A Revisit to the Markup Practice of Dynamic Pricing

Transcription

1 A Revisit to the Markup Practice of Dynamic Pricing Yifeng Liu and Jian Yang Department of Management Science and Information Systems Business School, Rutgers University Newark, NJ s: and March 2012; revised, December 2012, July 2013, March 2014 Abstract In this article, we focus on the markup practice of dynamic pricing in which a firm uses real-time inventory information to decide the most opportune time to raise its sales prices. In our model, demand arrives as a Poisson process in which the instantaneous rate is a product of two terms, one dependent on current price and the other on present time. Through a mixed use of mathematical induction and sample-path arguments, we establish the optimality of threshold policies. Under such policies, the firm would base its price-switching decision on a comparison made between the present time and the threshold level corresponding to the present price and inventory level. Feng and Xiao [5] have studied a stationary version of this problem along with the opposite markdown case. Though the earlier work has made dramatic advances in dynamic pricing and has pioneered many relevant techniques, we believe its treatment of the markup case warrants some revision. In particular, we find that the markup case is in possession of the erstwhileunknown complementarity between price flexibility and inventory, a property whose counterpart is not true for the markdown case. Our development also leads to an efficient policy-computing algorithm. Key words: Dynamic Pricing; Markup Practice; Threshold Policy 1

2 1 Introduction Dynamic pricing is mainly concerned with how a firm can maximize its revenue by charging the right prices over a fixed time horizon, given that demand arrives in a random and yet price-sensitive fashion. This paper focuses on the case where the concerned firm is in the markup mode, so that it continuously charges a sequence of progressively raised prices. Airlines and hotels often practice markup to sell tickets to exploit the difference in cost sensitivity between leisure and business travelers. Obviously, a firm may also exercise markdown, or, when there is no predetermined direction in which prices can go, reversible pricing. In this study, we analyze a firm that faces Poisson demand and sell a finite initial inventory over a fixed time horizon. It is allowed to charge prices from a given set { p k k = 0, 1,..., K} in the increasing order p 0, p 1,..., p K over time. The Poisson arrival rate of demand follows the product form λ k (t) = ᾱ k β(t): the overarching time-dependent term β(t) spells out demand s oscillation over time, and the actual arrival rate is proportional to both this term and the price-dependent factor ᾱ k. We establish the optimality of a threshold policy, with the latter representable by time points τn k that depend on both prices k and inventory levels n and satisfy τm k τn k for m n. The firm should start to charge the next higher price when the decline of its inventory level has pushed the threshold point associated with the current price and inventory level behind the current time. The above can be viewed as a generalization of the markup portion of Feng and Xiao s [5] result concerning the price-sensitive but time-independent demand pattern. Using almost symmetric arguments, these authors also established the optimality of threshold policies for the opposite markdown practice, in which a firm charges prices in decreasing order over time. As an equally important contribution of ours, we point out that the symmetric treatment is problematic. For the markup case, we note the following two main points based on our own derivation: (i) The value function vn(t) k has decreasing differences between k and n (Proposition 3). (ii) The above point is crucial for the proof of the optimal policy, which was previously ignored in Feng and Xiao [5]. A counter example of a technical result in Feng and Xiao [5] is given in Appendix D. For the seemingly symmetric markdown case, the counterpart to (i), that vn(t) k has increasing differences between k and n, is neither true nor needed in the derivation involving its threshold policies. We also show that the threshold policy is not necessarily monotone in k; that is, the seller may prefer to skip certain price levels. In an online supplement, we work out details for the markdown case. There, efficient algorithms for computing threshold levels are developed as well. Now let us briefly recount earlier research on dynamic pricing. Gallego and van Ryzin [6] started treating dynamic pricing from the optimal control perspective. They examined the problem of dynamically pricing inventories with stationary, stochastic demands, and showed the optimality of a monotone pricing policy as well as the asymptotic optimality of fixed-price heuristics. Dealing 2

3 with a similar model with a finite number of price choices, Feng and Xiao [4] justified the earlier assumption that the revenue-generation rate increase and be concave in the demand arrival rate. They derived the optimality of a comparable monotone pricing policy and offered a procedure for computing this policy. When demand is time-varying, Zhao and Zheng [8] identified the decline over time of customers willingness to pay premiums as a sufficient condition for the monotonicity of the optimal pricing policy. This result generalizes the claim made by Bitran and Mondschein [1] that a monotone policy would be optimal when the ratio of arrival rates at any two given prices remains the same over time. For a multi-product/resource version of the problem, Gallego and van Ryzin [7] gave asymptotically optimal heuristics that are based on the solution to the problem s deterministic control counterpart. Feng and Gallego [2] initiated irreversible pricing control with their study of a problem involving two price choices and an arrival pattern that is dependent on price only. For this problem, they showed the optimality of a threshold-like policy. Feng and Xiao [5] generalized the above result to a case with multiple price switches in the same direction. Feng and Gallego [3] moved on to the more general problem where both paths of prices and arrival rates can be time-varying, and demand may be dependent on sales-to-date. For this case, however, optimal policies were not shown to necessarily come from the threshold-like family. The current paper can be considered as both a generalization to the product-demand form and a correction to the markup portion of Feng and Xiao [5], which in both irreversible pricing cases arrived at the same conclusion on the threshold-like policy shape. Our derivation, however, sheds more light on the distinction between the markup and markdown cases. As a property, complementarity between price flexibility and inventory is not possessed by the markdown case; yet, it is essential to the derivation for the markup case. Feng and Xiao s [5] treatment of the markup case seems to have missed this point. Methodologically, we introduce a mixed use of mathematical induction and sample-path arguments to facilitate the derivation of the complementarity property. Additionally, we show that overall τn k is not decreasing in k for the two irreversible pricing cases, though for the reversible pricing case this trend, known as time monotonicity, exists (Zhao and Zheng [8]). The rest of the note is organized as follows. In Section 2, we provide the problem s formulation. We propose a procedure for constructing value functions and threshold levels in Section 3, and prove the optimality of the thus constructed threshold policy in Section 4. We then discuss merits of our current approach in Section 5. The note is concluded in Section 6. We have relegated technical details to appendices in the end. More developments regarding the markdown case and algorithms, as well as numerical results are contained in the online supplement. 3

