A Handout for Economics 201C - Menghan Xuxumhandy.com/wp/wp-content/uploads/2015/11/Handout.pdf · A Handout for Economics 201C " ... that there must be a general principle underly-ing

A Handout for

Economics 201C

”...This result(Revenue Equivalence Theorem) was first proved in the

early spring of 1979. Bill Saumuelson and I became convinced, by

a host of examples, that there must be a general principle underly-

ing them. I well remember the excitement of the moment when ev-

erything fell into place on a rainy afternoon in Newton Highlands,

Massachusetts...”

By John Riley

Essential Microeconomics

page 453

Wish You Find Your Own Excitement Someday

and Good Luck in Exams!

Menghan Xu

Spring 2014

Econ 201C Menghan Xu

Index

Week 1 2

Sealed First Price Auction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Single Crossing Property and Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Trembling Hand Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Week 2 10

All Pay Auction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Price Discrimination - Cellphone Plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Week 3 19

Buyer Equivalence and 2nd Price Auction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Signalling Game and SCP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Monopoly and Product Quality (COMP Fall ’09) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Week 4 27

BNE vs PBE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

PBE and Intuitive Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Signalling model with continuous types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Week 5 35

Signalling (COMP Spring ’13) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Sealed-bid Auctions (COMP Spring ’12) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

Week 6 42

Mechanism Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Mechanism Design Continuous Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Week 7 48

Ironing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

Efficient Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Week 8 55

Dominant Strategy Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Efficient Mechanism Design (COMP Spring ’12) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Sales of Indivisible Good . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Week 9 62

Optimal Auctions with Finite Types (COMP Spring ’13) . . . . . . . . . . . . . . . . . . . . . . . 62

Choosing Optimal Quantity of a Public Good . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Fixed Supply and Marginal Revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

-1-


Week 1

Sealed First Price Auction

Consider an auction with 1 item to sell. There are multiple buyers each of whom submits a sealed bid. The

highest bidder is the winner. If there is more than one highest bidder, then the winner is selected at random.

The winner pays his bid. The set of buyers’ type is Θ = {θ1, θ2...θT }. Type is private information of bidders.

Denote the value of the winner by vt = v (θt), so the winner’s ex-post utility is

v (θt)− bt

if win and zero otherwise. Ex-antely, for any bidder, his utility can be represented by two variables, bidding

price bt and winning probability wt. Notice that given strategy profile and beliefs, bt determines wt. So

choosing bt means choosing corresponding winning probability wt. Under this setting, one’s expected utility

is

ut (wt, bt; θt) = wt · (v (θt)− bt)

Example with 2 bidders under 2 types

Consider a situation in which there are only 2 bidders i = 1, 2. Each can be of either high or low type -

{θL, θH}. The probability distribution of types is {f (θH) , f (θL)}. Both probabilities are positive and add

up to one. In terms of value, we assume 0 < v (θL) < v (θH). The seller announces that the minimum price

that will be accepted is p0 = v (θL). As a part of ice-breaking condition, we assume that a low type bidder

always quits. Due to the assumption, when a high type bids 0, if the other player is of low type, the high

type will win for sure.

Lemma 1. There exists no pure strategy BNE

Proof. Assume not and denote (b1, b2) are each player’s bid if one is of high type. Consider the following 2

cases.

1. b1 6= b2 - It is trivial, skipped.

2. b1 = b2 = b

• If b ≥ v (θH), the expected utility is less or equal to zero. By deviating to b′ = v (θH)−ε, a bidder

can win only when the other player is of low type. His expected payoff is

ui (wi, bi; θi) = ui (f (θL) , b′; θH) = f (θL) · ε > 0

therefore, b ≥ v (θH) cannot be BNE.

• If b < v (θH), then one’s expected utility equal to

ui

(w(b), b; θH

)=

[f (θL) +

1

2f (θH)

](v (θH)− b

)

-2-


If player i deviates to b′ = b+ ε, he will win with probability one,

ui

(1, b+ ε; θH

)= v (θH)− b− ε

Obviously, when ε is small enough, the deviation is absolutely profitable.

By the lemma, the high type bidder must play a mixed strategy in equilibrium. Denote the mixed

strategy equilibrium strategy profile be bi (θH) , i = 1, 2 and Gi (b; θH) be the c.d.f of bj (θH). The following

several results can help us characterize the shape of G. Before we start, recall that when a player implements

a mixed strategy, he must be indifferent among all actions with positive density. The indifferent principle is

important in our analysis and we will use it repeatedly.

Lemma 2. 1There is no mass point on bi (θH), i = 1, 2.

Proof. First we can easily rule out the mass point on b = v (θH). Indifference means that the expected payoff

from the auction is zero. If so, by indifference principle, the expected payoff of the mixed strategy is zero.

For any other bid b with positive density, it must be true. However, a high bidder always wins if the other

player is of low type. Therefore, the expected payoff is at least f (θL) ·(v (θ)− b

)> 0.

If the mass point b∗ is between [v (θL) , v (θH)), to find a profitable deviation, we define the following new

CDF

Gε (x; b) =

G (x) if x < b∗

G (x)−G (b∗) + limy→b∗− G (y) if x ∈ [b∗, b∗ + ε)

G (x) if b∗ + ε ≥ x

Essentially Gε represents a distribution in which the mass point at b∗ has been shifted rightward by the

amount of ε. Also notice that the pdf functions of G and Gε are the same everywhere but b∗ and b∗ + ε.

Our purpose is to show that, given the other bidder plays G, the expected payoff by deviating to Gε is

larger than that staying at G even if ε converges to zero. The following graph illustrates how G and Gε are

different.

1In this lemma, we assume both bidders are using same mixed strategy. In principle, it is not necessary. You can try toprove the lemma but relaxing the assumption of symmetry.

-3-


The expected payoff under G is

V (G;G) ≡∫∈[v(θL),v(θH)]\{b∗,b∗+ε}

(f (θL) + f (θH) G (x)

)(v (θH)− x)G′ (x) dx

+(f (θL) + f (θH) G (b∗)

)Pr (x = b∗|G) (v (θH)− b∗)

+(f (θL) + f (θH) G (b∗ + ε)

)Pr (x = b∗ + ε|G) (v (θH)− b− ε)

and the expected payoff of deviating to Gε is

V (Gε;G) ≡∫b∈[v(θL),v(θH)]\{b,b+ε}

(f (θL) + f (θH) G (x)

)(v (θH)− x)G′ε (x) dx

+(f (θL) + f (θH) G (b)

)Pr (x = b|Gε) (v (θH)− b)

+(f (θL) + f (θH) G (b+ ε)

)Pr (x = b+ ε|Gε) (v (θH)− b− ε)

where

G (x) ≡ lims→x−

G (s) +1

2

[G (x)− lim

s→x−G (s)

]is the true winning probability of biding b. If G is smooth at b, then G = G. If there is a mass point at

b, the winning probability of bidding b consists of two parts. First, one will win if, the other player bids

strictly less than b, whose probability is lims→x− G (s). Second, if the other player bids b, whose probability

is G (x)− lims→x− G (s), one’s winning probability is the half of it.

Notice that first, the integrals in the two terms are identical since the pdf’s are identical on the set

[v (θL) , v (θH)] \ {b∗, b∗ + ε}. Second, Pr (x = b∗ + ε|G) = Pr (x = b|Gε) = 0. Therefore, we have

V (Gε;G)− V (G;G) =(f (θL) + f (θH) G (b+ ε)

)Pr (x = b+ ε|Gε) (v (θH)− b− ε)

−(f (θL) + f (θH) G (b)

)Pr (x = b|G) (v (θH)− b)

By definition of Gε, we have Pr (x = b+ ε|Gε) = Pr (x = b|G) > 0. Thus, we need to only check

V (Gε;G)− V (G;G)

Pr (x = b|G)=(f (θL) + f (θH) G (b+ ε)

)(v (θH)− b− ε)−

(f (θL) + f (θH) G (b)

)(v (θH)− b)

Letting ε converge to zero, the above term converges to

f (θH) · limε→0

[Gj (b+ ε)− Gj (b)

]· (v (θH)− b)

= f (θH) · 1

2· Pr (x = b|G) · (v (θH)− b) > 0

That is, for ε small enough, such shifting is strictly profitable. In conclusion, a BNE does not allow the

existence of any mass point!

Putting the tedious math aside, the proof actually has some economics intuitions. When a mass point

exists in buyer j’s mixed strategy, player −j has only half chance to win if he bids at the mass point, too.

Remember that the bidder’s utility is determined by (b, w). The shift will change b by only ε but double w.

The trade-off between increasing tiny in b and lowering huge w implies that such deviation is profitable.

-4-


Again, in this proof, we assume symmetry in G. The above intuition, however, does not require it. If

interested, you can prove the lemma assuming that Gi and Gj can be different.

Next lemma shows that G is strictly increasing by arguing that there exists no flat range in it.

Lemma 3. There is no flat range in G.

Proof. Suppose that there is a range [bl, bu] over which Gj is constant. (We can assume the interval is closed,

since we just proved that there is no mass point.) For some bl− εl, bu + εu ≥ 0 at which Gj is increasing, by

indifference principle, we have

(f (θL) + f (θH)Gj (bl + εl)) · (v (θH)− bl + εl) = (f (θL) + f (θH)Gj (bu + εu)) · (v (θH)− bu − εu)

as εl and εu converge to zero, we have

(f (θL) + f (θH)Gj (bl)) · (v (θH)− bl) = (f (θL) + f (θH)Gj (bu)) · (v (θH)− bu)

Gj (bu) = Gj (bl) gives us

v(θH)− bl = v(θH)− bu

bl = bu

Contradictory!

The last step is to show that the lower bond of suppG is exactly v(θL).

Lemma 4. The lower bound of the support of G is v (θL).

Proof. Suppose the lower bound for player i is bi > v (θL), for i = 1, 2. WLOG, when b1 ≤ b2, player 1’s

winning probability is just f (θL) if he bids b1. So the expected payoff is

f (θL) (v (θH)− b1)

So, he can be better-off by just change b1 to v (θL). Then the expected payoff changes to

f (θL) (v (θH)− v (θL))

If v (θL) = b1 < b2, then player 1 will put all density below b2 to v (θL), which violates the lemma of

no-mass-point.

The above results characterize the shape of CDF. In summary, G is a smooth and strictly increasing

CDF function with lower bound equal to low type agent’s value. The following proposition can be treated

as a summary of the above results

Proposition 1. There exists a symmetric equilibrium in which high value buyer bids according to a mixed

strategy b(θH) with support[v (θL) , b

]and CDF G (b; θH) in C1 and is strictly increasing.

-5-


Given the proposition, one’s winning probability of bidding b is

w (b) = f (θL) + f (θH)G (θH , b)

so expected payoff of player i equals to

ui (b; θH , G) = [f (θL) + f (θH)G (b; θH)] · [v (θH)− b]

To solve for closed form of G. We have two tasks - (i) we first need to pin down b; (ii) we need to compute

G (b) for all b on[v (θH) , b

]. By indifference principle, bidding b and bidding v (θL) should be indifferent to

both players. Bidding v (θL) gives player i

Ui (v (θL) ; θH , G) = f (θL) [v (θH)− v (θL)]

Bidding b generates

Ui(b; θH , G

)= v (θH)− b

Thus, we have

v (θH)− b = f (θL) [v (θH)− v (θL)]

b = f (θH) v (θH)− f (θL) v (θL)

We can follow the same logic to solve for G, by rearranging

f (θL) [v (θH)− v (θL)] = [f (θL) + f (θH)G (b; θH)] · [v (θH)− b]

At this point, we solve for a symmetric mixed strategy BNE with two bidders and two types. Here are

several remaining questions you might be interested in.

Question. If we relax the assumption that low type never bids, shall we get a different equilibrium?

