Online Supplement: Price Commitments with Strategic Consumers: Why it can be Optimal to Discount More Frequently...Than Optimal

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1 Online Supplement: Price Commitments with Strategic Consumers: Why it can be Optimal to Discount More Frequently...Than Optimal A Proofs Proof of Lemma 1. Under the no commitment policy, the indifferent consumer solves xfx)dx ) = c. 1) µ As the left-hand-side LHS) strictly decreases with and the right-hand-side RHS) is constant, there either exists a unique [, 1] which solves 1), or, if ) v h xfx)dx > µc, there does not exist a which solves 1), in which case = 1. Proof of Lemma 2. First, note that the expected sales function is given by S ˆ / = xfx)dx + F and that S ) = ds/) d = 2 F. Differentiating R h s ) with respect to, we get: ζ s ) = drh s ) d = S/) + S /)) µc ˆ / = xfx)dx µc. R h s ) is concave because ζ s ) is decreasing in. The optimal s may be 1 a corner solution) if ζ s 1) or interior, in which case solving the first-order condition ζ s ) = gets the desired result. Note that s, because we assume that > c. Proof of Lemma 3. Under the discount-frequently policy, the indifferent consumer solves ˆ δ/f ) xfx)dx = µc. 2) 1

2 As the left-hand-side LHS) strictly decreases with and the right-hand-side RHS) is constant, there either exists a unique [, 1] which solves 2), or, if ˆ δ/f ) xfx)dx > µc, there does not exist a which solves 2), in which case f = 1. Proof of Theorem 1. It is useful to thin of f as a function of δ. f can be rewritten in terms of by comparing 2) with the definition of. There are three regimes that specify the relationship: δ/ f = 1 δ/ δ < vh / δ 1 1 > / Whenever f < 1, ) δ/ f xfx)dx = µc holds and whenever f = 1, ) δ xfx)dx > µc holds. Define ˆδ as follows: 1 ˆδ = < / v vh h / <. Thus, if the firm chooses δ [ /, ˆδ], then f = δ/. But if the firm chooses δ [ˆδ, 1], then f = 1. The revenue function can be written in terms of δ and. As is fixed, the revenue function can be expressed exclusively in terms of δ without an implicit function defining another term): ) R f δ; ) /δ F + xfx)dx + v δ h F δ / δ ˆδ = F δ) + δ xfx)dx + F ) ˆδ < δ 1. Differentiate the revenue function: dr f dδ = ) /δ 2 f δ) xfx)dx δ ˆδ < δ 1 / δ ˆδ It is immediately clear that δ > ˆδ is not optimal - in this case the firm is maring down more frequently than optimal, but there is no benefit in terms of increased high type demand they are already all visiting). It is also apparent that it is optimal to choose δ = ˆδ. Proof of Theorem 2. To search for the optimal pricing policy, start by fixing, the high-type s search strategy. For a given, find a set of prices, one for each possible demand realization, such that the firm s revenue is maximized and is the optimal strategy for high-type consumers. The strategy is optimal if the high-types expected value of search equals their cost of search conditional that fraction of high-type consumers visit. Next, inspect the set of chosen prices to confirm that the set can be implemented with discount-frequently. Finally, if discount-frequently is optimal for any given, then it must be the optimal policy overall. 2

3 Begin with some definitions. There exists a threshold demand realization, = /, such that for each x <, high-type demand, x, is strictly less than capacity. For x <, let sr, x) be the sales function for demand realization x and price r: For x, the firm sells units for every r. r sr, x) = 3) x < r Let P = {p : p : R + R + } be the class of price functions that maps each demand realization x to its price px). The firm s objective is to choose a function p P that maximizes revenue conditional on a search constraint that stipulates that the expected value a high-type receives from search is at least as great as the cost of search: max p P s.t. Rp) = max p P px)spx), x)fx)dx + px)) xfx) µ dx + The search constraint 4) can be rewritten as: where px)xfx)dx + g) = S px)) fx) µ px)fx)dx g), µc. px)fx)dx dx c 4) Note that g ) is independent of the chosen prices. Define the slac in the search constraint as: g) px)xfx)dx px)fx)dx. An increase in any price p x) has two effects: i) it increases revenue and ii) it decreases the slac in the constraint. Thus, an optimal p x) can be found by continuously increasing the set of prices so as to maximize the ratio of the marginal increase in revenue to the marginal decrease in the slac. Define p l P as the constant function px) = x. With this policy the firm generates in revenue, which is a lower bound on the revenue that can be achieved with the optimal policy. Given, if the search constraint is not satisfied with this pricing policy, then cannot be the equilibrium search strategy in the optimal policy because it does not generate at least in revenue). Thus, it is sufficient to consider values of such that the search constraint is satisfied with p l : i.e., it must be that is sufficiently small so that )S µc. In other words, with the pricing strategy p l, there must be some slac in the search constraint. Starting with p l, we next increase prices for some values of x so as to increase revenue while not violating the search constraint. For the most part, revenue increases and the slac decreases smoothly in price for each x, except for the very first increase in the price above. The first incremental price increase above yields a discontinuous decrease in revenue because all low-type shoppers abandon their purchases). Thus, for all x, the first incremental price increase is particularly costly - it decreases the slac without increasing revenue. In fact, the firm generates the same revenue with price as it does with price ˆpx) = /x). 3