4 2 Formulation Over some time interval [0, T ], we suppose that the concerned firm is to sell N items of a product and that it is allowed to consecutively charge a sequence of K + 1 prices p 0, p 1,..., p K, where 0 < p 0 < p 1 < < p K. Demand comes to the firm as a time-varying Poisson process whose instantaneous rate λ k (t) depends on both the current price p k charged by the firm and the overall market condition at time t. We also suppose that the demand rate function is of the product form. We place strictly positive constants ᾱ 0, ᾱ 1,..., ᾱ K and strictly positive and continuous function β( ) on [0, T ], so that λ k (t) = ᾱ k β(t), k = 0, 1,..., K, t [0, T ]. The time multiplier β( ) may be a consequence of seasonality of the product being sold, the fact that the concerned product is newly introduced or is gradually phasing out of the market, or other factors such as predictable changes over time of the competitive firms aggregate behavior. The price multiplier ᾱ k indicates the relative attractiveness of the product under price p k. We assume that: (S1) for the revenue generation rates, p 0 ᾱ 0 > p 1 ᾱ 1 > > p K ᾱ K. (S1) lets lower prices be associated with higher revenue generation rates, since there would be no need to consider a lower price if it did not generate revenue faster than a higher price. In case prices p form a continuum and the arrival rate function is λ( ), the counterpart to (S1) would be (p λ(p)) < 0, which could be rewritten as λ (p) > 1. (1) λ(p)/p In economics terms, the assumption states that demand for the concerned product is elastic. Clearly, (S1) would lead to the weaker (S1 ) for the demand arrival rates, ᾱ 0 > ᾱ 1 > > ᾱ K. We take the view that potential buyers arrive at the rate of ᾱ 0 β(t) at time t, and among all those that arrive, only a ᾱ k /ᾱ 0 portion will purchase the product when the price being charged is p k. We now define λ, so that λ = ᾱ 0 sup β(t). (2) t [0,T ] By the continuity of β( ) and the compactness of [0, T ], we know that λ is a strictly positive and finite constant. By (S1 ), we know that λ is the highest instantaneous arrival rate that can ever be achieved. 4

5 We denote a threshold policy by τ = (τ k n k = 1, 2,..., K, n = 1, 2,..., N) ( N ) K, where N [0, T ] N is defined through N = {(τ 1, τ 2,..., τ N ) 0 τ N τ N 1 τ 1 T }. (3) For the current markup case, adopting such a policy entails that the firm should raise its price from p k 1 to p k when its inventory level drops to n ahead of the threshold level τ k n. Throughout, we define all random elements on a probability space (Ω, P, F) that is equipped with a filtration (F t t [0, T ]). Each price choice k is associated with a point process (N k (t) t 0) that has intensity process (ᾱ k β(t) t [0, T ]) and is adapted to the given filtration. For any s < t, the increment N k (t) N k (s) is independent of F s while for n = 0, 1, 2,..., P [N k (t) N k (s) = n] = exp( ᾱ k ˆβ(s, t)) (ᾱk ˆβ(s, t)) n. (4) n! In (4), we have taken the conventions 0 0 = 1 and ˆβ(s, t) = t s β(u) du. Let vn(t) k be the maximum revenue the firm can make in time interval [t, T ] when it starts the interval with price p k and inventory level n. As the highest price p K is the last price for the firm to charge, we have v K n (t) = p K E[(N K (T ) N K (t)) n]. (5) For k = K 1, K 2,..., 0, the firm still has the chance to increase its price to p k+1 when it is currently charging p k. Hence, vn(t) k = sup E[ p k {(N k (τ) N k (t)) n} + v k+1 (τ)], (6) (n N τ T (t) k (τ)+n k (t)) + where T (t) is the set of stopping times later than t. 3 A Constructive Procedure We define the infinitesimal generator Gn(t) k corresponding to price p k, inventory level n, and time t, so that when it is applied to a well-defined function vector u = (u n (t) n = 0, 1,..., N, t [0, T ]), Gn(t) k u = d t u n (t) + ᾱ k β(t) (u n 1 (t) u n (t)). (7) We think it is an abuse of notation to call the left-hand side G k u n (t), because knowing u n (t) at the particular n and t alone would not help one get to the right-hand side. For the case where the time multiplier β( ) is stationary, Theorem 1 of Feng and Xiao [5] offers sufficient conditions for a vector of functions to be the value functions vn(t). k But this result can be easily generalized for the case where β( ) is time-variant. In the same spirit of this theorem, we find the following. 5

6 Proposition 1 For any k = K 1, K 2,..., 0, a function vector u (u n (t) n = 0, 1,..., N, t [0, T ]) that is uniformly bounded and absolutely continuous in t for every n will be v k (v k n(t) n = 0, 1,..., N, t [0, T ]), if it satisfies the following: (i) u n (t) v k+1 n (t) for every n = 0, 1..., N and t [0, T ], (ii) u n (T ) = 0 for every n = 0, 1,..., N and u 0 (t) = 0 for every t [0, T ], (iii) for n = 1, 2,..., N and t [0, T ], u n (t) = v k+1 n (t) implies G k n(t) u + p k ᾱ k β(t) 0, (iv) for n = 1, 2,..., N and t [0, T ], u n (t) > v k+1 n (t) implies G k n(t) u + p k ᾱ k β(t) = 0. Proposition 1 offers a hint as to how the value functions v k n(t) and threshold levels τ k n can be constructed. First, due to (5) and the fact that v K 0 (t) = 0, we can establish all the vk n (t) values in an n-loop. Then, for any k = K 1, K 2,..., 0, suppose vn k+1 (t) is known for all n and t. We can now go through an n-loop to find all the vn(t) s. k First, we can let v0 k (t) = 0 as suggested by (ii) of the proposition. Second, suppose vn 1 k (t) is known for all t and some n = 1, 2,..., N. Then, we have vn(t k ) = 0 by (ii) of the proposition. Next, we may weave vn(t) k for ever smaller t values by solving the differential equation G k n(t) v k + p k ᾱ k β(t) = 0 as indicated by (iv) of the proposition. We stop at the t when vn(t) k is to sink below vn k+1 (t), which is not allowed by (i) of the proposition. For time t earlier than this t, which we mark as τn k+1, we let vn(t k ) be v k+1 (t ). According to (iii) of the proposition, we still need Gn(t k ) v k + p k ᾱ k β(t ) 0 for t < τn k+1 for the thus constructed vn( ) k to be the true value function. Nevertheless, let us proceed with the construction procedure thus outlined. Not knowing whether what shall be constructed are the true value functions, we call them u s instead of v s. Formally, here is the iterative procedure for constructing function vector u = (u k n(t) k = 0, 1,..., K, n = 0, 1,..., N, t [0, T ]) and point vector τ = (τ k n k = 1, 2,..., K, n = 1, 2,..., N). First, let Then, for n = 1, 2,..., N and t [0, T ], let u K n (t) = ᾱ K u k 0(t) = 0, k = K, K 1,..., 0, t [0, T ]. (8) T t β(s) ( p K + u K n 1(s)) exp( ᾱ K ˆβ(t, s)) ds. (9) Next, we go over an outer loop on k = K 1, K 2,..., 0 and an inner loop on n = 1, 2,..., N. At each k and n, first let Then, let u k n(t) = ᾱ k T t β(s) ( p k + u k n 1(s)) exp( ᾱ k ˆβ(t, s)) ds, t [0, T ]. (10) τ k+1 n = inf{t [0, T ] u k n(t) > u k+1 n (t)}, (11) with the understanding that τ k n = 0 when the concerned inequality is always true and τ k n = T when it is never true. Finally, let u k n(t) = u k+1 n (t), t [0, τ k+1 n ]. (12) 6