Buyer Equivalence

One very important result of auction theory is that in both 1st price and 2nd price auctions, buyer’s expected

payoffs are the same in equilibrium. From the above example, we can easily see that the expected payoffs in

both auctions equal to 2

f (θH) · f (θL) [v (θH)− v (θL)]

The result requires buyers being risk neutral and is not limited within these two types of auctions. We will

discuss it in the future.

Question. If buyers are risk-averse, how will equilibria of 1st and 2nd price auction be different?

2Recall that in 2nd price auction, buyers will bid their true value

-6-


Single Crossing Property and Monotonicity

Recall the above problem, a bidder’s utility function can be represented as

ui (w, b; θ) = w (v (θ)− b)

And the slope of indifference curve is the marginal rate of substitution. (MRS)

MRSi (w, b; θ) ≡ db

dw

∣∣∣∣u

= −∂Ui/∂w∂Ui/∂b

=v (θ)− b

w

In general, the marginal utility of quantity is on numerator and that of price is on denominator. We can see

that, for the same (w, b), higher v gives higher MRS and hence higher slope. For any different types, each

of their indifference curves cross each other at most once.

Definition 1. Single crossing property is the condition of utility that higher indexed types have stronger

preference for q. To be specific, type θt has stronger preference for q than θs if for any (q, r) and (q, r) with

q < q, we have

(q, r) �s (q, r)⇒ (q, r) �t (q, r)

Lemma 5. Preference rankings under SCP

Suppose q < q, if the SCP holds, then (i) (q, r) �s (q, r) ⇒ (q, r) �t (q, r) for all t > s; (ii) (q, r) �t(q, r)⇒ (q, r) �s (q, r) for all s < t.

Proof. Condition (i) is just by definition. For (ii), assume not, i.e.

(q, r) �s (q, r)

since q < q, we must have

(q, r) �t (q, r)

Contradictory!

In the auction framework, q, quantity can be treated as expected number of items bought (of course it

is smaller than one), which is equivalent to winning probability. Assume low type weakly prefers (b, w) than

(w′, b′), i.e.

w (v (θL)− b) ≥ w′ (v (θL)− b′)

or

(w − w′) v (θL) ≥ wb− w′b′

obviously, we can see that if w > w′

(w − w′) v (θH) > wb− w′b′

That is, high type strict prefers (w, b) if w > w′. According to the following graph, (b, w) and (b′, w′) is

-7-


indifferent to low type but high type strictly prefers the one with higher winning probability.

In next few weeks, you will see many utility functions that satisfies single crossing property. Classical

examples include Education Signalling (Spence), Insurance (Rothschild-Stiglitz), etc. Under single crossing

property, the most important quality of equilibrium is monotonicity.

Lemma 6. If (wt, bt) is the best response for type t and (ws, bs) is the best response for type s j. If vs < vt,

then ws ≤ wt.

Proof. Assume not, i.e. ws > wt. Best response means that

wt (vt − bt) ≥ ws (vt − bs)

ws (vs − bs) ≥ wt (vs − bt)

The first inequality can be modified

wt (vt − bt)− wt (vt − vs) > ws (vt − bs)− ws (vt − vs)

wt (vs − bt) > ws (vs − bs)

Contradictory!

-8-


Trembling Hand Equilibrium

Definition 2. Consider a game with finite strategy sets. A strategy profile s is a trembling hand perfect

equilibrium if there exists a sequence of totally mixed strategy profiles {sn} where sn converges to s such that

for all j we have

uj(sj , s

n−j)≥ uj

(sj , s

n−j)

Notice that, to make a strategy profile THP, we only need to find one sequence of {sn}. The strategies

must be totally mixed. Also, we need to fix player j’s strategy at sj and tremble other player’s strategy.

The intutition of the refinement is that, one still want to follow the strategy even when others are making

small mistakes. Let’s consider a simple modification of Battle-of-Sexes

Boy

O M

Girl O 1 , 1 2 , 0

M 0 , 0 2 , 2

Obviously, (O,O) and (M,M) are both pure NE. But are both THP? Start from (O,O). If the boy deviates

with (1− ε, ε), the girl’s expected payoff of playing O is (1 + ε) and that of playing M is 2ε. When ε is small

enough, playing O dominates playing M . On the other hand, if the girl plays (1− ε, ε), the expected payoff

of the boy for playing O is 1− ε and that of playing M is 2ε. Same conclusion. So (O,O) is THP. In terms

of (M,M), assume the boy plays (ε, 1− ε), the girl’s expected payoff of playing O is ε+ 2 (1− ε) and that

of playing M is 2 (1− ε). No matter how small ε is, playing O is strictly better. Hence, (M,M) is not THP.

-9-


Week 2

All Pay Auction

So far, we have studies several types of auctions. In the lecture, you have seen how to solve for sealed first

price auction under continuous type model. In this handout, let’s discuss All Pay Auction, in which, as the

name suggests, each agent submits a bid; the highest bidder wins the item and every agent pays their bid.

Model

There is one item to be sold to j = 1, 2, 3...J risk neutral buyers. Each buyer can of type θ drawn from a

continuous distribution with support Θ = [α, β]. The corresponding CDF and PDF are F and f , respectively.

To solve for the equilibrium strategy profile, which is a bidding function bj(θ), we first assume they follow

certain properties, solve for the bidding function and then verify the assumptions.

Assumption. We assume that

• The equilibrium bidding function bj(θ) is strictly increasing and continuously differentiable.

• Equilibrium is symmetric: bj(θ) = b(θ),∀j.

Value Functions

Given the assumptions above, the type and bidding price are of a monotonic bijection. So the winning

probability in equilibrium becomes

w(θi) = Pr(θj ≤ θi|i 6= j) = F J−1(θi)

The equilibrium payoff of type θi buyer is

U(θi) = w(θi) · θi − b(θi)

The derivative of the equilibrium payoff function is

U ′(θi) = w(θi) + w′(θi) · θi − b′(θi) (1)

In order to characterize equilibrium we need to define the utility function that someone deviates from θ.

Define

u(b(θ′); θ) ≡ w(θ′) · θ − b(θ′)

be the utility function that type θ buyer bids the price that is supposed to be used by type θ′ buyer according

to b

Notice that by the assumption that b is strictly increasing, bidding b(θ) is equivalent to reporting that

your type is θ. Therefore, the above utility function can be simplified as

u(θ′; θ) = w(θ′) · θ − b(θ′)

-10-


Obviously,

U(θ) = u(θ; θ)

The setting of U and u will be used repeatedly when you are solving these type of models, so please define

them clearly.

Equilibrium winning probability

By the above settings, we can achieve our first result - in equilibrium, the winning probability of type θ

equals to marginal utility of the type,

w(θ) = U ′(θ)

To achieve the result, there are two approaches. First, given b(θ), one’s utility function is

u(θ′; θ) = w(θ′) · θ − b(θ′)

BNE requires that reporting true type is optimal for each type, i.e.

∂u(θ′; θ)

∂θ′

∣∣∣∣θ′=θ

= w′(θ) · θ − b′(θ) = 0

Put it back to equation (1) above, we have U ′(θ) = w(θ). Alternatively, we can get the same argument

graphically or by envelope condition. To see this, let’s consider a player deviates his strategy to always

pretending to be type θ′, b(θ) = b(θ′). Then his winning probability is always w(θ) since other players stick

to b. So his utility function is

u(θ; θ′) ≡ u(θ′; θ) = w(θ′) · θ − b(θ′)

Though this utility function appears exactly the same as u, now θ′ is the parameter and θ is the independent

variable of the function. Since U is the equilibrium payoff, u should be below U everywhere but at θ′ = θ.

It is true for every θ. Or, consider the function δ(θ; θ′) ≡ U(θ)− u(θ′; θ). It should be non-negative for all θ

and equal to zero at θ′. Which means that

∂δ(θ; θ′)

∂θ

∣∣∣∣θ=θ′

= 0

-11-


Therefore, the slope of U(θ) should be the same as that of u(θ; θ′) at θ = θ′. The slope of u is exactly w, so

we draw the same conclusion. The following graph demonstrates the facts stated above.

Notice that since w(θ) = F J−1(θ), U ′(θ) is increasing or U(θ) is convex.

Bidding Function

Next step is to solve for the optimal bidding function b. By w(θ) = U ′(θ), we have

U(θ) =

∫ θ

α

U ′(x)dx =

∫ θ

α

w(x)dx

By definition of U , we have b(θ) = w(θ) · θ − U(θ), the optimal bidding function is

b(θ) = w(θ) · θ −∫ θ

α

w(x)dx = w(θ) · θ −∫ θ

α

F J−1(θ)dx

At this point, if we know the functional form of F , the question is solved. Conventionally, in most cases, we

do one step further - integrating the above term by parts, we obtain

b(θ) = θw(θ)− xw (x)|θα +

∫ θ

α

xw′(x)dx

= αw(α) +

∫ θ

α

θw′(θ)dθ

=

∫ θ

α

θw′(θ)dθ

= (J − 1)

∫ θ

α

θF J−2(θ)f(θ)dθ

Notice that we crucially assumed that the lowest type α cannot win, which is true when b is strictly increasing.

By the expression above, we can see that it is indeed the case.

b′(θ) = (J − 1)F J−2(θ)f(θ)dθ > 0

-12-


So we are done.

Buyer Equivalence

Recall that in both 1st and 2nd price auction, the expected buyer payoff in the symmetric equilibria are

equal for each type. It is also true for all pay auction. The reason is simple. Recall that, in all three types

of auctions we’ve seen, the following is always true

U ′(θ) = w(θ) = F J−1(θ)

So by taking integral, we can find the equilibrium payoff. Further, since the item is always sold to the buyer

with highest value, the social surplus is maximized. Thus, seller’s revenues should be equivalent, too.

The following problem is a practise of the continuous type auction problem. Now we have more than one

item to be sold.

Question. Sealed 2nd price auction

Based on all the information in all pay auction example, now consider sealed 2nd price auction.

1. Show that the equilibrium payoff function U(θ) satisfies U ′(θ) = w(θ)

2. Show that the equilibrium bidding function is

b(θ) = (θ) = θ − 1

w(θ)

∫ θ

0

w(x)dxcvh

3. Given the type is distributed uniformly on [0, 1], find the functional form of equilibrium bidding function

b and payoff function U(θ).

Question. Continuous value, 2 items and 3 players

Assume that there are 2 identical item to be sold. There are 3 players whose type is iid draw from

distribution F (θ) on [α, β]. Each of the two highest bidders wins an item and pays their own bids.

1. Show that winning probability of type θ is w(θ) = 1− (1− F (θ))2.

2. Show that the equilibrium payoff function U(θ) satisfies U ′(θ) = w(θ)

3. Show that the equilibrium bidding function is

b(θ) = θ − 1

w(θ)

∫ θ

0

w(x)dx

4. Given the type is distributed uniformly on [0, 1], find the functional form of equilibrium bidding function

b and payoff function U(θ).

-13-


Price Discrimination - Cellphone Plan

In two-part pricing example, consumers pick a plan with certain entry fee K and choose how much to

consume q given unit price p. In this example, let’s consider that each consumer sees a list of plans (q,R) in

which q is the service quantity and R is the lump-sum cost. Each consumer is of a type θ with probability

f(θ). We use p(θ, q) to represent type θ consumer’s marginal utility. Assume that

∂p

∂θ> 0 and

∂p

∂q< 0

The benefit of consuming q equals to

B(q; θ) ≡∫ q

0

p(θ, x)dx

And the utility function of type θ choosing package (q,R) is

u(q,R; θ) = B(q; θ)−R

Single Crossing Property

To conduct the analysis, first we need to check whether single cross property holds or not in this example.

