√gE (xE)− gE (µE).
The next lemma provides an upper bound on the difference between E [fE (C(E), µ)] and E [fE (C(E), µ∗)]for each assortment E ∈ E . Before we proceed to the lemma, we need the following definition (see Haussler(1992) for more details).
Definition 1. For any T ⊂ Rd, T is full if there exists some translation of T that intersects all 2d orthantsof Rd, that is, there exists x ∈ Rd such that {sign(x+ y) : y ∈ T} = {0, 1}d where for each vector z ∈ Rd,sign(z) = (sign(z1), . . . , sign(zd)) and sign(·) is the usual sign function which is 1 if the argument is positiveand zero otherwise.
Definition 2. Let F be a family of functions from a set Z to R. For any sequence z = (z1, . . . , zd) of pointsin Z, let F|z = {(f(z1), . . . , f(zd)) : f ∈ F} ⊂ Rd. The pseudo-dimension of F is the largest d such thatthere exists a sequence z of d points in Z and F|z is full.
For our application, we will be interested in the collection of functions GE defined as follows:
GE ≡{f(·, xE) : {0, i, j} → R+
∣∣ xE ∈ RC ∩ [−M,M ]C},
where M is the constant given in Assumption 2. Note that each function in GE is indexed by a C-dimensionalvector xE . The following lemma shows that the pseudo-dimension of GE is at most C + 1.
Lemma C.2. For each E ∈ E, the pseudo-dimension of GE is at most C + 1.
45
Proof. To establish this result, it suffices to show that for any (z1, z2, . . . zC+2) ∈ ({0} ∪ E)C+2, the set ofvectors {(
f(z1, xE), f(z2, xE), . . . , f(zC+2, xE),)
: xE ∈ RC ∩ [−M,M ]C}⊂ RC+2
cannot be full; that is, no translation of this set can intersect all orthants in RC+2. To prove this, note thatsince the set {0}∪E has C+2 elements, at least two of the zs’s must coincide. By relabeling if necessary, wecan assume without loss of generality that z1 = z2 = ` for some ` ∈ {0}∪E. However, since z1 = z2, we havethat f(z1, xE) = f(z2, xE) for all xE ∈ RC . Thus, the set
{(f(z1, xE), f(z2, xE)
): xE ∈ RC ∩ [−M,M ]C
}forms a line in the plane and no translation of this set can possibly intersect all four orthants in R2. Thus,the set cannot be full.
The following result follows from Theorem 11 in Haussler (1992).
Lemma C.3. Let F be a collection of functions from Z into [0, B] with pseudo-dimension d. Let W1,W2, . . . ,WT
be T independent and identically distributed random variables drawn according to any distribution from theset Z. Then, for each ε > 0,
Pr
{supf∈F
∣∣∣∣∣EW [f(W )]− 1T
T∑i=1
f(Wi)
∣∣∣∣∣ ≥ ε}≤ 8
(32eBε
ln32eBε
)de−ε
2T/64B2.
where W denotes a random variable having the same distribution as Wi’s.
We will now apply Lemma C.3 to our problem and establish the following result.
Lemma C.4. Under Assumption 2, for each E ∈ E and ε > 0,
Pr {gE (µE)− gE (µE) ≥ ε} ≤ 8(
64eBε
ln64eBε
)C+1
e−ε2T/256B2
,
where B = 2M + ln(C + 1) and the random variable µE denotes the maximum likelihood estimate definedin Equation (5) based on a sample of T customers, and the probability is taken with respect to the randomvariables C1(E), . . . , CT (E).
Proof. Let E ∈ E be given. To apply the result of Lemma C.3, we will consider the following collection offunctions
GE ≡{fE(·, xE) : {0} ∪ E → R+
∣∣ xE ∈ RC ∩ [−M,M ]C}.
Note that since xE ∈ RC ∩ [−M,M ]C , it follows that for each ` ∈ {0} ∪ E,
fE (`, xE) = ln
∑k∈{0}∪E
exk−x`
≤ ln(1 + Ce2M
)≤ ln
(e2M (C + 1)
)= 2M + ln(C + 1).
46
Note that
gE (µE)− g (µE) =
(E [fE(C(E), µE)]− 1
T
T∑t=1
fE (Ct(E), µE)
)
+
(1T
T∑t=1
fE (Ct(E), µE)− 1T
T∑t=1
fE (Ct(E), µE)
)
+
(1T
T∑t=1
fE (Ct, µE)− E [fE (C(E), µE)]
)
≤
(E [fE (C(E), µE)]− 1
T
T∑t=1
fE (Ct(E), µE)
)
+
(1T
T∑t=1
fE (Ct(E), µE)− E [fE (C(E), µE)]
)
≤ 2 supxE∈RC∩[−M,M ]C
∣∣∣∣∣E [fE (C(E), xE)]− 1T
T∑t=1
fE (Ct(E), xE)
∣∣∣∣∣ ,where the first inequality follows from Equation (5), which implies that
1T
T∑t=1
f (Ct(E), µE) = minxE∈RC∩[−M,M ]C
1T
T∑t=1
f (Ct(E), xE) .