7 The procedure we have outlined mainly goes through a k-loop to weave out the value functions. This is similar to what Feng and Xiao [5] have done. It is interesting to note that the procedure has no use for the differentiation operator, as it might be bad for numerical implementation. This presents a contrast to Feng and Xiao s procedure, which in step k involves the differentiation of u k+1 n (t); see their Theorem 2. In obtaining the integral forms (9) and (10), we have relied on an ordinary differential equation (ODE) and its known solution. Given continuous functions a( ) and b( ) defined on the interval [0, t], consider the following differential equation: d s f(s) = b(s) a(s) f(s), s (0, t). (13) This ODE has a unique solution f( ), so that for any s [0, t], f(s) = f(0) exp( s 0 a(u)du) + s 0 b(u) exp( s u a(v)dv) du = f(t) exp( t s a(u)du) t s b(u) exp( u s a(v)dv) du. (14) 4 Optimality and k-monotonicity We now move on to show that the above procedure indeed leads to the true value functions and an optimal pricing policy. First, through a sample-path argument that is reminiscent of Zhao and Zheng s [8] treatment of the same property for reversible dynamic pricing, we show that the marginal value of any additional item is diminished at a higher inventory level. Proposition 2 For any fixed k = 0, 1,..., K and t [0, T ], the value function v k n(t) is concave in n. We can use this result and more involved sample-path analysis combined with mathematical induction to show the added benefit that more flexibility in pricing can bring to the firm when there are more items in stock. Proposition 3 For any fixed t [0, T ], the value function v k n(t) has decreasing differences between k and n. Proposition 3 would play a critical role in the proof of the next key result. Its proof boils down to an induction process that for n = 0, 1,..., N 1, shows vn+1(t) k vn(t) k vn+1 k+1 (t) vk+1 n (t), k = 0, 1,..., K 1, t [0, T ]. (15) When proving (15) for each individual n, we borrow the sample-path idea from Zhao and Zheng [8]. In addition, both Proposition 2 and the same induction hypothesis (15) at inventory levels strictly below n would find their use in the proof. We believe the property presented in Proposition 3 is 7

8 previously unknown; for instance, it is never mentioned in Feng and Xiao [5]. Interestingly, the seemingly symmetric markdown case does not include the counterpart to this property. Indeed, our computational study has shown that the markdown case does not hold complementarity between price flexibility and inventory. We further demonstrate the optimality of a threshold policy for the markup case. For convenience, we have let the yet undefined τn K+1 = 0 for n = 1, 2,..., N and vn K+1 (t) = u K n (t) for n = 0, 1,..., N and t [0, T ]. Theorem 1 The u and τ as constructed from (8) to (12) satisfy the following for k = K, K 1,..., 0: (a[k]) for any n = 1, 2,..., N, (G k n) + (τ k+1 n ) v k+1 + p k ᾱ k β(τ k+1 n ) 0; (b[k]) G k n(t) v k+1 /β(t) is increasing in t for n = 1, 2,..., N and t (0, T ); (c[k]) u k n(t) = v k n(t) for any n = 0, 1,..., N and t [0, T ]; (d[k]) for any n = 1, 2,..., N, we have vn(t) k = vn k+1 (t) and Gn(t) k v k + p k ᾱ k β(t) 0 for t (0, τn k+1 ), and Gn(t) k v k + p k ᾱ k β(t) = 0 for t (τn k+1, T ); (e[k]) τn k+1 is decreasing in n; (f[k]) on top of vn(t) k having decreasing differences between n and t, it is further true that d t v k n(t)/β(t) is decreasing in t for n = 1, 2,..., N and t (0, T ). As a consequence, τ provides an optimal policy for the firm; under this policy, the firm should switch to price p k+1 when its inventory level drops to n before time τ k+1 n while its price level is p k. In (a[k]) of Theorem 1, the operator (G k n) + (t) is merely G k n(t) defined in (7) with d t replaced by the right derivative d + t. The proof of Theorem 1 depends very much on the complementarity property provided by Proposition 3. As an induction process over k = K, K 1,..., 0, it uses the validity of (a[k + 1]) to (f[k + 1]) to show the validity of (a[k]) to (f[k]) for each k. When proving the essential property (e[k]), that τn k+1 is decreasing in n, one logical step we take is from that vn(t) k vn k+1 (t) > 0 to that vn+1 k (t) vk+1 n+1 (t) > 0. What facilitates this step is Proposition 3, that vn(t) k has decreasing differences between k and n. Finally, regarding the k-monotonicity of the optimal policy, we gather the following result directly from the definition (11) and Theorem 1. Proposition 4 Suppose τ k n < T for some k = 1, 2,..., K 1 and n = 1, 2,..., N. Then, we have τ k+1 n τ k n if and only if v k 1 n (t) v k n(t) > 0 would lead to v k n(t) v k+1 n (t) > 0. The condition specified above is almost that of v k n(t) being quasi-concave in k. However, as prices and rates are only regulated by assumption (S1), there is no reason to believe this would be the general case. Looking at it from another angle, we see that the triggering event for any price switch is quantum in nature, as it involves the departure of an entire item of which the firm has only a finite stock. Meanwhile, p k+1 might be only a tiny bit above p k, and ᾱ k+1 might be only a 8

9 tiny bit below ᾱ k, with p k+1 ᾱ k+1 < p k ᾱ k ensured. But after the selling of the one item at price p k, the now more confident firm might consider p k+1 not high enough for its more precious remaining stock and ᾱ k+1 not low enough to sufficiently slow sales. Naturally, it would seek the next higher price p k+2, or even higher prices. Indeed, as confirmed by one of our computational studies, the threshold level τn k for the markup case is not necessarily decreasing in k. Hence, this case may have a leapfrog phenomenon: when it is time to switch to the price level p k+1, it may also be the time to switch to the next higher price p k+2, and so on and so forth. Therefore, when it is time to make the price switch from p k, the ultimate target should be some p k + (k,n), where k + (k, n) is not necessarily k + 1. For each n = 1, 2,..., N, we can use the following iterative procedure to find ( k + (k, n) k = 0, 1,..., K 1): for k = K 1 down to 0 let l = k + 1; while l K 1 and τn l+1 τn k+1 do let l = k + (l, n); let k + (k, n) = l. 5 Discussion For the markdown case, we can again use a constructive procedure to attain an optimal threshold policy, which is not necessarily k-monotone. For the sake of brevity, we choose to present details in the online supplement. Our product-form demand pattern satisfies the sufficient condition specified in Zhao and Zheng [8] for a k-monotone (time monotone in that context) threshold policy to be optimal for the reversible-pricing case. We can therefore see that the irreversible- and reversiblepricing cases clearly differ on whether optimal policies possess the k-monotone property. More importantly, our study points out that Feng and Xiao s [5] treatment of the markup case has minor problems. Their Lemma 1 claimed that the increase of Gn k 1 (t) v k in t and n alone, without other benefits that might come from the vn( ) s k being truly value functions of a markup problem, would lead to optimal threshold levels τ k n that necessarily satisfy 0 τ k N τ k N 1 τ k 1 T. While its counterpart for the markdown case is correct, we show in Appendix D a counter example for the markup case. In other words, while their conclusions on both markup and markdown cases are verifiable, Feng and Xiao s [5] approach to the markup case needs to be rectified. Therefore, more than offering a generalization of their results, our work contributes to recording a correct derivation for the markup case, which is fundamentally asymmetric to that for the markdown case. In our approach, Proposition 3 has been set aside for the sole purpose of illustrating the complementarity between price flexibility and inventory, a property that is both previously unknown and not possessed by the seemingly symmetric markdown case. In addition, property (a[k]) 9