Single crossing property, as it is called, indicates that the indifference curves of difference types cross once

and only once. The following is the definition of SCP under the setting of the cellphone plan example mode.



q < q, we have

(q,R) �s(q, R

)⇒ (q,R) �t

(q, R

)Consider about two types θH > θL and certain level of utility uH and uL, respectively. Then the

indifference curves are

R = B(q; θ)− uθ =

∫ q

0

p(θ, x)dx− uθ

Then the slope of indifference curve is equal to

∂R

∂q= p(θ, q)

which is increasing in θ. Therefore, for any q, the slope of high type consumer should be steeper. Hence, as

long as the two indifference curves cross each other, there can be only one interaction3.

uH − uL =

∫ q

0

p(θH , x)− p(θL, x)dx

By the definition of SCP, let’s consider about two plans (R1, q1) and (R2, q2) which are indifferent to

low type. In addition, we have q1 < q2. To check SCP, we need to show that the high type strictly prefers

3To be strict, we need to show an interaction indeed exist, in this example, we can simply assume that p has good propertieswhich make it indeed the case

-14-


(R2, q2).

B(q1; θL)−R1 = B(q2; θL)−R2

Hence

R2 −R1 = B(q2; θL)−B(q1; θL)

By definition of B, we have

R2 −R1 =

∫ q2

0

p(x; θL)dx−∫ q1

0

p(x; θL)

=

∫ q2

q1

p(x; θL)dx

Since p is increasing in θ, we have

R2 −R1 <

∫ q2

q1

p(x; θH)dx

=

∫ q2

0

p(x; θH)dx−∫ q1

0

p(x; θH)

B(q1; θH)−R1 < B(q2; θH)−R2

Therefore, the model does satisfy SCP. The following graph illustrates the relationship. The green line, high

type’s indifference curve is steeper for every q than the low type’s. So when the two packages are indifferent

to low type (both are on the blue line), then, the high type agent strictly prefers (q2, R2) whose quantity is

larger.

Why the single crossing property is important? One of the reasons is that SCP guarantees monotonicity.

In the language of the cellphone plan model, in BNE, higher type must purchase the package with higher

quantity. It can be summarized in the following lemma.

Lemma 7. If (q1, R1) and (q2, R2) are the equilibrium packages designed for type θ1 < θ2, then q2 ≥ q1.

Proof. The proof can be done by directly using SCP. Skipped.

Notice that, the above lemma can be implemented in not only two type case, but also discrete type or

-15-


even continuous type situation. More detail regarding single crossing property will be introduced in future

classes.

Now we can characterize equilibria given that there are only two types θH > θL. You can treat the

model as a sequential game. First, the monopolistic seller designs contracts. Then, the consumer chooses a

contract based on his own type. Of course, a consumer can choose nothing. The seller can either design only

one contract or a list of contracts. Which strategy will be chosen depends on the parameters of the game,

such as type distribution f(θ), functional form of B and so on. In the following sessions, we discuss the two

situations one by one.

Equilibrium with 1 contract

If there is only ONE contract offered, we have two situations. First, both types join. Second, high type takes

the offer and low type chooses nothing. The constraint (R, q) that both types join is

B(q; θL)−R ≥ 0

B(q; θH)−R ≥ 0

The above equation is solved by the constraints. That is, if we want both types to join, we only need

low type’s participation constraint to be satisfied. Further, in order to maximize profit, firm can bind the

constraint, B(q; θL)−R = 0. Firm’s profit function is

Π = maxq,R

R− cq = B(q; θL)− cq

First order condition tells us that

∂B(q; θL)

∂q= p(q; θL) = c

That marginal utility equals to marginal cost implies that the quantity is efficient for low type. However, it

is lower than high type’s efficient quantity. Such strategy will take all surplus from low type but high type

will earn a positive net payoff.

Next, if the contract is only attractive to high type, the high type’s participation constraint should be

binding for profit optimization purpose. In this case, firm’s expected payoff is f(θ)(B(q; θH)− cq) and first

order condition is∂B(q; θH)

∂q= p(q; θH) = c

The contract has the efficient quantity for high type consumers. But the contract is too expensive to low

type.

Equilibrium with 2 contracts

Now let’s consider the situation in which the firm designs 2 contracts to do indirect price discrimination.

Denote (qL, RL) and (qH , RL) to be the contracts designed for high and low type consumers, respectively.

To be optimal, the contracts have to satisfy two groups of constraints. Firstly, both types should be willing

-16-


to participate. The term used in literature is Individual Rationality Constraint (IR).

B(qH ; θH)−RH ≥ 0 (IRH)

B(qL; θL)−RL ≥ 0 (IRL)

Secondly, upon participating, each type should (weakly) prefer the contract designed for them. We call this

type of constraint Incentive Compatibility Constraint (IC).

B(qH ; θH)−RH ≥ B(qL; θH)−RL (ICH)

B(qL; θL)−RL ≥ B(qH ; θL)−RH (ICL)

Therefore, firm’s problem is

max(qL,RL)(qH ,RH)

f(θL)(RL − cqL) + f(θH)(RH − cqH)

subject to ICH, ICL, IRH and IRL. This program has four constraints so is impossible to solve by hand

without simplification. By doing the following exercise, you can show that ICL and IRH are redundant.

Question. To simplify the program,

1. Show that if ICH and IRL are satisfied, then IRH is satisfied.

2. Show that if ICH and IRL are binding, then ICL is satisfied.

By the above results, we have only two constraints left - ICH and IRL. If the firm is profit maximizing,

this two constraints must be binding4, which tells us that

RL = B(qL; θL)

RH = B(qH ; θH)−B(qL; θH) +B(qL; θL)

Plug them into firm’s objective function, we have

maxqL,qH

f(θL)(B(qL; θL)− cqL) + f(θH)(B(qH ; θH)−B(qL; θH) +B(qL; θL)− cqH)

By taking FOC, we have

p(qH ; θH) = c

p(qL; θL) = c+f(θHf(θL)

· (p(qH ; θH)− p(qL; θH) + p(qL; θL))

The above question is asking that, in such package design problem, if we consider the per unit price of

the good, who should enjoy a better price? Higher type or lower type?

4not, the firm can simply increase payment R without violating any other constraints. It is why we can impose the bindingassumptions for the question above

-17-


Conclusive Points

We have been through a long analysis, so let’s do a summarize.

1. The cellphone plan design problem satisfies SCP, which indicates monotonicity.

2. If firm designs only one package, it can either rule out low type or attract both types.

3. If firm designs 2 contracts, ICH and IRL are binding. Low type earns zero payoff, while high type earn

positive payoff which is called informational rent.

4. High type’s contract has optimal quantity while low type’s contract has quantity lower than optimal

level.

-18-


Week 3

Buyer Equivalence and 2nd Price Auction

Buyer Equivalence indicates that as long as an auction satisfies the followings

1. The number of items to be sold are fixed.

2. Players’ type distributions are independent and identical.

3. The items will be given to players with highest bid.

4. Reserve price is fixed5.

then a buyers’ expected payoffs should be the same. The key to the theorem is that we have w(θ) = U ′(θ),

so

U(θ) =

∫ θ

α

w(x)dx (1)

In this section, we use this result to derive the bidding function of second price auction. Denote the function

as b(θ). Then type θ bidder’s expected payoff equals to

U(θ) = w(θ)(θ − E[b(θ)|θmax = θ]

where θ is the second highest type according to the rule of 2nd price auction.. In order to derive the expression

of the utility function, we first consider distribution function of θ|θmax. Notice that θ|θmax ∈ [α, θ]. Denote

the CDF as F (x|θ),

F (x|θ) ≡ Pr(θ ≤ x|θmax = θ)

=Pr(1 buyer has type θ AND J − 1 players has type lower than x)

Pr(θmax is θ)

=f(θ) · F (x)J−1

f(θ) · F (θ)J−1

=w(x)

w(θ)

Therefore, the expectation of θ|θmax can be written as

E[b(θ)|θmax = θ] =

∫ θ

α

b(x)d

(w(x)

w(θ)

)Therefore, the equilibrium utility function can be written as

U(θ) = w(θ)

[θ −

∫ θ

α

b(x)d

(w(x)

w(θ)

)]5Optimal reserve price problem will be discussed later.

-19-


Integrating by part, we have

U(θ) = w(θ)

[θ − b(x)

w(x)

w(θ)

∣∣∣∣θα

+

∫ θ

α

b′(x)w(x)

w(θ)dx

]

= w(θ)θ − w(θ)b(θ) +

∫ θ

α

b′(x)w(x)dx

Combining equation (1), we have

w(θ)θ +

∫ θ

α

b′(x)w(x)dx =

∫ θ

α

w(x)dx+ w(θ)b(θ)

Taking derivatives of both sides w.r.t. θ, we get exactly b(θ) = θ.

-20-


Signalling Game and SCP

Briefly speaking, Signalling game is a circumstance in which one player (signaller) with private information

(type) can send signal about his/her information to other players(responders) who pay the signaller according

to the signal received and beliefs.

In the lecture, the model is described as follows.

Phase One - Signaller’s Action

Signaller draws his type from Θ = [α, β] according to distribution function f(θ). Then, he decides whether

to participate or not. Denote S ∈ Θ as the set of participation - if θ ∈ S, signaller will move on. Otherwise,

the game stops.

If the signaller decides to continue, he chooses a costly signal quality q(θ). In general, the cost function

is increasing in q and decreasing in θ.

Phase Two - Responder’s Action

Observing the signal quality, responders update their belief on signaller’s type by Bayes Rule, and determine

how much r to pay to the signaller. The payment scheme is model dependent. For instance, if there is

only one responder, it will behave as a monopolist. If there are multiple responders, the structure could be

complicated. One of the options is to implement Bertrand price competition - Responders pay the signaller

his expected marginal productivity. (See the following example)

Generally speaking, the signaller’s payoff can be expressed as u(θ, q, r).

Single Crossing Property

Now we discuss SCP of the model. Recall the following definition from week 1’s handout.



q < q, we have

(q, r) �s (q, r)⇒ (q, r) �t (q, r)

In the lecture, we imposes two assumptions on u(θ, q, r).

Assumption. The utility function satisfies

(i) Signaller prefers higher r.∂u

∂r> 0

(ii) The slope of indifference curve is decreasing in θ

∂

∂θ

(−∂u/∂q∂u/∂r

)< 0

Alternatively, it means that the marginal rate of substitute (MRS) is increasing in θ.

-21-


Based on the two assumptions, we can conclude that higher type signallers have higher preference on q.

Notice that, we only assume u is increasing in r but nothing is imposed on q. The following two graphs

illustrate that no matter whether u is increasing or decreasing in q, ”Higher type has stronger preference on

q” always holds.

Left graph illustrate a circumstance similar to Education Signalling model in which education is costly but

better worker can achieve a degree more easily. Second graph shows a two-good trade off situation (e.g.

U = log r + θ log(q)).

So far, since we are discussing signalling model, we always assume that u is increasing in r. Suppose u

is decreasing in r, for instance, insurance model, how the slopes of indifference curves change across types?

Remember that the principle that ”Higher type has stronger preference on q” always holds.

A Simple Example

Now we discuss a simple model with two types of workers (signallers) θ ∈ 1, 2, which denote the marginal

productivity. Also, the cost of getting a degree (signal) is costly

C(θ, q) =q

θ

The employers (responders) hire the worker and pay him wage r. Thus, the utility function is u(θ, q, r) =

r − q/θ. Suppose employers use Bertrand wage competition, that is, the payment is exactly the expected

marginal productivity of the worker, then in a separating equilibrium, employer with type θ’s utility function

will be

U(θ) = θ − q(θ)

θ

However, under pooling equilibrium, the high type will be paid blow his marginal productivity while low

type will be over-paid. That’s why high type has incentive to distinguish himself from low type, and low

type wants to mimic high type.

Now let’s consider the separating BNE and PBE of the game, respectively. Recall that BNE only requires

belief on the equilibrium path while PBE requires that on all the possible off-equilibrium paths, the beliefs

and actions have to be consistent.

To solve for equilibrium, we first write down all the constraints. Participating constraints requires that

U(θ) = θ − q(θ)

θ≥ 0

-22-


for both types, since at least all workers can earn 1. Incentive Compatibility constraints are as follows

2− q(2)

2≥ 1− q(1)

2

1− q(1)

1≥ 2− q(2)

1

As long as the above constraints are satisfied, the strategy profile is BNE. For instant, (q(1) = 0.5, q(2) = 2)

is one of them. Notice that BNE does not consider off-equilibrium path situation, so given the equilibrium

(q(1) = 0.5, q(2) = 2), we never ask ”what if a signal q = 1 shows up?” under BNE.