4 Thus, for all x, the optimal policy either charges, or some price ˆpx) px). In that range, additional increases in price generate smooth increases in revenue and smooth decreases in the slac. In particular, there are two cases to consider: Case 1: x, ˆpx) px). The marginal increase in revenue is Rp)/ px) = xfx) and the marginal decrease in the slac with respect to price is xfx). Thus, the increase in revenue per unit of decrease in slac is: xfx)/xfx) =. Case 2: < x. The marginal increase in revenue is Rp)/ px) = fx) and the marginal decrease in the slac with respect to price is /)fx). Thus, the increase in revenue per unit of decrease is slac is: fx)//)fx) =. To repeat, for all x, the first incremental price increase above actually decreases revenue and slac. Hence, any price increase above should first be done in the < x demand states. In these states, all price increases generate the same constant increase in the ratio of revenue to slac case 2), the optimal price is the maximum price so long as the search constraint is satisfied. Therefore, starting with the highest demand states in the range < x, increase the price from to the maximum price,, until either the search constraint binds i.e., all slac is consumed) or the price is increased in all of these demand states. This yields the following price function: px) = x x otherwise for some x. If x >, then no slac remains and the above pricing strategy is the optimal solution. If x =, then some slac remains in the search constraint and price increases in the x can be considered, which is done next. For x, an increase in price from to ˆpx) increases revenue by xp )fx) and decreases slac by p )xfx). The relative increase in revenue to slac consumed is the ratio: xp )fx) p )xfx) = p v ) l p x which is increasing in x. This implies that, for x, if the price is increased above, then it should be increased for the highest demand state with the price still at. Furthermore, because the marginal increase in revenue to consumed slac equals for all price increases above ˆpx) case 1) and the initial price increase from to ˆpx) is costly, if a price is increased above ˆpx), then it should be increased all the way to. This leads to a pricing policy in which the firm charges either or, and the search constraint is binding. In particular, x x px) = otherwise where x x. The above can be implemented as a discount-frequently policy. Therefore, for a given, discount-frequently maximizes revenue, which implies it is an optimal policy. Proof of Theorem 3. 1) Conditional on observing, the firm s revenue is maximized by discounting to 4

5 when realized high-type demand is less than )/ ), otherwise the firm charges the high price. If the firm follows this policy and all high-type consumers are myopic i.e., they surely visit the firm), then the firm s profit is Π m = + S) S )) c. Differentiate Π m with respect to : η) = dπ m /d = + F ) c. Π m is quasi-concave because η) = c >, lim η) = c < and η) is decreasing when η) = which implies there is a unique such that η) = ). To demonstrate the latter, note that there is a unique solution to η)/ F ) = because η)/vh F ) is increasing in : η)/ F ) = ) c 2) Differentiate the profit function, Π = R c with respect to : dπ d = + vh F ) ) ) ) S S) c < + F ) c. ) When <, Π ) is linear and increasing if c < + S vh S). If, Π ) is quasiconcave with a unique solution given by m see part 1)). Therefore, = max } {/, m, if c < c and otherwise. 3) If F /s ), the firm charges and will choose s = because c >. Assume that F /s ) >, so the firm charges p s >. Differentiate the profit function, Π s = R s c with respect to : dπ s d = { vh F ) c < F ) c. Observe that for < the function is linear and increasing if c < F s ). If, dπ s )/d is decreasing and hence Π s ) is concave, which guarantees a solution exists, is unique and is given by F ) = c / if > and by otherwise. 4) Differentiate the profit function Π f = R f c with respect to : + ) dπ F ) c ) f d = + F ) c < vh / + F ) c vh / < From the < part, it follows that f > if and only if c < c f. dπ f /d is continuous and decreasing strictly decreasing ) for. This implies that there exists a unique capacity level that maximizes profits. Solving + F ) c =, we get the first candidate. if c < c f and comparing it with m, we get the desired result. Proof of Theorem 4. 1) Proof of s < m : Let τ s = F ), τs ) = F ), and τm ) = + F ). If τs c, then m > s =. Suppose now that τ s > c. Since τ s ) < τ m ) and τ s) < and τ m) <, it must be that s < m. Note that s = F 1 c / ) < since τ s > c. Proof of m f : This immediately follows from Theorem 3. If m, then f = m. Otherwise, f > m. 2) If c = c, then m = / = = = f. Thus, for c c we have m = = f. For c > c, > m and is decreasing in c. Thus, f = < / =. 6) 5