Problem Set 3: Suggested Solutions

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1 Microeconomics: Pricing 3E00 Fall 06. True or false: Problem Set 3: Suggested Solutions (a) Since a durable goods monopolist prices at the monopoly price in her last period of operation, the prices must be falling over time and in the first period, the price is above the price of the model where sales are only possible in a single period. Solution. False (partly true). Suppose there are two periods. The monopolist will always set the monopoly price on the remaining part of the demand in the last period. Using the backward induction, the price in the first period cannot be lower than that in the second period, otherwise everyone would buy in the first period already. So, the price in the first period must be the same or higher. In general, if a consumer is impatient, he will buy earlier at a relatively higher price, whereas a patient consumer buys later at a relatively low price. Hence, it is true that the prices must be falling over time. However, it is not usually the case that the price in the first period is higher than that in a single period model. (See, the example in the lecture notes.) Especially if the discount rate is small, i.e. the consumers are very patient, then the price in the first period converges to the monopoly price of the last period as the number of periods of operation tend to infinity. In a sense, in order to induce the consumers to buy now rather than later, the monopoly has to decrease the price from that of a single period price. (b) If the buyers are irrational and the sellers know this, then the seller can always exploit the irrationality and make a higher profit than in a model with rational buyers. Solution. False. What if irrational buyers do not even trade in a situation where the rational ones do? (c) A first-price auction brings in more revenue to the seller than a second-price auction since in the second-price auction, the winner only has to pay the second-highest bid.

2 Solution. False. Remember the Envelope Theorem result derived in the lecture notes: equivalent revenues in FPA and SPA. (d) Suppose n bidders are shown a jar full of coins. Their task is to bid in an auction for the jar. Because of the winner s curse, the bidders will be more conservative in their bidding if n increases. Because of this, the revenue to the auctioneer goes down as n increases. Solution. False. As the number of bidders increases, the value of the nth order statistic, which equals the value of the winning bid, increases. Intuitively, you can think how an increase in the number of bidders makes the auction more competitive, so that very little surplus is left to the buyers. Hence, the revenue to the seller increases as the number of bidders go up.. Consider a model of peak-load pricing where the inverse demand function in season i {s, w} is given by: p i = 0 β i q i. (a) Draw the inverse demand curves for the case where β s = 3, β w = and draw also the marginal revenue curves in the same picture. Which of the two seasons is more likely to be the peak season? Solution. We can compute that the marginal revenue from season i = s, w equals MR i = 0 β i q i. Since for all price levels p, the demand in season s is less than the demand in season w, season w is the peak season. (b) Suppose that the only cost for the monopolist is the cost of building capacity. Let this cost be given by fk for k units of capacity. Find the optimal level of capacity and the optimal price if the monopolist must set the same price for the two seasons. Solution. Note that the monopolist has to set the same price for the both seasons. If capacity binds in the peak season w, we will have q w = k. We can solve the demands from the inverse demand functions and write profit maximization in terms of the price: max p p β w (0 p) + p (0 p) f (0 p) βs βw

3 The first-order condition with respect to p yields 0( β w + β s ) ( β w + β s )p + f β w = 0 p = f. The optimal capacity is thus k = f. Alternative approach: Note that the monopolist has to set the same price for the both seasons. If capacity binds in the peak season w, we will have q w = k, and the the common price for both seasons is p w,s = 0 β w q w = 0 k. Consequently, q s = 0 pw,s β s profit maximization will be: = 0 (0 k) 3 = k 3. Thus, the max π (k) k = q w (k) p w,s (k) + q s (k) p w,s (k) fk = k (0 k) + k 3 (0 k) fk. The first-order condition with respect to k yields 0 k k f = 0 k = f. The optimal price is thus p w,s = 0 k = f. (c) Suppose next that the monopolist can choose different prices in the two seasons. For what values of f is the capacity constraint binding in the high season only? Solution. If the capacity constraint binds only in the peak season, we have q w = k, and p w = 0 k. Thus, the monopolist maximizes max {(0 k) k fk} for the peak season. This yields k = 5 f k. For the low season, the monopolist maximizes max {q s (0 3q s )}, q s and hence q s = 5 3 and ps = 5. In order for the solution to be valid, we must have q s k f 0 3 (d) Solve the optimal capacity and supply levels for the two seasons both in the case where the capacity constraint binds only in one season and where it binds in both seasons. Solution. From Part (c) we know that if the capacity constraint binds only in one season, the supply during the peak season is k = 3