However, is (q(1) = 0.5, q(2) = 2) a PNE? No! In a PBE, we have to take care of all the possible

deviations. Consider the worst belief in which as long as a signal other than 0.5 or 2 is observed, one will be

treated as low type. Then, given the strategy profile (q(1) = 0.5, q(2) = 2), low type has incentive to send a

signal zero.

Therefore, in any separating PBE, q(1) = 0.

Question. Show that any strategy profile ”q(1) = 0, q(2) = q, q ∈ [1, 2]” and belief that ”q 6= 0or q ⇒ θ = 1”

consist a PBE.

-23-


Monopoly and Product Quality (COMP Fall ’09)

Each consumer purchases either 0 (not participate) or 1 unit of commodity. A type t buyer’s utility gain

from paying a price r for a unit of quality q, ui(q, r), is strictly increasing in q and strictly decreasing in r.

Moreover, we assume that for all t > s

−∂ut∂q

/∂ut∂r

(q, r) > −∂us∂q

/∂us∂r

(q, r)

at any point (q, r). Thus, the SCP holds. Denote the fraction of the population of type t is ft, t = 1, 2, ...T .

The cost of producing a unit of good of quality q is cq. Any offer made to one customer must be made to

all customers - no directly price discrimination.

Let r = Rs(q) be the indifference curve for type s through (q′, r′) and (q′′, r′′) where q′′ > q′.

(a) With the help of a graph, explain why any type t > s strictly prefers (q′′, r′′) and any type t < s

strictly prefers (q′, r′).

Solution: Skipped.

(b) Prove the statement in part (a).

Solution: This part requires us to interpret the graphical implication mathematically. We need to

consider two cases - t > s and t < s. For case one, it is equivalent to show that if type t > s is

indifferent between (q′, r′) and (q′′, r′′t ), then r′′t > r′′. Notice that

R′s(q) = −∂us∂q

(q,R(q))/∂us∂r

(q,R(q))

So for type s, we have

r′′ = r′ +

∫ q′′

q′R′s(q)dq

Correspondingly, for type t, we have

r′′t = r′ +

∫ q′′

q′R′t(q)dq

where Rt is the indifference curve of type t and goes through (q′, r′). And we want to show that rt > r.

Notice that it is not that obvious since we cannot simply claim that R′t(q) > R′s(q) for all q ∈ (q′, q′′)

(Why?)

Here is the strict proof. Assume the opposite, r′′t < r′′. Since at (q′, r′), we have R′t(q′) > R′s(q

′)

since both indifference curves go through (q′, r′). Thus, when the quality q∗ is closed enough to q′, by

continuity, on the interval [q′, q∗], R′t(q) > R′s(q), then we have

r∗s = r′ +

∫ q∗

q′R′s(q)dq < r∗t = r′ +

∫ q∗

q′R′t(q)dq

Hence, by Intermediate value theorem, there must exist a point q∗∗ such that

∫ q∗∗

q′R′s(q)dq =

∫ q∗∗

q′R′t(q)dq

-24-


But by the same argument, on a small interval [q∗∗, q∗∗+ε], the slope of Rt is larger everywhere. Thus,

we never could reach r′′t < r′′, which gives us contradiction.

Question. We just show that r′′t < r′′ is impossible. What if we assume that r′′t = r′′? (Hint: the

approach is similar)

By exatly the same method, we can show case two - t < s.

(c) Explain briefly why, for the direct mechanism {(qt, rt)}Tt=1 to be inventive compatible, {qt}t must be

increasing.

Solution: The statement essentially argues that with stronger preference in q, higher type must receive

higher q in equilibrium. The proof is simple. Assume the opposite, that is, there exist a pair of s < t

but qs > qt. By IC constraint, we have

(qs, rs) �s (qt, rt)

Since qs > qt, by SCP, we have

(qt, rt) �t (qs, rs)

But it violates type t’s IC constraint. Contradiction!

(d) Consider any direct mechanism {(qt, rt)}Tt=1 which is increasing in qt and for any t > 1, we have

(qt, rt) ∼t (qt−1, rt−1)

so called local downward constraints are binding. Show that the mechanism is incentive-compatible.

Solution: The proof consists of two parts6. First, each type t will not want to deviate upward to

any t+ s. Second, each type t has no incentive to deviate downward to any t− s.

The logic of the proof is that, first, we use local constraints to show

(qs−2, rs−2) �s−2 (qs, rs)

Then, by the same method, we can show t− 3...t− k iteratively.

Since the local downward constraints are binding, for type s, we have

(qs, rs) ∼s (qs−1, rs−1)

By monotonicity of qt and SCP we have

(qs, rs) ≺s−1 (qs−1, rs−1) (2)

Also, we have

(qs−2, rs−2) ∼s−1 (qs−1, rs−1)

6I will show the first part and the second part is for your practise.

-25-


which tells us that

(qs−1, rs−1) ≺s−2 (qs−2, rs−2)

Assume type s− 2 prefers (qs, rs) to (qs−2, rs−2), then

(qs, rs) �s−1 (qs−2, rs−2)

By the locally binding constraint of type s− 1, we conclude that

(qs, rs) �s−1 (qs−1, rs−1)

It is contradictory to (1). Hence, we conclude that any type t has no incentive to mimic type t + 2.

Next step, we use the same approach to show that any type t has no incentive to mimic type t+3...t+k.

Then we finish the proof.

(e) If the seller is a monopolist and there are only two types {1, 2}, solve for the profit-maximizing quality

choice (q1, q2) if (f1, f2) = (3/4, 1/4), ut(q, r) = θtq−0.5q2−r, where (θ1, θ2) = (10, 30) and production

cost c = 2.

Solution: This question is almost the same as what we’ve discussed on week 2. Follow the steps and

solve for it.

• Write down the seller’s problem subject to necessary constraints.

• Show low type’s IR constraint is binding.

• Show high type’s IC constraint is binding.

• Substitute rt by (q1, q2).

• Solve for optimal q. (Hint: (q1, q2) = (4/3, 28)).

-26-


Week 4

BNE vs PBE

In this section, let’s use a simple example to see the difference between Bayesian Nash Equilibrium and

Perfect Bayesian Equilibrium.

Definition 5. A Bayesian game consists of

• A set of players I;

• A set of actions for each player i: Si;

• A set of types for each player i: θi ∈ Θi;

• A payoff function for each player i: ui(s, θ), where s = (s1, s2, ..., sI), θ = (θ1, θ2, ..., θI);

• Probability distribution fi(θ) for player i’s type. For simplicity, we assume that the distributions are

independent across players.

Definition 6. The strategy profile s is a (pure strategy) Bayesian Nash Equilibrium if for all i ∈ I and

for all type, we have that

si(θi) = arg maxs′i∈Si

∑θ−i

f−i(θ−i)ui(s′i, s−i(θ−i), θi, θ−i)

In words, BNE says that given other’s strategy, each player maximizes his/her expected utility. Notice

that the definition of BNE is based on a static game with incomplete information. Although it can be used

on sequential games, there are many limitations. Consider the following signalling game with two types and

try to find some BNE’s.

Example - Signalling Model w/ 2 Types

There are two types of workers with private type set Θ = {θ1, θ2} = {1, 2}. A worker with type θ can send a

signal q at cost q2/(2 + θ). There is a competitive firm who observes worker’s signal and pays worker r ≥ 0.

Thus, the utility function of the worker is

u(θ, q, r) = r − q2

2 + θ

In principle, the firm is also a player and has its own utility function. But here, for simplicity, assume that

a firm’s BEST response is to pay the worker R(θ, q) = θ + 2q, where θ is firm’s belief on the worker’s type

based on signal. But we have not defined ”belief” yet, you can treat θ to be a premium paid to a worker

when observing q. So we can denote R(θ, q) = R(q) = θ(q) + 2q. Also, worker’s outside option is u0(θ) = 1.

(a) In such game, what’s strategy profile of the game? Show that (q1, q2) = (3.5, 5) can be part of a BNE

strategy profile.

Solution: The strategy for the worker is simply the signal quality q given his type θ, q∗(θ). The

firm’s strategy is that by observing each q, a payment r∗(q) needs to be chosen. Thus, the strategy

-27-


profile of the game is

(q∗1 , q∗2 , r∗(q))

According to definition of BNE, given other’s strategy, one is maximizing his/her own expected utility.

To show (q1, q2) = (3.5, 5) is part of of BNE, we need to find a payment function of firm r∗(q).

According to BNE definition, firm needs only to respond optimally at q = 3.5, 5. For the rest of q’s,

whether firm is best responding or not is not required by BNE. BNE only restricts that workers has

no incentive to deviate from q∗(θ). One of most naive strategies of a firm is.

r(q) =

1 + 2 · 3.5 if q = 3.5

2 + 2 · 5 if q = 5

0 otherwise

That is, for any other q, the responses the firm simply pays zero. If so, obviously, no worker has

incentive to deviate to those off-equilibrium path signals. The only possible signals that show up are

3.5 and 5. The next step is to check whether each type has incentive to deviate across 3.5 and 5.

Basically, we need only to check two IC constraints. I will leave the computation work to you.

Here is an analogy. When we study Nash equilibrium, we know that it cannot rule out a non-credible

threat in a entry/exit game, since the game is no longer static. So we introduce the concept of subgame perfect

equilibrium. In incomplete information case, similar issue arises - BNE does not impose any restriction on

off-equilibrium path responses. To fix the imperfection, we need a equilibrium concept for sequential game

with incomplete information (dynamic Bayesian game) to rule out such naive Bayesian Nash equilibria.

Beforehand, let’s first formally define what’s Dynamic Bayesian Game.

Definition 7. In addition to the element in normal Bayesian game, a Dynamic Bayesian Game also

have

• A sequence of histories at each stage of the game.

• An information partition, which determines which of the histories assigned to a player are in the same

information set.

In the signalling model framework, the history is the signal a firm observes. The information partition is

simple. Since all signal that is observed can be chosen by each type, all points are in the same information

set.

In addition, we need to define belief system. In words, it is an assignment of probabilities to every node

in the game such that the sum of probabilities in any information set is one. A belief system is consistent

for a given strategy profile if and only if the probability assigned by the system to every node is computed

by Bayes’ Rule.

One more concept to be introduced is sequential rationality. A strategy profile is sequentially rational

at a particular information set for a particular belief system if and only if the expected payoff of the player

whose information set it is (i.e. who has the move at that information set) is maximal given the strategies

played by all the other players. A strategy profile is sequentially rational for a particular belief system if it

satisfies the above condition for every information set7.

7The concepts are copied from Wikipedia, but it is enough.

-28-


In the context of signalling game, consistent belief means that given a strategy profile (q1, q2), the firm’s

belief function B(q) at least has two points B(q1) = θ1, B(q2) = θ2. Sequential rationality means that, given

a belief θ = B(q), the firm is always best responding - r = R(B(q), q).

Definition 8. Perfect Bayesian Equilibrium is a combination of strategy profile and a belief system

such that, given the strategy profile, the belief is consistent; given the belief system, the strategy profile is

sequentially rational.

Hence, PBE rules out the possibility that firm naively pays zero, since it is not sequentially rational to

any belief.

Example (cont’d)

In addition to the information given in previous part, since we have formally defined ”belief”, now we can

define firm’s best response - R(q, θ) = q + 2θ.

(b) Confirm that, in any PBE, (q1, r1) = (3, 7).

Proof: To show the statement, we prove by contradiction. First, we assume there is a PBE who

violates the statement. If we can show that for any belief, there is a profitable deviation, then we find

the contradiction. Here is a formal proof.