10 in Theorem 1, concerning right derivatives of the value functions, has no counterpart in earlier literature. Moreover, Feng and Xiao [5] used the increase of Gn k 1 (t) v k in n (their Lemmas 1 and 2), a property not necessarily found to be true by our computational test. Furthermore, our numerical tests confirmed the following: (a) the markdown case does not have complementarity between price flexibility and inventory, a property that is essential for the markup case; cases; (b) earlier treatment of the markup case needs corrections; (c) k-monotonicity is, in general, not true for the threshold policies of the irreversible-pricing (d) heeding the time-variability of β(t) helps reap huge benefits; and, (e) optimal policies for arrival patterns more general than the current product form are not necessarily threshold-like. For the sake of brevity, we choose to present these details in the online supplement. The validity of (a) and (b) highlights the asymmetry between the two irreversible-pricing cases and their needs for different treatments; the validity of (c) confirms that prices insufficiently distinguished from the current price might be skipped upon demand arrivals; the validity of (d) signifies the importance of the current extension to the product-form time-varying demand; finally, the validity of (e) warns that the threshold-like policy form should not be taken for granted. 6 Concluding Remarks We considered the markup case of dynamic pricing, where a firm has to charge an increasing sequence of predetermined prices. When the demand arrival pattern is of the product form, we found that a threshold policy can be optimal. Besides being a generalization of Feng and Xiao [5] on the same subject but concerning stationary demand, this paper corrects errors made in the earlier work. It relies on a complementarity property between price flexibility and inventory, an erstwhile-unknown property that sets the markup case apart from the markdown case. References [1] G. R. Bitran, and S. V. Mondschein, Periodic Pricing of Seasonal Products in Retailing, Management Science, 43 (1997), [2] Y. Feng, and G. Gallego, Optimal Starting Times for End-of-Season Sales and Optimal Stopping Times for Promotional Fares, Management Science, 41 (1995), [3] Y. Feng, and G. Gallego, Perishable Asset Revenue Management with Markovian Time Dependent Demand Intensities, Management Science, 46 (2000),

11 [4] Y. Feng, and B. Xiao, A Continuous-Time Yield Management Model with Multiple Prices and Reversible Price Changes, Management Science, 46 (2000a), [5] Y. Feng, and B. Xiao, Optimal Policies of Yield Management with Multiple Predetermined Prices, Operations Research, 48 (2000b), [6] G. Gallego, and G. J. van Ryzin, Optimal Dynamic Pricing of Inventories with Stochastic Demand over Finite Horizons, Management Science, 40 (1994), [7] G. Gallego, and G. J. van Ryzin, A Multiproduct Dynamic Pricing Problem and Its Applications to Network Yield Management, Operations Research, 45 (1997), [8] W. Zhao, and Y.-S. Zheng, Optimal Dynamic Pricing for Perishable Assets with Nonhomogeneous Demand, Management Science, 46 (2000),

12 Appendices A. Proof of Proposition 2: We can use a sample-path argument to prove that vn(t) k vk n 1 (t) + vk n+1 (t). (16) 2 We allow four pools of inventories, termed 1, 2, 1, and 2, to start at time t with the same price p k and experience the same sample path over the interval [t, T ]. These pools have different starting inventory levels and may exert different price controls, though. Pools 1 and 2 start with n + 1 and n 1 initial items and pools 1 and 2 with n items. Besides applying optimal time-increasing stopping-time pricing policies to pools 1 and 2, we apply the higher of the two prices for pools 1 and 2 to pool 1 and the lower of the two prices to pool 2, until the first moment, say s, when pool 1 is to have generated one more demand arrival than pool 1. After s, we let pool 1 follow pool 1 s decisions and pool 2 follow pool 2 s decisions. As both the minimum and maximum of two decreasing functions are decreasing functions themselves, pools 1 and 2 can be considered as adopting time-increasing stopping-time pricing policies as well. Suppose the moment ever occurred, i.e., s [t, T ). Then, it has already been shown by Zhao and Zheng that the total sales revenue made by pools 1 and 2 amounts to the same as that by pools 1 and 2. Suppose the moment never occurred, i.e., s = T. Then, it has been shown by Zhao and Zheng that pools 1 and 2 can generate as high a total sales revenue as pools 1 and 2. So regardless, on every sample path, pools 1 and 2 can generate as high a total revenue as pools 1 and 2. Thus, we have proved (16). B. Proof of Proposition 3: We use both induction and a sample-path approach to show (15) for any n = 0, 1,..., N 1. Let us first prove v k 1(t) v k 0(t) v k+1 1 (t) v k+1 0 (t), k = 0, 1,..., K 1, t [0, T ]. (17) For every possible k and t, we introduce four pools of inventories, 1, 2, 1, and 2, that experience identical sample paths. At time t, pools 1 and 1 are with price index k + 1, and pools 2 and 2 are with price index k. Also, pools 1 and 2 have one item, while pools 1 and 2 are out of stock. Apparently, pools 1 and 2 will continue to hold zero inventory. On the other hand, pool 2 can immediately raise its price to p k+1 and then match actions taken by pool 1. Hence, (17) is true. Now, for some n = 1, 2,...N 1, suppose it is true that, for any m = 0, 1,..., n 1, We now prove that vm+1(t) k vm(t) k vm+1 k+1 (t) vk+1 m (t), k = 0, 1,..., K 1, t [0, T ]. (18) vn+1(t) k vn(t) k vn+1 k+1 (t) vk+1 n (t), k = 0, 1,..., K 1, t [0, T ]. (19) 12