4 q w = 5 f and the supply in the season s is given by qs = 0 6. We know that this solution is valid if and only if f 0 3. If the capacity constraint binds in both seasons, the problem of the monopolist equals max {k (0 k) + k (0 3k) fk}. k The first-order condition yields k = 0 f 8, which is the solution if f When the public sector or private firms purchase goods or services, a procedure called reverse auction or procurement auction is often used. In such auctions, the buyer specifies a service to be procured, and the providers submit bids indicating the lowest price at which they would be willing to produce the service. In standard formats of these auctions, the producer submitting the lowest bid wins. (a) Formulate first-price and second-price procurement auctions as Bayesian games. Solution. A Bayesian game consists of players, their action sets, type sets, a probability distribution over the types, and a payoff function (a mapping from types and actions to utilities). Here we have N players, whose actions are their bids (A i = R (+) ). Here, types are not specified. In part (b), Θ i = [0, 00] =: C and the the c.d.f. F i : C [0, ] is given as F (c i ) = ci 00. Payoff function in FPA. u i (c i, t) = (t i c i ) (ti smallest). Payoff function in FPA. u i (c i, t) = (t () c i ) (ti smallest), where t () stands for the second lowest bid. (b) Do these games have solutions in dominant strategies? How would you compute the Bayes Nash equilibria in these games if the N producers have privately known costs c i [0, 00] and these costs are independent private values drawn from the uniform distribution? Solution. Exactly the same logic works here as in the normal case with bidders as buyers: a SPA has a dominant strategy to bid c i and a FPA does not have a dominant strategy. SPA: t i = c i. FPA: max P r(wins)(t i c i ) = P r(t(c j ) t i ) N (t i c i ). 4

5 FOC: (N )P r(t(c j ) t i ) N (t i c i ) P r(t(c j) t i ) t i + P r(t(c j ) t i ) N = 0. An equilibrium bidding strategy must be increasing in c i (why?). Therefore, P r(t(c j ) t i ) = P r(c j c i ) = F (c i ) in the equilibrium. Now we can write the FOC as ( F (c i )) N (( F (c i ))t (c i ) f(c i )(t i c i ) = 0 ( F (ci )) N dc i t i = c i ( F (c i )) N = c i + (00 c). N The solutions follows by solving the differential equation the same way as in the lecture notes. (c) A problem with these auctions is that sometimes producers submit unrealistically low bids in order to win. The idea is then to engage in bargaining over the final price to be agreed in later negotiations. To prevent this the following procedure has been proposed: Discard the lowest 0% of the bids as unrealistically low and award the contract to the lowest remaining bid at the price of that bid. Can you find a Bayes Nash equilibrium for this game? (Believe it or not, a variant of this auction has been in use for EU procurements). Solution. Now, everybody bidding 00 (or even more!) is a BNE. No one wants to deviate downwards since her bid would then be one of the unrealistically low ones and get discarded. Can you find any equilibria with lower bids? (Harder) 4. In the lectures, we have seen that all bidders in the second-price auction have a strategy that is a best response to all possible bids by other bidders as long as the auction is one with private values. Investigate if the same is true for other possible auction formats. For all of these cases (except d), you may take all bidders to have privately observed private valuations v i drawn from the uniform distribution on [0, ]. (a) All-pay auction where all bidders (winning and losing) pay their own bid b i and the bidder with the highest bid wins the object and gets v i. If two (or more) bids are tied for the highest bid, then the object is allocated according to a symmetric lottery to one of the highest bidders. 5