Assume not, i.e. there exist a PBE in which, q1 = q 6= 3. Then, r1 = 1 + 2q 6= 7 automatically. Denote

the belief function of the PBE as B(q). Notice that at least, B(q) = θ1. Therefore, the equilibrium

payoff for low type is

u(θ1) = 1 + 2q − q2/3

It is easy to verify that

1 + 2q − q2/3 < 1 + 2 · 3− 32/3 ≤ B(3) + 2 · 3− 32/3

The first inequality argues that, if the belief on q = 3 is θ1, then, low type has incentive to choose

q = 3. Second inequality claims that if the belief on q = 3 is some type higher 8, the incentive of

deviating to q = 3 is even larger. Thus, we conclude that, for any belief function B(q), playing q 6= 3

is worse than q = 3. Then, q 6= 3 cannot be part of any PBE - contradiction! �

(c) Find all PBE strategy profiles.

Solution: To claim that a strategy profile is part of PBE, we need only to find one belief function

which is consistent with the strategy profile. As shown in part (b), (q1, r1) = (3, 7) must be true.

Then, what we need is to find all possible (q2, r2). To make type 2 has no incentive to deviate, the

belief should be the ”worst” - as long as equilibrium q2 is not observed, one is to be believed as a low

type worker9

B(q) =

{θ2 if q = q2

θ1 if q 6= q2

8this model, the only possibility is θ2, but we can easily generalize it.9It is not a good belief, but PBE does not say anything about it.

-29-


The criteria for the PBE are, ∀q′ 6= q2,

2 + 2q2 −q22

4≥ 1 + 2q′ − q′2

4

1 + 2 · 3− 32

3≥ 2 + 2q2 −

q22

3

The first inequality is the incentive compatible constraint for high type. The second constraint is the

IC constraint for low type.

Question. In class, I wrote the following constraints.

2 + 2q2 −q22

4≥ 1 + 2 · 3− 32

4

1 + 2 · 3− 32

3≥ 1 + 2q2 −

q22

3

Unfortunately, they are wrong. Why?

Notice that, the term 1 + 2q′ − q′2

4 takes maximum equal to 5 at q′ = 4. Therefore, the two criteria

restrict the range of q2 within [3+√

3, 6]. All strategy profile [(3, 7), (q2, 2+2q2)] where q2 ∈ [3+√

3, 6]

will be part of the PBE given the ”worst” belief.

-30-


PBE and Intuitive Criterion

In previous section, we’ve found all the (pure) PBE strategy profiles. However, they are all under the ”worst”

belief, which is odd in terms of common sense. We need further refinement of PBE. The main idea of the

refinement is that, given a PBE, if some player deviates to an strategy which is not part of the strategy

profile, he will be treated as a ”credible” type in stead of simply the worst type10. Given a PBE, let U(θ) be

the equilibrium payoff and q be a strategy that is never played by any type in the equilibrium. Also, define

function

u(θ, θ, q)

be the payoff of a type θ player who chooses q and is believed to be type θ.

Definition 9. A belief function B is credible if there exist a type θ∗ such that

(i) B(q) = θ∗

(ii) u(θ∗, θ∗, q) > U(θ∗)

(iii) u(θ, θ∗, q) ≤ U(θ) for all θ 6= θ∗.

The PBE fails Strong Intuitive Criterion if there exists a credit belief.

Example (cont’d)

(d) Show that a PBE with (q2, r2) = (5, 12) fails S.I.C.

Proof: To show the statement, we need to fine a credible belief. Consider belief B in which B(3+√

3) =

θ2. Then we can easily see that11

u(θ2, θ2, 3 +√

3) > U(θ2)

u(θ1, θ2, 3 +√

3) ≤ U(θ1)

Thus, such PBE fails credible constraint.

(e) Show that all PBE with q2 > 3 +√

3 fails S.I.C.

Proof: For your practise.

With S.I.C, we rule out all the PBE’s except the one with binding local upward constraints. In addition,

such PBE is Pareto Efficient because higher type costs the least for education.

10Since professor Riley does not plan to extend such concept to general framework, we only focus on definitions of signallinggame version.

11Compute by yourself

-31-


Signalling model with continuous types

In this section, we review the signalling model with continuous types. In general, the model consists of the

following parts.

• A firm(employer) pays a worker with unknown type θ by observing a signal q. Since we are only

interested in PBE, the firm is always best responding to his belief and signal. Denote the best response

function as R(θ, q), where θ is the firm’s belief based on the signal.

• Each worker draws his type from the set Θ = [α, β] according to i.i.d F (θ).

• Each worker has a cost function of generating signal q,

C(θ, q) ≡ A(q)/φ(θ)

where A is increasing in q and φ is increasing in θ.

• Worker’s outside option is u0(θ).

There are 6 assumptions provided in Prof Riley’s lecture notes. Review what they are. In the following

analysis, we take the assumptions as given. Denote the utility function on equilibrium as

U(θ) = R(θ, q(θ))−A(q(θ))/φ(θ)

and utility of type θ who reports to be θ′ as

u(θ′; θ) = R(θ′, q(θ′))−A(q(θ′))/φ(θ)

By first order condition, we must have∂u

∂θ′

∣∣∣∣θ′=θ

= 0

Hence, we have

R′(θ, q(θ)) =A′(q) · q′(θ)

φ(q)

The marginal utility on equilibrium is

U ′(θ) = R′(θ, q(θ))− A′(q) · q′(θ)φ(q)

+A(θ)

φ(θ)2· φ′(θ)

By the first order condition we derived above, we have

U ′(θ) =A(θ)

φ(θ)2· φ′(θ)

Also, A(q(θ))/φ(θ) = R(θ, q(θ))− U(θ) by definition, we have

U ′(θ) = [R(θ, q(θ))− U(θ)] · φ′(θ)

φ(θ)

-32-


Rearrange the above term, we have

φ(θ)U ′(θ) + φ′(θ)U(θ) = R(θ, q(θ))φ′(θ)

It is obvious that the LHS of the above term is the derivative of φ(θ)U(θ). Thus, we can find a necessary

condition of the equilibrium,

φ(θ)U(θ) =

∫ θ

α

R(x, q(x))φ′(x)dx+K

So the equilibrium payoff takes the form

U(θ,K) =

∫ θαR(x, q(x))φ′(x)dx+K

φ(θ)

To find K, we need to specify outside option u0(θ), which depends on model. In general, u0(θ) should be

increasing in θ. See the next example.

PBE with Zero Participation

We use an example to show that, given certain sort of outside option, the unique PBE will be that all type

choose not to participate. Assume Θ = [0, 10], φ(θ) = θ2, R(θ, q) = θ and A(q) = q.

(a) Show that a necessary condition for PBE is that

U(θ,K) =2/3θ3 +K

θ2

The method is exactly the same as above.

Remark: Before moving on the part (b), notice that the full information equilibrium utility should

be V (θ) = θ since signalling is unnecessary. In any PBE, the equilibrium payoff cannot exceed the full

information utility V (θ) = θ.

(b) Assume the outside option is u0(θ) = 34θ. Show that the unique PBE is that S = ∅, where S is the set

of participation.

Proof: Assume not, i.e. there exists a K which induces a non-empty participation set. Denote the

lower bound of participating type is θ. Then, two conditions has to be satisfied. First, type θ is

indifferent between participating or not,

U(θ,K) = u0(θ)

Second, for all θ > θ, we have

U(θ,K) ≥ u0(θ)

From the first criterion, we get that k = θ3/12, which indicates that K must be positive. Define

∆(θ,K) = U(θ,K)− u0(θ)

-33-


Then we have ∂∆/∂θ < 0 when K > 0. It means that we can never satisfy the second requirement.

Therefore, the unique PBE is that no type participates.

-34-


Week 5

Signalling (COMP Spring ’13)

(a) Write down a simple Spence signalling model with T types.

Solution: Key elements of the model should include

• Signaller with private type θt ∈ {θ1, θ2...θT } with probability ft.

• Signaller can send a signal q with cost q/φ(θ), where φ(θ) is increasing θ.

• Responder’s best response is to pay (expected) marginal productivity to a signaller. R(θ) = m(θ).

m(θ) is increasing in θ.

• The outside option of all signallers is zero.

• Utility function of signaller is

u(r, q, θ) = r − q

φ(θ)

Remark: If time is limited, you may use m(θ) = θ and φ(θ) = θ.

(b) Consider T = 2. Characterize the complete set of separating PBE and Pooling PBE.

Remark: Notice that PBE is defined by a combination of strategy profile and a belief system. For

one strategy profile, there could be infinitely many consistent beliefs. Here, the requirement of the

question is just to find all the strategy profiles that exist in some PBE. Alternatively speaking, as long

as a strategy profile has one consistent belief, it is one of our targets.

Solution: First we discuss separating PBE. To judge whether a strategy profile (q1, q2) is part of a

PBE, we need only to check whether the ”worst belief” can support it or not. The worst belief is a

function B(q) in which B(q2) = θ2 and B(q) = θ1,∀q 6= q2. In any separating PBE, q1 = 0, since

low type agent will be always be believed as its own type except for reporting q2. Now consider q2.

Incentive compatibility requires that

m(θ1) ≥ m(θ2)− q2

φ(θ1)

m(θ1) ≤ m(θ2)− q2

φ(θ2)

The above inequalities imply that when q2 ∈ [q′, q′′], where q′ = [m(θ2) − m(θ1)] · φ(θ1) and q′′ =

[m(θ2)−m(θ1)] · φ(θ2), (0, q2) will be a separating PBE strategy profile.

As for Pooling PBE, there is only one q∗ on the equilibrium path. Again, we can use the ”worst belief”

- if q∗ is not observed, one will be believed as low type. The equilibrium payment of pooling equilibrium

is

E[m(θ)] = f1m(θ1) + f2m(θ2) ∈ [m(θ1),m(θ2)]

The ”worst belief” supports q∗ if and only if we have

E[m(θ)]− q∗

φ(θ1)≥ m(θ1)

E[m(θ)]− q∗

φ(θ2)≥ m(θ1)

-35-


Because the deviation will be treated as low type, the most profitable deviation is to report zero. Thus,

the right hand side of both constraints should be simply m(θ1). Obviously, the first inequality implies

the second one. Therefore, the range of pooling equilibrium signal is

0 ≤ q∗ ≤ [E[m(θ)]−m(θ1)] · φ(θ1)

(c) What is the Strong Intuitive Criterion? Which PBE satisfy this criterion? Explain carefully.

Remark: The strong version is the only one you’ve learnt. So no worry about the weak one.

Solution: Given a PBE, let U(θ) be the equilibrium payoff and q be a strategy that is never played

by any type in the equilibrium. Also, define function

u(θ, θ, q)

be the payoff of a type θ player who chooses q and is believed to be type θ.

Definition 10. A belief function B is credible if there exist a type θ∗ such that

(i) B(q) = θ∗

(ii) u(θ∗, θ∗, q) > U(θ∗)

(iii) u(θ, θ∗, q) ≤ U(θ) for all θ 6= θ∗.

The PBE fails Strong Intuitive Criterion if there exists a credit belief.

First, let’s find all the separating PBE that fail S.I.C. It is better for you to memorize that the only

PBE that does not fail S.I.C is the one with binding local upward constraint. From part (b), we know

that q2 ∈ [q′, q′′]. Consider PBE with q2 > q′ and under the ”worst” belief. To check whether such

PBE fails S.I.C, we need to find a credible believe. Define a belief B(q) in which B(q′) = θ2. In this

case,

m(θ2)− q′

φ(θ2)< m(θ2)− q2

φ(θ2)

m(θ1) = m(θ2)− q′

φ(θ1)

The first inequality says that under the new belief, high type is better off if report q′. Second equation

implies that low type has no incentive to deviation under the belief. Thus, the belief B(q) is credible

- the PBE fails S.I.C. The only separating PBE that survives the credibility check is the one with

q2 = q′.