13 For any possible k and t, we rely on pools 1, 2, 1, and 2 that experience identical sample paths. At time t, pools 1 and 1 are with price index k + 1, and pools 2 and 2 are with price index k. Also, pools 1 and 2 both have n + 1 items, while pools 1 and 2 both have n items. For s [t, T ], let us use N i (s) to denote the inventory level of pool i at time s. For instance, we have N 1(t) = n and N 2(t) = n + 1. We let pools 1 and 2 execute their respective optimal decisions. For a certain period, we let pool 1 follow pool 1 s decisions and pool 2 follow pool 2 s decisions. Note that prices adopted by all pools are increasing over time. We let this certain period end at the first moment, say s [t, T ), when (a) pools 1 and 2 have reached the same price, (b) pools 2 and 2 have just experienced one more arrival than pools 1 and 1, or (c) pools 1 and 2 both have just one item left while pools 1 and 2 have no item left. We may denote the case where none of the above occurs by s = T. This is the case when by time T, at any moment the price taken by pool 2 has always been strictly lower than that taken by pool 1, yet all pools have admitted the same demand arrivals, and none of the pools have run out of stock. We caution that the opposite to (b) will not occur, since before its price catches up with that of pool 1, pool 2 always charges a strictly lower price than pool 1 and hence, by (S1 ), has more chance to realize sales. For all cases, pools 1 and 2 will have together generated the same revenue as pools 1 and 2 by time s. This also means that we are already done when s = T. When (a) ever occurs, we may let pools 1 and 2 both execute optimal decisions from time s on. Then, within the time interval [s, T ], pool 1 will behave exactly the same as pool 2, and pool 2 will behave exactly the same as pool 1. So, pools 1 and 2 will continue to together produce the same revenue as pools 1 and 2 do. When (b) ever occurs, denote the price taken by pool 1 at time s by k 1 and the price taken by pool 2 at time s by k 2. We have k 1 > k 2, and n + 1 = N 1 (t) N 1 (s) = N 1(s) + 1 = N 2(s) + 1 = N 2 (s) + 2. (20) Hence, from the induction hypothesis (18), we have But from Proposition 2, we also have Combining (21) and (22), we obtain v k 2 N 2(s) (s) vk 2 N 2 (s) (s) vk 1 N 2(s) (s) vk 1 N 2 (s)(s). (21) v k 1 N 2(s) (s) vk 1 N 2 (s) (s) vk 1 N 1 (s) (s) vk 1 N 1(s) (s). (22) v k 2 N 2(s) (s) vk 2 N 2 (s) (s) vk 1 N 1 (s) (s) vk 1 N 1(s) (s). (23) In view of the memorylessness property of the Poisson process, (23) means that, conditioned on the same sample path up to the provocation of (b), pools 1 and 2 will on average together earn more than pools 1 and 2 in the time interval [s, T ]. 13

14 When (c) ever occurs, denote the price taken by pool 1 at time s by k 1 and the price taken by pool 2 at time s by k 2. We have k 1 > k 2, and From the induction hypothesis (18), we have N 1 (s) = N 1(s) + 1 = N 2(s) = N 2 (s) + 1 = 1. (24) v k 2 N 2(s) (s) vk 2 N 2 (s) (s) vk 1 N 1 (s) (s) vk 1 N 1(s) (s). (25) In view of the memorylessness property of the Poisson process, (25) means that, conditioned on the same sample path up to the provocation of (c), pools 1 and 2 will on average together earn more than pools 1 and 2 in the time interval [s, T ]. In view of all the above, we see that (19) is true. process. Therefore, (15) is true. We have hence completed the induction C. Proof of Theorem 1: The proof has use of the following lemma, which was originated in Karlin (1968) and also used in Feng and Xiao [5]. Lemma 1 Let k > 0 and φ(t) = + t ρ(s) exp( k (s t)) ds. Then, φ(t) will be decreasing in t 0 if ρ(s) is decreasing in s 0. We prove the theorem through an induction on k. Let us focus on proving (a[k]) to (f[k]) first. Take n = 1, 2,..., N. From (5), we have n 1 vn K (t) = p K E[(N K (T ) N K (t)) n] = p K (n (n m) P [N K (t, T ) = m]), (26) which, by (4), amounts to v K n (t) = p K n p K exp( ᾱ K ˆβ(t, T )) Taking derivative over t, we find that, for t (0, T ), m=0 n 1 m=0 d t v K n (t) = p K ᾱ K β(t) exp( ᾱ K ˆβ(t, T )) Taking differences over n, we find that v K n (t) v K n 1(t) = p K (1 exp( ᾱ K ˆβ(t, T )) From (28) and (29), it can be checked that (n m) (ᾱ K ˆβ(t, T )) m. (27) m! n 1 m=0 n 1 m=0 (ᾱ K ˆβ(t, T )) m. (28) m! (ᾱ K ˆβ(t, T )) m ). (29) m! G K n (t) v K (t) + p K ᾱ K β(t) = 0, t (0, T ). (30) 14

15 Consulting (13) and (14), we obtain v K n (t) = ᾱ K T In view of the construction (9), we may see (c[k]), that t β(s) ( p K + v K n 1(s)) exp( ᾱ K ˆβ(t, s)) ds. (31) u K = (u K n (t) n = 0, 1,..., N, t [0, T ]) = v K = (v K n (t) n = 0, 1,..., N, t [0, T ]). (32) From (30), we may confirm (d[k]) with the understanding that τn K+1 = 0 for n = 1, 2,..., N. The convention for the τn K+1 s also leads directly (e[k]). By (30) and the convention on τn K+1 and vn K+1 (t), we may see that (a[k]) and (b[k]) are both true. To verify (f[k]), we can use the same method on (f[k]) for k = K 1, K 2,..., 0, which is presented near the end of this proof. Suppose for some k = K 1, K 2,..., 0, we have (a[k + 1]), that (G k+1 n ) + (τ k+2 n ) v k+2 + p k+1 ᾱ k+1 β(τ k+2 n ) 0, n = 1, 2,..., N, (33) (b[k + 1]), that G k+1 n (t) v k+2 /β(t) is increasing in t for n = 1, 2,..., N and t (0, T ), (c[k + 1]), that (d[k + 1]), that, for n = 1, 2,..., N, and u k+1 n (t) = v k+1 n (t), n = 0, 1,..., N, t [0, T ], (34) v k+1 n (t) = v k+2 n (t) and G k+1 n (t) v k+1 + p k+1 ᾱ k+1 β(t) 0, t (0, τ k+2 n ), (35) G k+1 n (t) v k+1 + p k+1 ᾱ k+1 β(t) = 0, t (τ k+2 n, T ). (36) (e[k+1]), that τn k+2 is decreasing in n, and (f[k+1]), that vn k+1 (t) has decreasing differences between n and t. Now we embark on showing (a[k]) to (f[k]). For n = 1, 2,..., N, by the definition of τ k+1 n through (11), we know that u k n(t) > v k+1 n (t), t (τ k+1 n, T ), (37) u k n(τ k+1 n ) = v k+1 n (τ k+1 n ), (38) and d + t uk n(τn k+1 ) d + t vk+1 n (τn k+1 ). (39) By (13) and (14), we may see that the construction (10) renders G k n(t) u k + p k ᾱ k β(t) = 0, t (τ k+1 n, T ). (40) Our construction through (10) to (12) also guarantees that u k n 1(τ k+1 n ) v k+1 15 k+1 n 1 (τ n ). (41)