The final inequality follows from the fact that both µE and µE are in RC ∩ [−M,M ]C . Therefore,
Pr {gE (µE)− gE (µE) ≥ ε}
≤ Pr
{sup
xE∈RC∩[−M,M ]2
∣∣∣∣∣E [fE (C(E), xE)]− 1T
T∑t=1
f (Ct(E), xE)
∣∣∣∣∣ ≥ ε
2
}
≤ 8(
64eBε
ln64eBε
)C+1
e−ε2T/256B2
,
where B = 2M + ln(C + 1). The last inequality follows from Lemma C.3 and the fact that the collection offunctions G has a pseudo-dimension of at most C + 1.
Finally, here is the proof of Theorem 5.
Proof. It follows from Lemma C.1 that
P
∑k∈{0}∪E
∣∣∣θk (E)− θk (E)∣∣∣ ≥ ε
≤ Pr{gE (µE)− g (µE) ≥ ε2
4
}≤ 8
(256eBε2
ln256eBε2
)C+1
e−ε4T/4096B2
where the last inequality follows from Lemma C.4.
D. Proof of Theorem 6
The proof of Theorem 6 makes use of the following result. Recall that
η = minE∈E
min{i,j}∈E:i 6=j
|θi(E)− θj(E)|
Under Assumption 1, we know that η > 0.
47
Lemma D.1. For each E ∈ E, if maxk∈E∣∣∣θk(E)− θk(E)
∣∣∣ ≤ η/3, then
maxi∈E,j∈E:i 6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ ≤ 6 max{wi, wj}
η2
∑k∈E
∣∣∣θk(E)− θk(E)∣∣∣
Proof. Let i ∈ E and j ∈ E be given. Since maxk∈E∣∣∣θk(E)− θk(E)
∣∣∣ ≤ η/3, it follows from the definition of
η that θi(E) 6= θj(E). Thus, by definition,∣∣∣I(i, j)− IE(i, j)∣∣∣ =
∣∣∣∣∣θi (E)wi − θj (E)wjθi (E)− θj (E)
− θi (E)wi − θj (E)wjθi (E)− θj (E)
∣∣∣∣∣≤ wi
∣∣∣∣∣ θi (E)θi (E)− θj (E)
− θi (E)
θi (E)− θj (E)
∣∣∣∣∣+ wj
∣∣∣∣∣ θj (E)θi (E)− θj (E)
− θj (E)
θi (E)− θj (E)
∣∣∣∣∣=
wi
∣∣∣θi (E) θj (E)− θi (E) θj (E)∣∣∣
|θi (E)− θj (E)| ·∣∣∣θi (E)− θj (E)
∣∣∣ +wj
∣∣∣θj (E) θi (E)− θj (E) θi (E)∣∣∣
|θi (E)− θj (E)| ·∣∣∣θi (E)− θj (E)
∣∣∣≤
2 max{wi, wj}(∣∣∣θi (E)− θi (E)
∣∣∣+∣∣∣θj (E)− θj (E)
∣∣∣)|θi (E)− θj (E)| ·
∣∣∣θi (E)− θj (E)∣∣∣
≤6 max{wi, wj}
(∣∣∣θi (E)− θi (E)∣∣∣+∣∣∣θj (E)− θj (E)
∣∣∣)η2
where the last inequality follows from the fact that since∣∣∣θi (E)− θi (E)
∣∣∣ and∣∣∣θi (E)− θi (E)
∣∣∣ are at most
η/3,∣∣∣θi (E)− θj (E)
∣∣∣ ≥ η/3.
Here is the proof of Theorem 6.
Proof. Note that by our construction I(i,0) = I(i,0) = wi for all i. By the union bound,
P{
maxE∈E
{max
{i,j}⊆E:i6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ ,max
i∈E
∣∣∣θi (E)− θi (E)∣∣∣} > ε
}≤
∑E∈EP{
max{i,j}⊆E:i 6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ > ε OR max
i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > ε
}≤
∑E∈EP{
max{i,j}⊆E:i 6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ > ε OR max
i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > min{ε, η/3}
}
≤∑E∈EP
{∑i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > min
{ε,η
3,
εη2
6 max` w`
}}
≤∑E∈EP
{∑i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > εη2
6 max` w`
},
where the last inequality follows from the fact that max` w` ≥ 1. Note that the third inequality follows fromthe fact that the event
max{i,j}⊆E:i 6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ > ε OR max
i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > min{ε, η/3}
is a subset of the event ∑i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > min
{ε,η
3,
εη2
6 max` w`
}
48
To see this, note that if maxi∈E∣∣∣θi (E)− θi (E)
∣∣∣ ≤ min{ε, η/3} and max{i,j}⊆E:i6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ > ε,
then it follows from Lemma D.1 that
ε < max{i,j}⊆E:i 6=j
∣∣∣I(i, j)− IE(i, j)∣∣∣ ≤ 6 max` w`
η2
∑i∈E
∣∣∣θi (E)− θi (E)∣∣∣ ,
which gives the desired result.