To check whether a pooling equilibrium with q∗ fails S.I.C, we need to find a q and corresponding

credible belief B(q). Anyway, we need to assign B(q) = θ2 since all the pooling PBE q∗’s are based on

-36-


”worst belief”. To make B credible, the following constraints must be true,

m(θ2)− q

φ(θ2)> E[m(θ)]− q∗

φ(θ2)

m(θ2)− q

φ(θ1)≤ E[m(θ)]− q∗

φ(θ1)

The two inequalities pin down a non-empty set for q,

[m(θ2)− E[m(θ)]] · φ(θ2)) + q∗ ≤ q < [m(θ2)− E[m(θ)]] · φ(θ2)) + q∗

Thus, there always exists a credible belief. In conclusion, pooling PBE always fails S.I.C.

(d) For large T does the pooling equilibrium satisfy the Strong Intuitive Criterion?

Solution: In addition to existing {θ1, θ2}, let’s add one more type θm = 12 (θ1 + θ2).

To find a credible belief, let’s guess and verify. Denote q∗ as the equilibrium signal in a pooling PBE.

Assume off-equilibrium q exists and the belief on the q is θ2, the highest possible type. The belief is

credible if and only if the following inequality system holds

m(θ2)− q

φ(θ2)> E[m(θ)]− q∗

φ(θ2)

m(θ2)− q

φ(θ1)≤ E[m(θ)]− q∗

φ(θ1)

m(θ2)− q

φ(θm)≤ E[m(θ)]− q∗

φ(θm)

Then, the range of q is

[m(θ2)− E[m(θ)]] · φ(θm) + q∗ ≤ q < [m(θ2)− E[m(θ)]] · φ(θ2) + q∗

Recall that, if there are only two types, then the range of q∗ is

[m(θ2)− E[m(θ)]] · φ(θ1) + q∗ ≤ q < [m(θ2)− E[m(θ)]] · φ(θ2) + q∗

which is larger than that in 3-type case. You can expect that as the number of types increases, the

range will become smaller and smaller. Intuitively, S.I.C requires that under credible belief, only one

type becomes strictly better-off, while all the others cannot benefit from deviating to q. However, when

the density of types become larger, if one type is strictly better-off, other types who are closed to it

are more likely to be better-off, too. So as T increases, it is more and more difficult to find a credible

belief. The extreme case can be seen in next part, the continuous case.

Now consider a continuous version or the model where types have support Θ = [α, β].

(e) Are there multiple pooling PBE? Explain. If there is a unique PBE, discuss whether it satisfies the

Strong Intuitive Criterion. If there are multiple pooling PBE, pick any one and discuss whether it

satisfies the Strong Intuitive Criterion.

-37-


Remark: This question does not specify what should be the outside option. For simplicity, let

assume it is a constant over types u0(θ) = u0.

Solution: Let S be the set of participating in a pooling PBE. If θ ∈ S, the signaller sends a unique

q∗. Other signal will be treated as lowest type α. The payoff of type θ ∈ S equals to

U(θ) = E[m(θ)|θ ∈ S]− q∗

φ(θ)

The term is strictly increasing in q. If S 6= Θ, S has to take the form [θ, β]. Also, we need every type

outside S has no incentive to participate, which means that

u0 = E[m(θ)|θ ∈ S]− q∗

φ(θ)

Notice that E[m(θ)|θ ∈ S] is increasing in θ. For each q∗, you will have an unique θ. There should a

continuum of pooling equilibrium depending on q∗ chosen.

To check whether pooling equilibrium in continuous type model satisfies S.I.C or not, assume there is

a pooling PBE in which S = [α, β], full participating. Let q∗ be the equilibrium signal. Assume that

there exists an off-equilibrium signal q 6= q∗. And the belief on q is B(q) = θ. If the belief is credible,

it has to satisfy

m(θ)− q

φ(θ)> E[m(θ)]− q∗

φ(θ)

and for all θ 6= θ,

m(θ)− q

φ(θ)≤ E[m(θ)]− q∗

φ(θ)

However, consider a type of agent who is closed enough to θ, θ = θ + ε. When the ε is small enough,

given all the functions such as m and φ are continuous, θ + ε must satisfy that

m(θ)− q

φ(θ + ε)> E[m(θ)]− q∗

φ(θ + ε)

which is contradictory to credit belief. Thus, pooling equilibrium in continuous type never fails S.I.C

since we cannot find credible belief.

(f) The signalling cost function is C(θ, q) = qA(θ) . A type θ ∈ [0, 1] worker has a marginal production

m(θ) = θ. Characterize a separating signalling equilibrium as a solution to a differential equation.

Solution: Denote {q(θ)} as the equilibrium signals. By first order condition, we have,

U ′(θ) =A′(θ)

A2(θ)· q(θ)

By definition, U(θ) = θ − q(θ)A(θ) , so the above equation becomes

U ′(θ) =A′(θ)

A(θ)· (θ − U(θ))

The separating equilibrium payoff function should be the solution to the above ODE. In addition U(θ),

-38-


we can easily achieve equilibrium signals q(θ) by

q(θ) = (θ − U(θ)) ·A(θ)

(g) Suppose that an innovation halves cost of the education so that the new cost is C(θ, q) = q

A(θ)= q

2A(q) .

Draw a conclusion about the difference in equilibrium payoffs.

Solution: There won’t be any change of U(θ) by the differential equation found in (f) if we simply

multiply A by a constant.

-39-


Sealed-bid Auctions (COMP Spring ’12)

A single item is to be sold by sealed bid auction. Let bH be the highest bid submitted and let bS be the

second highest bid submitted. The high bidder wins and pays αbH + (1 − α)bs where α ∈ [0, 1]. There are

n buyers. Buyer i, i = 1, ...n has a value θi which is an independent draw from a distribution with support

[0, 1], c.d.f F (θ) and p.d.f. f(θ).

(a) Solve for the equilibrium expected payoff V (θ). As far as possible prove every claim that you make.

Remark: Recall that when we are solving a continuous type auction model, we first assume an

increasing bidding function and then verify it.

Solution: Assume B(θ), which is strictly increasing in θ is the equlibrium bidding function. If all

other buyers bid according to it, buyer 1, WLOG, wins by bidding B(x) if all θj < x, j = 2, ...n. Then

the winning probability is

w(x) = Fn−1(x)

Her expected payoff is therefore

u(θ, x) = w(x) · θ − r(x)

where r() is the equilibrium expected payment. In equilibrium,

θ = arg maxx

u(θ, x)

Let V (θ) be the equilibrium payoff, we have

V (θ) = w(θ)θ − r(θ)

Then, FOC implies that

w′(θ)θ = r′(θ)

and so

V ′(θ) = w′(θ)θ + w(θ)− r′(θ) = w(θ)

Therefore, the equilibrium payoff should be

V (θ) =

∫ θ

0

Fn−1(x)dx+K

Since lowest type has zero value on the item, V (0) = 0, so K = 0.

(b) Draw a conclusion as to the effect on both buyer’s expected payoffs and expected seller revenue as the

parameter α varies.

Solution: By

V (θ) =

∫ θ

0

Fn−1(x)dx

we know that buyer’s expected payoff is independent of α. For buyer, anyway the item will be sold to

the highest type and generate social surplus of θ. Therefore, seller’s expected payoff is unaffected as

well. Alternatively, by r(θ) = w(θ)θ − V (θ), we can get the same conclusion.

-40-


(c) The seller announces a reserve price(minimum bid) of θ. If there is only on bidder, that bidder will

pay αb+ (1− α)θ. Do the conclusions of (b) continue to hold? Explain carefully.

Solution: Revenue equivalence principle still holds here. With the reserved price, an agent enters if

and only if θ ≥ θ. Then type θ buyer should be indifferent between in and out. So his V (θ) equals to

zero. We have

V (θ) =

∫ θ

θ

Fn−1(x)dx

which is independent of α, too.

(d) If n = 2 and F (θ) = θ, obtain a differential equation for the equilibrium bid function. Confirm that

for some k, B(θ) = kθ is an equilibrium bid function.

Solution: First, by the distribution function, winning probability w(θ) = θ. Therefore,

V (θ) =

∫ θ

0

w(x)dx =

∫ θ

0

xdx =1

2θ2

Also, V (θ) = F (θ)θ − r(θ) gives us that r(θ) = 12θ

2.

Since there are only two players, if one wins, he will pay a linear combination of his own bid and the

other player’s bid. We have

r(θ) = [E[αB(θ) + (1− α)B(x)|x < θ]]·w(θ)

The first part E[αB(θ)|x < θ] = αB(θ) = αB(θ). The second part is

E[(1− α)B(x)|x < θ] = (1− α)

∫ θ

0

B(x)

F (θ)dx

Therefore, take the derivative of r, we have

r′(θ) = αB′(θ)θ +B(θ)

By r = 12θ

2, we have the following differential equation

αB′(θ)θ +B(θ) = θ

The hint of the question mentions that B(θ) = kθ. Put it into the ODE, we have

αB′(θ)θ +B(θ) = αkθ + kθ = θ

The ODE holds if k = 11+α .

-41-


Week 6

Mechanism Design

In this section, we introduce the basic framework of mechanism design. In general, mechanism designer’s

job is to create an ideal game which encourages agents to participate and achieve certain goals. Examples

include price discrimination by monopolistic producer, public good provision by government and so on.

In the former example, the producer is maximizing profit while in the latter one, government is trying to

optimizing social welfare. The key challenge of designing such mechanism is information asymmetry - agents

have unobservable preferences(types) on the good.

Description of the model

There are two players. Player 0, the designers, chooses a game with strategy set S0 for himself and S1 for

the next mover, player 1.

Player 1’s utility is based on 3 elements - his own preference θ, which his private information, quantity

q consumed and total payment r.

Assumption. The utility function u(θ, q, r) satisfies

• u is quasi-linear in terms of transfer.

u(θ, q, r) = B(θ, q)− r

• The utility function is increasing in θ and q.

• The utility function satisfies single crossing property

−∂u/∂q∂u/∂r

is increasing in θ.

The assumptions imply that higher type has stronger preference in q, which will be the base of many

further conclusions.

Knowing player 1’s preference distribution, the designer’s strategy set for himself is the set of all possible

combination of q and r. It is equivalent to choosing a game since no matter what the game looks like, finally

the payoff is in terms of q and r. Thus, choosing q and r is the most general way to define the game.

If there is totally T types of agents, then there is no point for player 0 to design more than T combinations

of (q, r). Therefore, a mechanism is defined as

M = {(q(θt), r(θt)}θt∈Θ

Seeing the mechanism, player 1 with type θ will choose the most preferred one or outside option. An

incentive compatible mechanism must satisfy that θ ∈ arg maxs∈Θ u(θ, q(s), r(s)).

The following question is just a warm-up two type example. Answer will be provided. Please solve by

yourselves and confirm it.

-42-


Question. Assume there two types of buyers with θ ∈ {θ1, θ2} = {6, 8}. The benefit of consuming quantity

q is

B(θ, q) ≡∫ x

0

p(θ, x)dx

where p(θ, q) ≡ θ − q.Let the two types are equally likely. There is a monopolistic seller who can design

mechanism on q, r. The cost of production is C(q) = q.

(a) Write down the optimization problem of the seller. Please include objective functions and necessary

constraints.

(b) Show that the necessary condition for a profit maximizing separating equilibrium is that ICH and IRL

are binding.

(c) Solve for the mechanism.

Answer: q1 = 3, q2 = 7.

Assume that the seller can only use two-part pricing instead of choosing q, r directly. Denote k and p

are entry fee and per-unit price, respectively.

(d) Given (k, p) solve for type θ’s optimal q(k, p) and total payment r(k, p)

Answer: q = θ − p, r = p(θ − p)+k

(e) Solve for optimal two part pricing mechanism {k(θ), p(θ)}. Is it equivalent to choosing q and r directly?

Answer: p1 = 3, p2 = 1. In this example, two-part pricing generates same result as choosing r, q.

-43-


Mechanism Design Continuous Types

Now assume buyers are continuously distributed. θ ∈ [α, β]. The pdf and cdf are f(θ) and F (θ), respectively.