16 Combining (38), (39), (40), and (41), we obtain Thus we have (a[k]). (G k n) + (τ k+1 n ) v k+1 + p k ᾱ k β(τ k+1 n ) 0. (42) Note that (a[k + 1]) means (33). This, (b[k + 1]), and (d[k + 1]) together lead to the fact that Gn k+1 (t) v k+2 /β(t) is increasing in t and below p k+1 ᾱ k+1. From the definition of u k+1 n (t) for t [0, τn k+2 ] through (12), (c[k + 1]), and (e[k + 1]), we may see that G k+1 n (t) v k+1 β(t) n (t) v k+2, t (0, τn k+2 ). (43) β(t) = Gk+1 Hence, in view of the above and (b[k + 1]) again, we may see that Gn k+1 (t) v k+1 /β(t) is increasing in t and below p k+1 ᾱ k+1 when t (0, τn k+2 ), and is flat at p k+1 ᾱ k+1 for t (τn k+2, T ). Now, note that G k n(t) v k+1 G k+1 n (t) v k+1 β(t) = (ᾱ k ᾱ k+1 ) (vn 1 k+1 (t) vk+1 n (t)), (44) which, by (S1 ) and (f[k + 1]), is increasing in t. This and the just proved result together lead to the increase of G k n(t) v k+1 /β(t) in t. Hence, we have (b[k]). From (a[k]) and (b[k]), we obtain, for n = 1, 2,..., N, G k n(t) v k+1 + p k ᾱ k β(t) 0, t (0, τ k+1 n ). (45) Our construction (12) and (c[k + 1]) dictate that u k n(t) = v k+1 n (t), t [0, τ k+1 n ]. (46) By its construction, u k n(t) is uniformly bounded by NλT ; it is also Lipschitz continuous in t with coefficient Nλ, and hence absolutely continuous in t. By (37), (40), (45), and (46), and as well as the fact that vn k+1 (T ) = 0 for every n, we may see that u k (u k n(t) n = 0, 1,..., N, t [0, T ]) satisfies the sufficient conditions (i) to (iv) stipulated in Proposition 1. Hence, we have shown (c[k]), that u k n(t) = v k n(t) for every n = 0, 1,..., N and t [0, T ]. From (40), (45), (46), and (c[k]), we easily have (d[k]). For n = 1, 2,..., N 1, we have, from (11), (c[k + 1]), and (c[k]), By Proposition 3, however, we have Combining (47) and (48), we obtain v k n(t) v k+1 n (t) > 0, t (τ k+1 n, T ). (47) v k n+1(t) v k+1 n+1 (t) vk n(t) v k+1 n (t). (48) vn+1(t) k vn+1 k+1 k+1 (t) > 0, t (τn, T ). (49) 16

17 But in view of (11), (c[k + 1]), and (c[k]), this leads to τn+1 k+1 τ n k+1. Therefore, we have (e[k]). Let us now turn to the proof of (f[k]). For convenience, we denote d t vn(t)/β(t) k by wn(t). k When t (0, τn k+1 ), which is when k = K, we have wn(t) k = wn k+1 (t) from (d[k]). Hence, following (f[k + 1]), we know that wn(t) k is decreasing in t. Let t (τn k+1, T ) then. By (d[k]), we know v k n(t) v k n 1(t) = p k + wk n(t) ᾱ k. (50) By the boundary conditions vn(t k ) = vn 1 k (T ) = 0, we therefore have Taking derivative on (50), it follows that w k n(t ) = p k ᾱ k. (51) G k n(t) w k = 0, t (τ k+1 n, T ). (52) In view of (13), (14), and (51), we can solve (52) to obtain, for t (τn k+1, T ), T wn(t) k = p k ᾱ k exp( ᾱ k ˆβ(t, T )) + ᾱ k β(s) wn 1(s) k exp( ᾱ k ˆβ(t, s)) ds, (53) t which, by the identity results in ᾱ k T t β(s) exp( ᾱ k ˆβ(t, s)) ds = 1 exp( ᾱ k ˆβ(t, T )), (54) T wn(t) k = p k ᾱ k + ᾱ k t β(s) ( p k ᾱ k + wn 1(s)) k exp( ᾱ k ˆβ(t, s)) ds. (55) Since ˆβ(0, ) is a strictly increasing function on [0, T ], we can define strictly increasing function θ( ) on [0, ˆβ(0, T )], so that ˆβ(t, θ(y)) = ˆβ(0, θ(y)) ˆβ(0, t) = y ˆβ(0, t), y [0, ˆβ(0, T )]. (56) This then leads to d y θ(y) = 1 d s ˆβ(0, s) s=θ(y) = 1 β(θ(y)). (57) In view of the above, we can bring a change of variables to the integral involved in (55), so that the latter becomes w k n(t) = p k ᾱ k + ᾱ k ˆβ(0,T ) ˆβ(0,t) ( p k ᾱ k + w k n 1(θ(y))) exp( ᾱ k (y ˆβ(0, t))) dy. (58) Suppose w k n 1 (t) is decreasing in t for t (0, T ), then since θ( ) is increasing, we know wk n 1 (θ(y)) is decreasing in y for y (0, ˆβ(0, T )). From (51) which also applies to w k n 1 (T ), we may see that p k ᾱ k + w k n 1(θ(y)) 0, y (0, ˆβ(0, T )), (59) 17

18 and p k ᾱ k + w k n 1(θ(( ˆβ(0, T )) )) = 0. (60) Hence, (58) can be rewritten as w k n(t) = p k ᾱ k + ᾱ k + ˆβ(0,t) ( p k ᾱ k + w k n 1(θ(y))) + exp( ᾱ k (y ˆβ(0, t))) dy. (61) By the decrease of ( p k ᾱ k + w k n 1 (θ(y)))+ in y 0, the increase of ˆβ(0, t) in t, and Lemma 1, we can get the decrease of wn(t) k in t on (τn k+1, T ). But combining with the earlier result on the other half interval, we may get the decrease of wn(t) k in t on the entire (0, T ) from that of wn 1 k (t). As w0 k(t) = 0 for all t (0, T ), we can therefore use induction on n to prove the decrease of wk n(t) in t (0, T ) for all n = 0, 1,..., N. For t (0, τn k+1 ), which is when k = K and a subset of (0, τ k+1 ) by (e[k]), we have, by (d[k]), n 1 vn(t) k vn 1(t) k = vn k+1 (t) v k+1 (t). (62) Hence, vn(t) k vn 1 k (t) is decreasing in t by (f[k + 1]). For t (τ k+1 n, T ), we can achieve the same property by (50) and the just proved decrease of wn(t) k in t. Thus, we have shown (f[k]). Note that, n 1 when k = K, the proof has no involvement of any [k + 1]-property. We have now completed the induction process. Therefore, (a[k]) to (f[k]) are all true for k = K, K 1,..., 0. From these, we see that, for any k = K 1, K 2,..., 0 and n = 1, 2,..., N, v k n(t) { = v k+1 n (t), t [0, τn k+1 ], > vn k+1 (t), t (τn k+1, T ). (63) Hence, we may see that each τn k+1 offers an optimal time by which the firm is to raise its price from p k to p k+1 when it has n remaining items. D. A Counter Example to a Claim of Feng and Xiao [5]: Consider an example with T = 1, K = 1, N = 2, and ᾱ 0 = 2. Let v 1 0 (t) = 0, v1 1 (t) = 2t + 2, and v1 2 (t) = t2 4t + 3. We can check that d t v1 1(t) = 2 and d tv2 1 (t) = 2t 4. These lead to G0 1 (t) v1 = 4t 6 and G2 0(t) v1 = 2t 2 + 6t 6. Hence, G1 0(t) v1 and G2 0(t) v1 are both increasing in t [0, 1]. In addition, G 0 2 (t) v1 G 0 1 (t) v1 = 2t 2 + 2t, which is positive for t [0, 1]. That is, G 0 n(t) v 1 is increasing in n too. On the other hand, we may let v 0 1 (t) = t2 t+2 and v 0 2 (t) = v1 2 (t) = t2 4t+3. Now, v1 0(t) v1 1 (t) = t2 + t, which is strictly positive for t (0, 1), and v2 0(t) v1 2 (t) = 0. Thus, it is not true that v1 0(t) v1 1 (t) v0 2 (t) v1 2 (t) on t [0, 1]. Among other violations, the last point consists of a violation of Proposition 3. So the possibility that these v k n( ) s form actual value functions for a markup problem has been ruled out. When we pretend that these v k n(t) s form value functions for a markup problem, however, we should have τ 1 1 as the smallest t such that v0 1 (t) v1 1 (t) > 0 and τ the smallest t such that

