Problem Set 3: Suggested Solutions

Transcription

1 Microeconomics: Pricing 3E Fall 5. 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. True/False. 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 factor 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. Think about the Revenue Equivalence Theorem. (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 winners 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 = 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 = β i q i. The inverse demand and the marginal revenue curves for the two seasons are shown in Figure. 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, and the the common price for both seasons is p w,s = β w q w = k. Consequently, q s = pw,s β s Thus, the profit maximization will be: = ( k) 3 = k 3.

3 Figure : Demand and marginal revenue curves for the model of peak-load pricing. max π (k) = q w (k) p w,s (k) + q s (k) p w,s (k) fk k = k ( k) + k 3 ( k) fk. The first-order condition with respect to k yields k k f = k = 5 3 f. The optimal price is thus p w,s = 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 = k. Thus, the monopolist maximizes max {( k) k fk} for the peak season. This yields k = 5 f k. Hence, the profits from the peak season equal p w q ( ). w fk = 5 f 3

4 For the low season, the monopolist maximizes max {q s ( 3q s )}, q s and hence q s = 5 3 and ps = 5, which means that the profits from the low season are 5 3. Thus, the total profits from the two season when the capacity constraint binds only in season w equals π wbid = ( ) 5 f = 3f 3f+75. In order for the solution to be valid, we must have q s < k f < 3 = 4 9. (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 = q w = 5 f and the supply in the season s is given by qs = 6. We 4 know that this solution is valid if and only if f < 9. If the capacity constraint binds in both seasons, the problem of the monopolist equals max {k ( k) + k ( 3k) fk}. k The first-order condition yields k = f. Plugging this into the objective function results to π wsbind = ( f ) ( f ) + ( ) ( f 3 ( )) f f ( ) f = ( ) f 4 = f 4f+4 6. π wbind > π wsbind if and only if 3f 3f+75 > f 4f ( 3f 3f + 75 ) > 3 ( f 4f + 4 ) f f + 7 > 3f f + 9f 5 >. 4

5 This is an upward opening parabola with roots 5 f, = ± 9. Thus, π wbind > π wsbind 5 if and only if f > f = 9, which is a contradiction. (Recall, in order for the solution to part c) to be 4 valid, we must have f < 9.) Thus, the monopolist is always better off by producing at full capacity in both seasons. 3. 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 [, ]. (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. Solution. In an all-pay auction, if you bid your own valuation, you get zero if you win the auction, and you must pay your own bid otherwise. Hence, your expected payoff is negative. If you bid more than your own valuation, your expected payoff is negative if you win the auction, and you must pay more than your own valuation otherwise. Thus, ovarbidding is clearly not optimal. Finally, if you bid less than your own valuation, your payoff is positive if you win the auction, and you have to pay less than your own valuation otherwise. Thus, it seems that underbidding is your optimal strategy and in the Appendix I show that this is indeed the case. (b) Second-price auction with a (possibly random) reservation price. Here the seller submits a (possibly random) bid b, but otherwise the rules are as in second-price auction (with b treated as one of the bids). Solution. By the same argument as in the case of a standanrd second-price auction, it is optimal strategy for all the bidders to bid their true valuation. 5

6 (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. Suppose all other buyers j i bid their true valuations. Consider the problem of bidder i. Let v = max {v j, j i} denote the highest valuation amongst the bidders other than i. Let v = max { {v j, j i}\{v } } denote the second highest valuation amongst the bidders other than i. Consider the situation in which v < v i < v. By bidding her true valuation, i will not win the object and ends up with zero payoff. By bidding b i > v, i wins the object and gets a payoff of v i v. Hence, bidding ones true valuation does not constitute equilibrium bidding behavior. The optimal bid is derived in the Appendix. 4. 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 [, ]. (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 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

7 (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 ) = ; p : p ( p ) =. Thus, p = 5 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. 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 7

8 xg (x) dx = x ( x) dx = [ x 3 x3] = [ ] 3 = 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 ) = ( v). The probability of case, that is, the probability that one of the valuations is below and one above is given by F ( [ ( ) F )] =.

9 Thus, the expected revenue of the seller is given by 4 vg ( v v ) dv + = 4 = 4 = 4 v ( v) dv + 4 [ [ ] x 3 x3] + [ [ ] ] 7 + = 4 ( ) = = 5. Appendix: Optimal Bidding Strategies All-pay Auction We will look for an equilibrium where each bidder uses a bidding strategy that is a strictly increasing, continuous, and differentiable function of his value. To do this, suppose that bidders j i use identical bidding strategies b j = b (v j ) with these properties consider the problem faced by bidder i. Bidder i s expected payoff, as a function of her bid b i and valuation v i is U (b i, s i ) = v i Pr [b i > b (v j ), j i] b i, where Pr [b i > b (v j ), j i] = Pr [ b j < b (b i ), j i ] = F n [ b (b i ) ], where n is the number of bidders and F is the uniform distribution. Thus, i chooses b to solve The first-order condition is max b i { vi F n [ b (b i ) ] b i } (n ) v i F n [ b (b i ) ] f [ b (b i ) ] b [b (b i )] =. In a symmetric equilibrium, b i = b (v i ). Plugging this into the objective function 9