The buyer’s utility function is defined as

u(θ, q, r) = B(θ, q)− r =

∫ q

0

p(θ, x)dx− r

The function B is increasing in q and θ, and satisfies SCP.

Notation. Denote Fi(x1, x2...xN ) is the partial derivative of function F w.r.t. ith variable xi. That is

Fi(x) =∂F (x1, x2...xN )

∂xi

If there is a the incentive compatible mechanism

M = {(q(θ), r(θ)}θ∈Θ

Then the equilibrium payoff function U(θ) satisfies

U ′(θ) = B1(θ, q(θ))

Before considering how the seller setting prices and quantities, we know that expected social surplus should

be equal to expected utility of a buyer and expected profit of the seller. That is

S = U + U0

The expected profit of the seller

U0 ≡∫ β

α

r(θ)− c(q(θ))dF (θ)

Notice that

r(θ) = B(θ, q(θ))− U(θ)

Therefore, U0 can be re-written as

U0 =

∫ β

α

B(θ, q(θ))− c(q(θ))dF (θ)︸︷︷︸Social surplus S

−∫ β

α

U(θ)dF (θ)︸︷︷︸Expected utility U

-44-


The social surplus term is a function of q(θ). The expected utility term requires further transformation.

∫ β

α

U(θ)dF (θ) =

∫ β

α

U(θ)d(−(1− F (θ))

= −(1− F (θ))U(θ)|βα +

∫ β

α

(1− F (θ))U ′(θ)dθ

= U(α) +

∫ β

α

(1− F (θ))B1(θ, q(θ))dθ

=

∫ β

α

1− F (θ)

f(θ)B1(θ, q(θ))dF (θ)

In the lecture notes, we’ve seen welfare function with linear combination of firm’s profit and buyer’s

utility.

W (γ) = U0 + γU

where γ is a parameter that controls how much the designer cares about buyer’s utility. By S = U + U0, the

objective function can be also expressed as W

W = S − (1− γ)U

Therefore, using the result above, we have

W =

∫ β

α

B(θ, q(θ))− c(q(θ))dF (θ)− (1− γ)

∫ β

α

1− F (θ)


=

∫ β

α

I(θ, q(θ))dF (θ)

where the function I is defined by

I(θ, q) ≡ B(θ, q)− c(q)− (1− γ)B1(θ, q)1− F (θ)

f(θ)

Now we have a objective function w.r.t. q(θ). Point-wise maximization requires that, for each q(θ)

∂W

∂q(θ)= 0

Alternatively, it is equivalent to have∂I(θ, q)

∂q= 0

By simple algebra, we get

B2(θ, q(θ)) = (1− γ)B12(θ, q(θ))1− F (θ)

f(θ)+ c′(q(θ)) (3)

Recall that B(θ, q) =∫ q

0p(θ, x)dx, thus, B2(θ, q(θ)) = p(θ, q(θ)) and B12(θ, q(θ)) = p1(θ, q(θ)). Equation (1)

now becomes

p(θ, q(θ)) = (1− γ)p1(θ, q(θ))1− F (θ)

f(θ)+ c′(q(θ)) (1’)

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Notice that when θ = β, (1’) becomes p(β, q(β)) = c′(q(β)), which implies that the quantity of highest type

q(β) is at social optimal level.

Notice that if we don’t impose any assumption on functions p or F , it is very likely that quantities q(θ)

derived by (1’) may violate monotonicity. If so, incentive compatibility cannot hold, too. In this case, we

need further modification of the quantity functions. We will leave the discussion next week.

On more comment - one of the most commonly used assumption that can lead to monotonicity is in-

creasing hazard rate which is defined by

λ(θ) ≡ f(θ)

1− F (θ)

For instance, uniform distribution satisfies the assumption.

Example. Now consider the following simple example. Assume p(θ, q) = θ−q. The per-unit cost is constant

c = 1. Let θ is uniformly distributed on [2, 3]. Then parameter γ = 0 - purely a profit maximization problem.

Given the expression of p, B(θ, q) = θq − 12q

2. Buyers’ outside options are zero.

If we plug all the information into formula (1’), we have

θ − q(θ) = (1− (θ − 2)) + 1

q(θ) = 2θ − 4

By the expression of q(θ), we know that B1(θ, q(θ)) = q(θ) = 2θ − 4. By FOC, we know that U ′(θ) =

B1(θ, q(θ)) in equilibrium. Therefore, we have

U(θ) =

∫ θ

2

2θ − 4dθ = θ2 − 4θ + 4

Easy to see that U(2) = 0 by boundary condition. What’s remaining is to solve for r(θ). It is simply by

r(θ) = B(θ, q(θ))− U(θ). By simple algebra, we have

r(θ) = (θ − 2)(6− θ)

Implementation

So far, we’ve solved for profit maximizing mechanism {q(θ), r(θ)}. But in the game, buyers choose

optimal package instead of reporting θ. Thus, we want to know what (q, r) relationship can implement the

mechanism we just solved. Consider type θ buyer’s optimization problem facing (q, r(q)). His problem is

maxqB(θ, q)− r(q)

FOC implies that he will choose q such that B2(θ, q) = r′(q) or p(θ, q) = θ − q = r′(q). The q chosen here

should be q(θ) solved before. Therefore, we have

r′(q) = θ − q (*)

By the result solved in the first part, we also have

q(θ) = 2θ − 4 (**)

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We can combine (*) and (**) to get rid of θ. It gives us

r′(q) = 2− 1

2q

By taking integration, we have r(q) = 2q − 14q

2 + K. The constant K has to be zero since the lowest type

θ = 2 will buy nothing and pay nothing. You can see that the final expression of r, q relationship is

r(q) = 2q − 1

4q2

which is a concave function. It implies that the per unit payment r/q is decreasing in q - the more you buy,

the higher discount you will get. The result in price discrimination is called non-linear pricing.

-47-


Week 7

Ironing

Last time, we’ve introduced how to derive the optimization conditions of a profit maximization mechanism.

Recall that, we’ve achieved the following objective function for firm

W =

∫ β

α

B(θ, q(θ))− c(q(θ))dF (θ)− (1− γ)

∫ β

α

1− F (θ)


=

∫ β

α

I(θ, q(θ)dF (θ)

where the function I is defined by

I(θ, q) ≡ B(θ, q)− c(q)− (1− γ)B1(θ, q)1− F (θ)

f(θ)

Point-wise maximization requires that, for each q(θ)

∂W

∂q(θ)= 0

Alternatively, it is equivalent to have∂I(θ, q)

∂q= 0

By simple algebra, we get

B2(θ, q(θ)) = (1− γ)B12(θ, q(θ))1− F (θ)

f(θ)+ c′(q(θ)) (4)

Recall that we define B(θ, q) ≡∫ q

0p(θ, x)dx. So B2(θ, q(θ)) = p(θ, q(θ)) and B12(θ, q(θ)) = p1(θ, q(θ)).

Equation (1) now becomes

p(θ, q(θ)) = (1− γ)p1(θ, q(θ)1− F (θ)

f(θ)+ c′(q(θ)) (1’)

Notice that when θ = β, (1’) becomes p(β, q(β)) = c′(q(β)), which implies that the quantity of highest type

q(β) is at social optimal level.

Notice that if we don’t impose any assumption on functions f or F , it is very likely that quantities q(θ)

derived by (1’) violates monotonicity. If so, incentive compatibility cannot hold, too. In this case, we need

further modification of the quantity function. The following example shows the problem.

Example. Let γ = 0. Assume buyer’s type θ is distributed on [0, 2], following the CDF function as below:

F (θ) =

{34θ if θ ∈ [0, 1)12 + 1

4θ otherwise.

Also, let p(θ, q) = θ − q and cost of production simply equals to zero. By (1’), we will have the following

-48-


optimal condition

q(θ) = θ − 1− F (θ)

f(θ)

By the functional form of F , we have

q(θ) =

{2θ − 4

3 if θ ∈ [0, 1)

2θ − 2 otherwise.

There are two problems of the curve. First, q(θ) is negative when θ < 2/3. Second, q(θ) is non-monotone,

which will violate incentive compatibility. The first problem is easy to solve - just let q = 0 at that part.

To fix the 2nd problem, we need to modify q(θ) to an increasing function. The modification consists of two

steps. First, we show that the best modification is to make the non-monotone part flat. Second, given it has

to be flat, we find the optimal flat level. For the first step, a formal statement is as follows,

Claim 1. The best increasing curve q(θ) that goes across (1, q∗) has the following structure,

q(θ) =

2θ − 4

3 if θ ∈ [0, θ1)

q∗ if θ ∈ [θ1, θ2)

2θ − 2 otherwise.

where q∗ = 2θ1 − 4/3 = 2θ2 − 2.

Graphically, given arbitrary increasing curve (orange) that goes across the point (1, q∗), the expected payoff

generated by the seller is less than that of choosing the flat line (red line). The proof is simply by the fact

that I(θ, q) is maximized at blue curve, so, when q is above the curve, ∂I/∂q < 0, which indicates that the

revenue I can be improved by decreasing q. Similarly, when q is below the blue curve, ∂I/∂q > 0, increasing

q can improve I. Since every curve has to go across (1, q∗) by assumption, the best increasing q(θ) is the

-49-


one defined in the claim. Since we make the middle part flat, the method is called ironing.

Recall that the above argument is based on fixing q∗, next step is to find which q∗ is the best. It is also

equivalent to finding optimal θ1 or θ2. By

q∗ = 2θ1 −4

3= 2θ2 − 2

We know that θ2 = θ1 + 1/3. So firm’s problem is

Rironing = maxθ1

∫ θ1

0

I(θ, q(θ))dF (θ) +

∫ θ1+ 13

θ1

I(θ, q(θ1))dF (θ) +

∫ 2

θ1+ 13

I(θ, q(θ))dF (θ)

In this example, we have I(θ, q) = θq− 12q

2− q 1−F (θ)f(θ) , so the remaining problem is trivial but tedious. If you

wish to make the computation easier, here is a possible simplification. Maximizing Rironing is equivalent to

minimizing ∆R ≡ R−Rironing where R is firm’s revenue generated by point-wise maximization,

R =

∫ 2

0

I(θ, q(θ))dF (θ)

Anyway, Rironing is smaller than R. Our purpose is to choose θ1 that can minimize the difference between

the two term. Then the problem becomes

∆R = minθ1

∫ θ1+ 13

θ1

I(θ, q(θ))− I(θ, q(θ1))dF (θ)

By taking derivative w.r.t. ∆R w.r.t. θ1, finally you will get

θ21 −

7

3θ1 +

23

18= 0

So θ1 = 7−√

36 .

-50-


Hint: By fundemental theorem of caculus, there is a (better) way to simplify ∂∆R∂θ1

.

∂∆R

∂θ1=

[I(θ1 +

1

3, q(θ1 +

1

3))− I(θ1 +

1

3, q(θ1))

]− [I(θ1, q(θ1))− I(θ1, q(θ1)]

+

∫ θ1+ 13

θ1

∂

∂θ1(I(θ, q(θ))− I(θ, q(θ1))) dF (θ)

Easy to see that, terms in ”[·]” are zero. So we have

∂∆R

∂θ1=

∫ θ1+ 13

θ1

∂

∂θ1(−I(θ, q(θ1))) dF (θ)

Notice that, F (θ) and q(θ) has difference forms when θ ∈ [0, 1) and [1, 2]. So we need to split it into

∂∆R

∂θ1=

∫ 1

θ1

∂

∂θ1(−I(θ, q(θ1))) dF (θ) +

∫ θ1+ 13

1

∂

∂θ1(−I(θ, q(θ1))) dF (θ)

Then, what’s left is to plug-in F and q solved above and compute ∂∆R∂θ1

. Recall that in this question,

I(θ, q) = θq − 1

2q2 − q 1− F (θ)

f(θ)

and

q(θ) = 2θ − 4

3

...