19 v2 0(t) v1 2 (t) > 0; also, each threshold level should be set at 0 when the corresponding strict positivity is always true and set at T = 1 when the corresponding strict positivity is never true. Therefore, we should have τ 1 1 = 0 and τ 1 2 = 1 for this example. It is therefore not true that τ 1 1 τ 1 2. But the latter is required for a threshold policy. 19

20 Supplemental Materials for A Revisit to the Markup Practice of Dynamic Pricing 7 The Markdown Case In the markdown case, the firm has to consecutively charge the sequence of decreasing prices p K, p K 1,..., p 0. Again, we aim at showing the optimality of a threshold policy τ = (τ k n k = 1, 2,..., K, n = 1, 2,..., N) ( N ) K. Under this policy, the firm should lower its price from p k to p k 1 when the threshold time τ k n corresponding to its current inventory level n is about to be passed. Let v k n(t) be the maximum revenue the firm can make in time interval [t, T ] when it starts time t with price p k and inventory level n. Since the lowest price p 0 is the last price the firm can charge before running out of stock, we have v 0 n(t) = p 0 E[(N 0 (T ) N 0 (t)) n]. (64) When the firm is charging any other price p k for k = 1, 2,..., K, it has yet to dynamically decide the time to switch to the next price p k 1. Hence, we have vn(t) k = sup E[ p k {(N k (τ) N k (t)) n} + v k 1 (τ)], (65) (n N τ T (t) k (τ)+n k (t)) + where T (t) is again the set of stopping times later than t. Similar to Proposition 1 in the note, we have the following markdown counterpart. Proposition 5 For any k = 1, 2,..., K, a function vector u (u n (t) n = 0, 1,..., N, t [0, T ]) that is uniformly bounded and absolutely continuous in t for every n will be the value-function subvector v k (v k n(t) n = 0, 1,..., N, t [0, T ]), if it satisfies the following: (i) u n (t) v k 1 n (t) for every n = 0, 1..., N and t [0, T ], (ii) u n (T ) = 0 for every n = 0, 1,..., N and u 0 (t) = 0 for every t [0, T ], (iii) for n = 1, 2,..., N and t [0, T ], u n (t) = v k 1 n (t) implies G k n(t) u + p k ᾱ k β(t) 0, (iv) for n = 1, 2,..., N and t [0, T ], u n (t) > v k 1 n (t) implies G k n(t) u + p k ᾱ k β(t) = 0. Proposition 5 offers a hint as to how the value functions v k n(t) and threshold levels τ k n can be constructed. First, due to (64), v 0 n(t) can be constructed in an n-loop.then, for any k = 1, 2,..., K, suppose vn k 1 (t) is known. We can then go through an n-loop to find all the vn(t) s. k First, we can let v0 k(t) = 0 as suggested by (ii) of the proposition. Second, suppose vk n 1 (t) is known for some n = 1, 2,..., N. Then, starting with t = T, we can equate vn(t) k to vn k 1 (t) for ever smaller t values until it is to occur that Gn(t) k v k + p k ᾱ k β(t) > 0. The latter is not allowed by (iii) and (iv) of the proposition. For time t earlier than this t, which we mark as τ k n, we let v k n(t ) be the solution to G k n(t ) v k + p k ᾱ k β(t ) = 0. 1

21 According to (i) of the proposition, we still need v k n(t ) v k 1 n (t ) for t < τ k n for the thus constructed v k n( ) to be the true value function. Nevertheless, let us go ahead with the construction procedure thus outlined. Not knowing whether what shall be constructed are the true value functions, we call them u s instead of v s. Formally, here is the iterative procedure. First, let Then, for n = 1, 2,..., N and t [0, T ], let u 0 n(t) = ᾱ 0 u k 0(t) = 0, k = 0, 1,..., K, t [0, T ]. (66) T t β(s) ( p 0 + u 0 n 1(s)) exp( ᾱ 0 ˆβ(t, s)) ds. (67) Next, we go over an outer loop on k = 1, 2,..., K and an inner loop on n = 1, 2,..., N. At each k and n, first let τ k n = inf{t [0, T ] G k n(t) u k 1 + p k ᾱ k β(t) 0}, (68) with the understanding that τ k n = 0 when the concerned inequality is always true and τ k n = T when it is never true. Then, let and when t [0, τ k n), u k n(t) = u k 1 n (t), t [τ k n, T ], (69) τ k u k n(t) = u k 1 n (τn) k exp( ᾱ k ˆβ(t, τn)) k + ᾱ k n β(s) ( p k + u k n 1(s)) exp( ᾱ k ˆβ(t, s)) ds. (70) t Now, we proceed to show the optimality of the construction. concavity result. First, we have the following Proposition 6 For any fixed k = 0, 1,..., K and t [0, T ], the value function v k n(t) is concave in n. For convenience, we have let the yet undefined τn 0 = T for n = 1, 2,..., N and vn 1 (t) = u 0 n(t) for n = 1, 2,..., N and t [0, T ]. Using Proposition 6, we can prove the following key result for the markdown case. Theorem 2 The u and τ as constructed from (66) to (70) satisfy the following for k = 0, 1,..., K: (a[k]) G k n(t) v k 1 /β(t) is decreasing in t for n = 1, 2,..., N and t (0, T ); (b[k]) u k n(t) = v k n(t) for any n = 0, 1,..., N and t [0, T ]; (c[k]) for any n = 1, 2,..., N, we have G k n(t) v k + p k ᾱ k β(t) = 0 for t (0, τ k n) and G k n(t) v k + p k ᾱ k β(t) 0 for t (τ k n, T ); (d[k]) τ k n is decreasing in n; (e[k]) more than having decreasing differences between n and t, v k n(t) satisfies the following for every t (0, T ): d t v k 1(t) 0, 2