10 (and dropping the subscripts) yields (n ) vf n (v) f (v) b (v) = b (v) = (n ) vf n (v) f (v) v b (x) dx = v v x (n ) F n (x) f (x) dx v }{{} = d dx F n (x) int. by parts b (v) b (v) = [ F n (x) x ] v v v v F n (x) dx b (v) b (v) = [v n v n ] n [vn v n ] v = b (v) = b (v) = n n vn Thus, the optimal bid is less than the true valuation of the bidder. Third-price Auction We again look for an equilibrium where each bidder uses a bidding strategy that is a strictly increasing, continuous, and differentiable function of his value. (Unlike with all-pay auction, in a third-price auction it is not necessarily always the case that the optimal bidding strategy is strictly increasing in value. As we will see below, the optimal bidding function is strictly increasing in value if F is log-concave. Since uniform distribution satisfies this, it is safe for us to focus on these strictly increasing strategies.) Suppose that bidders j i use identical bidding strategies b j = b (v j ) with the properties specified above. Consider then the problem faced by bidder i. Bidder i s expected payoff, as a function of her bid b i and valuation v i is U (b i, s i ) = {v i E [b (v ) b (v ) < b i ]} Pr [b i > b (v )], where v and v denote the highest and the second highest of the competing valuations. We have Pr [b i > b (v )] = Pr [ b j < b (b i ), j i ] = F n [ b (b i ) ], where n is the number of bidders and F is the uniform distribution. As to the expected value for the second highest bid, E [b (v ) b (v ) < b i ] = b (b i) b (x) f (n ) ( x v < b (b i ) ) dx,

11 where f (n ) (x) is the probability density function of the second highest valuation out of the n valuations lower than b (b i ). In order to obtain an expression for it, note that the event that v x when v < b (b i ) is the union of two events: ) v k x, for all k i, the probability of which is F n (x) F n [b (b i)] ; and ) v k x for all k i, j, and x < v j < b (b i ) for j i, the probability of which equals (n ) F n (x){f[b (b i)] F (x)} F n [b (b i)]. These two combined yield F (n ) ( x v < b (b i ) ) = (n ) F n (x) F [ b (b i ) ] (n ) F n (x) F n [b. (b i )] The associated probability density function, f (n ) ( x v < b (b i ) ), is given by the derivative of F (n ) ( x v < b (b i ) ) with respect to x: (n )(n )F n 3 (x)f[b (b i)]f(x) (n )(n )F n (x)f(x) F n [b (b i)] = (n )(n ){F[b (b i)] F (x)}f n 3 (x)f(x) F n [b (b i)] = (n ) { [ F n [b (b i)] F b (b i ) ] F (x) } f n (x), where we have used the observation that the probability density function of the highest valuation amongst the n valuations lower than b (b i ) is given by f n (x) = d [ F n (x) ] = (n ) F n 3 (x) f (v). () dv Thus, the expected value of the second highest bid is given by E [b (v ) b (v ) < b i ] = b (b i) b (x) (n ) { [ F n [b (b i)] F b (b i ) ] F (x) } f n (x) dv, and the objective of bidder i is max b i { = max b i v i F n [ b (b i) ] F n [ b (b i) ] b { v i F n [ b (b i) ] b (b i ) (bi ) b (x) (n ) { [ F n [b (b F b i )] (b ] i) F (x) } f n (x) dx b (x) (n ) { F [ b (b i) ] F (x) } f n (x) dx } } The first-order condition is given by

12 v i (n )F n [b (b i )]f[b (b i )] f[b b (b i )] (b i ) b (x) (n ) f b [b (b i )] b [b (b i )] n (x) dx =. In a symmetric equilibrium, b i = b (v i ). Again we can plug this into the objective function and drop the subscripts in order to get v(n )F n (v)f(v) b (v) = f(v) b (v) v b (x) (n ) f n (x) dx vf n (v) = v b (x) f n (x) dx Differentiating with respect to v gives F n (v) + (n ) vf n 3 (v) f (v) = b (v) f n (v) b (v) = F n (v)+(n )vf n 3 (v)f(v). f n (v) As a final step, plug the expression for f n (v) given in () into the bid function above in order to get b (v) = F n (v)+(n )vf n 3 (v)f(v) (n )F n 3 (v)f(v) = v + F (v) (n )f(v). () Given that the valuations are uniformly distributed on [, ], the equilibrium bidding function takes the form b (v) = n n v. In order to check that the obtained equilibrium bidding function is indeed strictly increasing, notice from () that this is the case when the ratio F ( ) f( ) is strictly increasing. This is the same as requiring that ln (F ) is a concave function, or equivalently, that F is log-concave. i.e. d f(v) dv F (v) <