From Theorem 5, we have that for each E ∈ E ,
P
{∑i∈E
∣∣∣θi (E)− θi (E)∣∣∣ > εη2
6 max` w`
}
≤ 8
(9216eB (max` w`)
2
ε2η4ln
9216eB (max` w`)2
ε2η4
)C+1
exp
{− ε4η8T
5308416B2 (max` w`)2
}
Since |E| ≤ 5(N/C)2, the desired result follows.
E. Proof of Proposition 8
Proof. Let r =(√
5− 1)/2 denote the golden ratio and let w∗ = max1≤`≤N w`. Since the classical Golden
Ratio search reduces the size of the target set by factor of r in each iteration, the maximum number ofiterations is at most dlnL/ ln(1/r)e. For k = 1, . . . , dlnL/ ln(1/r)e, let Gk denote the event that we makethe wrong decision in the kth iteration; that is, we made a decision that is different from we would have doneif we can evaluate the profit function f(·) exactly.
Consider the first iteration and suppose we are comparing assortments Ds and Dt for some s and t. Thecomparison is based on the sample average of WGRS samples for each assortment. Let Xs
1 , . . . , XsWGRS
denotethe observed profit associated with the selections of the WGRS customers who were offered the assortmentDs. Similarly, let Xt
1, . . . , XtWGRS
denote observed profits associated with Dt. By definition, we know thatE [Xs
i ] = f(Ds) and E [Xti ] = f(Dt) for all 1 ≤ i ≤ WGRS . Suppose that f(Ds) > f(Dt). Then, we can
bound the probability of error in the first iteration as follows
P {G1} = P
{1
WGRS
WGRS∑i=1
Xsi <
1WGRS
WGRS∑i=1
Xti
}
≤ P
{1
WGRS
WGRS∑i=1
(Xti − f(Dt)
)+
1WGRS
WGRS∑i=1
(f(Ds)−Xsi ) > f(Ds)− f(Dt)
}
≤ P
{1
WGRS
WGRS∑i=1
(Xti − f(Dt)
)+
1WGRS
WGRS∑i=1
(f(Ds)−Xsi ) > β
}
≤ P
{1
WGRS
WGRS∑i=1
(Xti − f(Dt)
)>β
2
}+ P
{1
WGRS
WGRS∑i=1
(f(Ds)−Xsi ) >
β
2
}
≤ 2 exp{−β
2WGRS
4w∗
},
where the last inequality follows from the classical Chernoff-Hoeffding Inequality (Hoeffding (1963)). Usinga similar argument, it is easy to verify that the probability of error P {Gk} in the kth iteration is boundedabove by 2k exp
{−β
2WGRS
4w∗
}.
49
To determine an upper bound on the regret after T periods, we note that the total number of samplesthat we used before we are left with a single assortment is at most
WGRS ·(⌈
lnLln(1/r)
⌉+ 3)≤ 4WGRS ·
⌈lnL
ln(1/r)
⌉,
where the factor of 3 reflects the fact that the standard Golden Ratio search starts with 4 points in thefirst iteration and adds one additional search point in subsequent iterations. The maximum expected regretincurred from these samples is at most 4w∗ ·WGRS ·
⌈lnL
ln(1/r)
⌉. Moreover, if we correctly identify the optimal
assortment at the end of the search, the regret will be zero thereafter. Thus, maximum expected regret afterwe concluded the search algorithm is at most
w∗T · P{GdlnL/ ln(1/r)e
}≤ 2w∗T
⌈lnL
ln(1/r)
⌉exp
{− β2WGRS
4 max` w`
},
where the right hand side follows from the upper bound on the error probability. Using the definition ofWGRS , we have the following upper bound on the expected regret after T periods:
4w∗ ·WGRS ·⌈
lnLln(1/r)
⌉+ 2w∗T
⌈lnL
ln(1/r)
⌉exp
{− β2WGRS
4 max` w`
}≤ 4w∗
(4w∗ lnTβ2
+ 1)(
lnLln(1/r)
+ 1)
+ 2w∗(
lnLln(1/r)
+ 1)
= 4w∗(
4w∗ lnTβ2
+ 1)(
ln(L/r)ln(1/r)
)+ 2w∗
(ln(L/r)ln(1/r)
)=
16(w∗)2 ln(L/r)β2 ln(1/r)
lnT + 6w∗ln(L/r)ln(1/r)
,
and the desired result follows from the fact that 1/r ≤ 2 and 1/ ln(1/r) ≤ 2.079.