22 and for n = 1, 2,..., N 1, d t v k n+1(t) d t v k n(t) ᾱ k β(t) (v k n 1(t) 2v k n(t) + v k n+1(t)), which is negative due to Proposition 6. Proof: We prove by induction on k. Let us first focus on proving (a[0]) to (e[0]). Take n = 1, 2,..., N. First, we have and (b[0]), that G 0 n(t) v 0 (t) + p 0 ᾱ 0 β(t) = 0, t (0, T ). (71) u 0 = (u 0 n(t) n = 0, 1,..., N, t [0, T ]) = v 0 = (v 0 n(t) n = 0, 1,..., N, t [0, T ]). (72) From (71), we may confirm (c[0]) with the understanding that τ 0 n = T for n = 1, 2,..., N. The convention for the τ 0 n s also leads directly (d[0]). From (64), we have v 0 1(t) = p 0 E[N 0 (t, T ) 1] = p 0 P [N 0 (t, T ) 1] < p 0. (73) As v0 0 (t) = 0 for any t [0, T ], we can apply (71) at n = 1 to get d t v 0 1(t) = ᾱ 0 β(t) (v 0 1(t) p 0 ), t (0, T ). (74) which is negative by (73). For n = 1, 2,..., N 1, we can apply (71) at both n and n + 1 to obtain d t v 0 n+1 d t v 0 n(t) = ᾱ 0 β(t) (v 0 n 1(t) 2v 0 n(t) + v 0 n+1(t)), t (0, T ). (75) So (e[0]) is satisfied as well. Finally, by our default definition of vn 1 (t) and (71), we know that (a[0]) is true. Suppose for some k = 1, 2,..., K, we have (a[k 1]), that Gn k 1 (t) v k 2 /β(t) is decreasing in t for n = 1, 2,..., N and t (0, T ), (b[k 1]), that (c[k 1]), that, for n = 1, 2,..., N, u k 1 n (t) = vn k 1 (t), n = 0, 1,..., N, t [0, T ], (76) G k 1 n (t) v k 1 + p k 1 ᾱ k 1 β(t) = 0, t (0, τ k 1 n ), (77) and G k 1 n (t) v k 1 + p k 1 ᾱ k 1 β(t) 0, t (τ k 1 n, T ), (78) (d[k 1]), that τ k 1 n is decreasing in n, and (e[k 1]), that, for t (0, T ), d t v k 1 1 (t) 0, t (0, T ), (79) 3

23 and for n = 1, 2,..., N 1, d t vn+1 k 1 d tvn k 1 (t) ᾱ k 1 β(t) (vn 1 k 1 (t) 2vk 1 n (t) + vn+1 k 1 (t)), t (0, T ). (80) Now we embark on showing (a[k]) to (e[k]). Take n = 1, 2,..., N. From the definition of τ k 1 n through (68), (b[k 1]), and (c[k 1]), we may see that for t (0, τn k 1 ), and Gn k 1 (t) v k 1 + p k 1 ᾱ k 1 β(t) = 0 < Gn k 1 (t) v k 2 + p k 1 ᾱ k 1 β(t), (81) Gn k 1 (t) v k 1 + p k 1 ᾱ k 1 β(t) = Gn k 1 (t) v k 2 + p k 1 ᾱ k 1 β(t) 0, (82) for t (τn k 1, T ). Therefore, Gn k 1 (t) v k 1 + p k 1 ᾱ k 1 β(t) = [Gn k 1 (t) v k 2 + p k 1 ᾱ k 1 β(t)] 0, (83) and hence by the strict positivity of β( ), Gn k 1 (t) v k 1 β(t) = [ Gk 1 n (t) v k 2 + p k 1 ᾱ k 1 ] 0 p k 1 ᾱ k 1. (84) β(t) Note that the above is even true for k = 1 due to the default definitions. Combining (a[k 1]) and (84), we obtain the decrease of Gn k 1 (t) v k 1 /β(t) in t. Now, note that Gn(t) k v k 1 Gn k 1 (t) v k 1 β(t) = (ᾱ k 1 ᾱ k ) (v k 1 n (t) vn 1 k 1 (t)), (85) which, by (S1 ) and (e[k 1]), is decreasing in t. This and the just proved result together lead to the decrease of G k n(t) v k 1 /β(t) in t. Hence, we have (a[k]). By the definition of τ k n through (68) and (b[k 1]), we have G k n(t) v k 1 + p k ᾱ k β(t) > 0, t (0, τ k n). (86) Due to the strict positivity of β( ) and (b[k 1]), the threshold τ k n also satisfies τ k n = inf{t [0, T ] Gk n(t) v k 1 β(t) + p k ᾱ k 0}. (87) But this and (a[k]) will lead to G k n(t) v k 1 β(t) + p k ᾱ k 0, t (τ k n, T ). (88) Combining (86), (88), and the strict positivity of β( ), we arrive to G k n(t) v k 1 + p k ᾱ k β(t) { > 0, t (0, τ k n ), 0, t (τ k n, T ). (89) 4

24 Our construction (69) and (b[k 1]) dictate that Also, we may see that the construction (70) renders u k n(t) = v k 1 n (t), t [τ k n, T ]. (90) G k n(t) u k + p k ᾱ k β(t) = 0, t (0, τ k n). (91) From the first half of (89), we have, for any n = 1, 2,..., N and t [0, τ k n), τ k vn k 1 (t) < vn k 1 (τn) k exp( ᾱ k ˆβ(t, τn)) k + ᾱ k n β(s) ( p k + vn 1 k 1 (s)) exp( ᾱk ˆβ(t, s)) ds. (92) t Using (70) and (92), as well as the fact that u k 0 (t) = vk 1 0 (t) = 0 for any t [0, T ], we can use induction over n to prove that u k n(t) > v k 1 n (t), n = 1, 2,..., N and t [0, τ k n). (93) By its construction, u k n(t) is uniformly bounded by NλT ; it is also Lipschitz continuous in t with coefficient Nλ, and hence absolutely continuous in t. By (89), (90), (91), and (93), as well as the fact that vn k 1 (T ) = 0 for every n, we may see that u k (u k n(t) n = 0, 1,..., N, t [0, T ]) satisfies the sufficient conditions (i) to (iv) stipulated in Proposition 5. Hence, we have shown (b[k]), that u k n(t) = v k n(t) for every n = 0, 1,..., N and t [0, T ]. From the second half of (89), (90), (91), and (b[k]), we easily have (c[k]). Now take n = 1, 2,..., N 1. By (S1 ), (e[k 1]), and Proposition 6, we have d t vn+1 k 1 (t) d tvn k 1 (t) ᾱ k 1 β(t) (vn 1 k 1 (t) 2vk 1 n (t) + vn+1 k 1 (t)) ᾱ k β(t) (vn 1 k 1 (t) 2vk 1 n (t) + vn+1 k 1 (t)) 0. (94) We therefore have the negativity of G k n+1 (t) vk 1 G k n(t) v k 1 = d t vn+1 k 1 (t) d tvn k 1 (t) + ᾱ k β(t) (2v k 1 n (t) vn 1 k 1 (t) vk 1 n+1 (t)). (95) But by the definition of τ k n and τ k n+1, this leads to τ k n+1 τ k n. Therefore, we have (d[k]). Let us turn to the proof of (e[k]). When t (τ1 k, T ), we have which is negative by (79). When t (0, τ1 k ), we have d t v k 1(t) = d t v k 1 1 (t), (96) d t v k 1 (t) = ᾱk β(t) (v k 1 (t) pk ) = ᾱ k β(t) [v1 k 1 (τ1 k) exp( ᾱk ˆβ(t, τ1 k)) + p k ᾱ k τ1 k t β(s) exp( ᾱ k ˆβ(t, s)) ds p k ] = ᾱ k β(t) exp( ᾱ k ˆβ(t, τ1 k )) (vk 1 1 (τ1 k) pk ), (97) 5