6 Solution. In an all-pay auction, all losers would like to deviate to either bidding 0 or winning the auction - no dominant strategies. (b) Second-price auction with a (possibly random) reservation price. Here the seller submits a (possibly random) bid b 0, but otherwise the rules are as in second-price auction (with b 0 treated as one of the bids). Solution. By the same argument as in the case of a standard secondprice auction, it is optimal strategy for all bidders to bid their true valuation. This can be easily seen by thinking the reserve price as an additional bidder. (c) Third-price auction. Here the highest bidder pays the third highest price. Non-winning bidders pay nothing. Ties are broken with a symmetric randomization. Solution. It cannot be a dominant strategy to bid ones own valuation: the second highest bidder would want to deviate to bid more than the highest. Other dominant strategies? No, can create similar situations when the bidder would like to increase (or decrease) her bid if everybody else bids their true valuation. 5. This problem shows that selling by auctions is in general a good idea. Consider a setting where a seller sells a single indivisible object to one of two bidders. The bidders have a demand for at most one unit of the good and their valuations are independent and drawn according to the uniform distribution on [0, ]. (a) Suppose that the seller sets a single price p for the good and the buyers either accept the price or not. If exactly one bidder accepts, then she is allocated the good. If both accept, then the good is allocated randomly to one of the bidders at price p. If neither bidder accepts, then the good is not sold. Compute the expected revenue maximizing p for the seller and the maximized expected revenue. Solution. The problem of the seller is 6

7 max {p [ Pr (v i < p, i)]} p { [ = max p F (p) ]} F uniform p { ( = max )} p p. p The first-order condition gives p = and the corresponding expected revenue equals (b) Suppose next that the seller makes a take-it-or-leave-it offer to one of the bidders at p. If the offer is accepted, then the good is sold at that price. If the offer is rejected, the seller makes a take-it-or-leaveit offer p to the other bidder. If this is accepted, the good is sold. If the second offer is also rejected, the good is left unsold. Compute the expected revenue maximizing p and p and the maximized expected revenue. Solution. The problem of the seller is { } max p [ Pr (v < p )] + p Pr (v < p ) [ Pr (v < p )] p, p { } = max p ( p ) + p p ( p ), p, p where the latter line follows from the fact that the valuations are uniformly distributed. The first-order conditions are p : p + p ( p ) = 0; p : p ( p ) = 0. Thus, p = 5 8 and p = and the expected revenue equals (c) Show that in a second-price auction, the seller gets an expected revenue of 3. Solution. The expected revenue of the seller equals the expectation of the second highest value. Thus, let us derive the distribution of the lower of the two bidders values. 7

8 Let g (x) denote the probability that the lower of the two bidders values is x. There are two possible ways this event can occur. One way is that bidder has value x and bidder has value greater than x. The probability of this event is f (x) [ F (x)] = x. The other way is that bidder has value greater than x and bidder has value x. This also occurs with probability x. Thus g (x) = ( x). Thus, the expected value of the lower of the two valuations is given by xg (x) dx = x ( x) dx 0 0 = [ x 3 x3] 0 = 3. (d) (Harder) Show that a second-price auction with reserve price yields 5 an expected revenue of and thus raises more revenue than any of the alternatives proposed. Solution. There are two cases to consider: ) The case where the valuations of the both bidders are above ; and ) The case where one of the valuations is above. All other cases yield zero payoff to the seller. Case : The probability that the valuations of both bidders are greater than is given by [ F ( )] = 4. Let g ( v v ) denote the probability that the lower of the two bidders values is v, conditional on both values being above. There are two possible ways this event can occur. One way is that bidder has value v and bidder has value greater than v. The probability of this event is f ( v v ) Pr ( v v v ) = f(v) F( ) [ F (v)] F( ) = 4 ( v). The other way is that bidder has value greater than v and bidder has value v. This also occurs with probability 4 ( v). Thus g ( v v ) = 8 ( v). The probability of case, that is, the probability that one of the valuations is below and one above is given by F ( [ ( ) F )] =. 8

9 Thus, the expected revenue of the seller is given by 4 vg ( v v ) dv + = 4 = 4 = 5. 8v ( v) dv + 4 [ 8 [ x 3 x3] ] + 9

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