-51-


Efficient Mechanism

In this section, we study how to design a mechanism that achieve efficiency. The key point of the mechanism

design is to align individual incentive with social surplus. The model is as follows. There is a designer and

I agents indexed by i. Agent i’s type set is Θi = [αi, βi]. The allocation rule is q = (q1, q2...qI). The cost

of production is C(q). Notice that, the structure of q is general. For example, in buyer-seller problem, qi

will be different; in auction, qi’s have to be add up to 1; in a public good provision problem, every agent

enjoys the same q. Agent’s benefit is Bi(θi, q), where q is the vector defined above. In addition, agent i’s

final payoff is quasi-linear

ui = Bi(θi, q) + ti

where ti is a monetary transfer from the designer to agent i. Social surplus is defined by

s(θ, q) =∑i

Bi(θi, q)− C(q)

Given a vector of θ, denote q∗(θ) be the social optimal allocation rule.

For simplicity, we consider a two-agent situation, which can be easily extended to I-agent case. Since

q∗(θ) ∈ arg maxq{s(θ, q)}

we have

(θ1, θ2) ∈ arg maxx∈Θ{s(θ1, θ2, q

∗(x1, x2))}

In particular, when person 2 reports the true, so does person 1:

θ1 ∈ arg maxx1

{s(θ1, θ2, q∗(x1, θ2))} (5)

for all θ2 ∈ Θ2.

To design an efficient mechanism, we propose the following transfer payment method ti(x1, x2) and then

show that reporting truth is the best response for both agents. Given agents’ reports (x1, x2), let

ti(x1, x2) ≡ B−i(x2, q∗(x1, x2))− C(q∗(x1, x2))

That is, given reported (x1, x2), the designer computes social optimal q∗ accordingly. The transfer to agent

i is the benefit generated by the other person −i minus the total production cost. Why such ti can induce

truth telling? If person 2 is telling the truth, person 1’s utility by reporting x1 is

u1(θ1;x1, x2) = B1(θ1, q∗(x1, θ2)) + t1(x1, θ2)

= B1(θ1, q∗(x1, θ2)) +B2(θ2, q

∗(x1, θ2))− C(q∗(x1, θ2))

= s(θ1, θ2, q∗(x, θ2))

Recall (5), s(θ1, θ2, q∗(x, θ2)) will be maximized at x1 = θ1. Thus, truth-telling is a best response.

-52-


A Public Good Example

A designer plans to produce a unit of public good at cost k. There are 2 agents, both have value distributed

on [α, β]. For simplicity, assume 2β > k and α + β < k. Based on the setting, q(θ1, θ2) = 1 if θ1 + θ2 ≥ k;

equal to zero otherwise. Recall that

ti(x1, x2) ≡ B−i(x2, q∗(x1, x2))− C(q∗(x1, x2))

So when x1 + x2 < k, q∗ = 0, everyone’s utility will be zero. If x1 + x2 ≥ k, q∗ = 1, so

t1(x1, x2) = x2 − k and t2(x1, x2) = x1 − k

The transfer rule will make truth-telling best response for both players. However, the revenue of the designer

is

−t1(x1, x2)− t2(x1, x2)− k = −x1 − x2 + k

In equilibrium, when both agents are telling the truth, the designer has to pay θ1 + θ2 − k which is equal to

the social surplus.

Although in the previous section we’ve already shown that truth-telling is a best response in such mech-

anism, there is no bad to check in this example. The game is symmetric, so only consider agent 1’s problem.

Given x2 = θ2, when θ1 + θ2 < k, true-telling will result in q = 0. Then player 1’s revenue is zero. If player

1 reports anything such that x1 + θ2 < k, the effect is the same. If he reports such that x1 + θ2 ≥ k, q = 1,

then t1(x1θ2) = θ2 − k and his payoff will be θ1 + θ2 − k < 0. Therefore, true-telling is one of the best

responses. When θ1 + θ2 ≥ k, if player 1 reports x1 such that x1 + θ2 < k, then q = 0 and he gets zero. If

x1 + θ2 ≥ k, his utility will be θ1 + θ2 ≥ k. Thus, reporting the true value is one of the best responses, too.

In conclusion, for all θ2, reporting true value is a best-response for agent 1.

Cheaper mechanism

In previous analysis, we’ve shown that to let the agents report true types, designer can compensate(charge)

them such that everyone’s final utility will be equal to maximized social surplus if telling truth. However, it

is too expensive from the perspective of the designer’s payoff. To make it cheaper, consider the modification

of t. Define

ti(x1, x2) ≡ B−i(x2, q∗(x1, x2))− C(q∗(x1, x2))− s(αi, x2, q

∗(αi, x2))

That is, we reduce the original transfer by the minimum social surplus that the market can achieve given

x−i. Plug this term into agent i’s utility function,

ui(θi;x1, x2) = s(θi, x−i, q∗(αi, x−i))− s(αi, x−i, q∗(αi, x−i))

We can see that, the new term does not affect agent’s incentive since it is determined by the other agent’s

choice. But ui has to be greater than zero to let the agent participate. So the largest reduction is

s(αi, x2, q∗(αi, x2)).

-53-


Question. Here are two exercises.

1. Show that the cheaper mechanism in a 2-buyer, 1-item auction problem is equivalent to 2nd price

auction.

2. Derive the cheaper efficient transfers under the two agent, discrete type public good provision problem.

Let θ1 = {5, 12} and θ2 = {0, 10}. The cost of the public good is 6. (Essential Micro, Exercise 12.4-1)

-54-


Week 8

Dominant Strategy Equilibrium

The reason why VCG is powerful is that reporting true type is not only a best response but also a dominant

strategy for each agent. The proof is similiar to that for showing best response.

Proposition 2. The efficient allocation q∗(θ) can be achieved as a dominant strategy equilibrium if, based

on the vector of announced values x, the allocation is q∗(x) and agent i is given a subsidy(possibly negative)

of

t∗i (xi, x−i) =∑j 6=i

Bj(x, q∗(xi, x−i))− s(αi, x−i; q∗(αi, x−i))

• Notice that the term∑j 6=iBj(x, q

∗(xi, x−i) includes the cost of production C(q∗(x)).

Proof. Notice that under the mechanism, agent i with type θi who reports xi will get his net contribution

to social surplus

Ni(θi, x−i) ≡ s(θi, x−i; q∗(xi, x−i))− s(αi, x−i; q∗(αi, x−i))

The second term is a constant to agent i’s report. As long as Ni ≥ 0, designer will try to increase the term

to make the mechanism cheaper. Therefore, the largest constant to agent i is the maximized social surplus

when he draws his lowest type αi. Then, showing that

ti(xi, x−i) =∑j 6=i

Bj(x, q∗(xi, x−i))

is a truth telling efficient mechanism is enough. Also, I will just show 2-agent case and assume cost is zero.

Extending to general case with finitely many agents is trivial.

The purpose of the proof is actually to show that, for agent 1, WLOG,

t1(x) = B2(x1, x2, q∗(x))

can induce truth telling. That is, it has to satisfy that, given x2, ∀x1,

U1(θ1, q∗(θ1, x2)) ≡ B1(θ1, q

∗(θ1, x2)) + t1(θ1, x2)

≥ B1(θ1, q∗(x1, x2)) + t1(x1, x2) ≡ U1(θ1, q

∗(x1, x2))

To see why the above inequality is true, consider the definition of q∗(x). We know that q∗(x) is the efficient

output vector given report x = (x1, x2). Thus, we have

s(θ1, x2; q∗(θ1, x2)) ≡ B1(θ1, q∗(θ1, x2)) +B2(x2, q

∗(θ1, x2))

≥ B1(θ1, q) +B2(x2, q) ≡ s(θ1, x2; q)

for any other output vector q. Also, for any x1, there exists a vector q such that

q = q∗(x1, x2)

-55-


Therefore,

B1(θ1, q∗(θ1, x2)) +B2(x2, q

∗(θ1, x2)) ≥ B1(θ1, q∗(x1, x2)) +B2(x2, q

∗(x1, x2))

for all x1.

Remark: Above statement is a clarification of ”key observation” in Prof Riley’s Efficient Mechanism I

handout (page 2). In the notes, it says that

s(θ; q∗(θ)) ≥ s(θ; q∗(x))

for all x. It is indeed true. But it might be better to say that

s(θ; q∗(θ)) ≥ s(θ; q)

And for all x, there exist some q such that q = q∗(x)

-56-


Efficient Mechanism Design (COMP Spring ’12)

Let B(θi, q) be the benefit to agent i, i = 1, 2, if the designer chooses q and the agent’s type is θi ∈ Θi =

[αi, βi]. The cost of this action to the designer is C(q). Let S(θ1, θ2, q) be social surplus if the designer

chooses q and let q∗ be the social surplus maximizing choice.

(a) Explain why θ1 = arg maxx1{S(θ1, x2, q

∗(x1, x2))}.

Solution: The proof is very similar to the proof of the proposition above. By the definition of

q∗(x1, x2), we know that

S(θ1, x2, q∗(θ1, x2)) ≥ S(θ1, x2, q)

Thus, for any x1, let q = q∗(x1, x2), we achieve the conclusion.

(b) The designer offers a transfer payment ti(x) to each agent and commits to choosing q∗(x) based on

the agents’ responses. What’s the transfer payment with the property that it is a dominant strategy

for agent i to reveal his true type?

Solution:

Based on the conclusion in part (a), truth telling is a dominant strategy if agent 1 gets final payoff of

S(θ1, x2, q∗(θ1, x2)). By definition,

S(θ1, x2, q ∗ (x1, x2)) ≡ B(θ1, q∗(x1, x2)) +B(x2, q

∗(x1, x2))− C(q∗(x))

Therefore,

t1(x) = B(x2, q∗(x))− C(q∗(x))

(c) Show that if the designer chooses transfer payments such that the participation constraint is binding,

then the designer’s payoff is

UD(θ) = S∗(α1, θ2) + S∗(θ1, α2)− S∗(θ1, θ2)

where S∗(θ) ≡ S(θ, q∗(θ)).

Solution: Given the transfer ti solved in part (b), we know that agent i’s payoff is S∗(θ) = S(θ, q∗(θ))

in equilibrium. Subtract off the minimum payoff so that the participant constraint is binding. Then,

u1(θ) = S∗(θ)− S∗(α1, θ2)

and similarly

u2(θ) = S∗(θ)− S∗(θ1, α2)

Correspondingly, the transfers in equilibrium are

t1(θ) = B(θ2, q∗(θ))− C(q∗(θ))

t2(θ) = B(θ1, q∗(θ))− C(q∗(θ))

-57-


Notice that, in this question, designer bears the cost of production. Therefore, the designer’s payoff

UD(θ) = −t1(θ)− t2(θ)− C(q∗(θ))

= S∗(α1, θ2) + S∗(θ1, α2)− S∗(θ1, θ2)

(d) Suppose that some amount q of a commodity is to be produced for agent 1 with benefit function

B(θ1, q) = (4 + θ1)q− 12q

2. The commodity will be produced by agent 2. Only she knows the constant

unit cost c = 2 − θ2. Both θ1 and θ2 ∈ [0, 1]. Show that q∗(θ) = (4 + θ1 − c) = (2 + θ1 + θ2) and

that maximized social surplus is S∗(θ) = 12 (2 + θ1 + θ2)2. Then appeal to (c) to examine whether the

mechanism is profitable for the designer.

Remark: Although the question asks you to appeal to (c), the model here is a little different from

that in previous parts. From part (a) to (c), we assume that the designer pays the cost, but here we

assume that agent 2 pays the cost. Then, designer’s equilibrium payoff will be just

UD(θ) = −t1(θ)− t2(θ)

Solution: Social surplus equals to

S(θ, q) = B(θ1, q)− C(θ2, q) = (4 + θ1)q − 1

2q2 − (2− θ2)q

Take the derivative of S w.r.t. q and let it equal to zero. The optimal quantity q∗(θ) = (2 + θ1 + θ2)

and the optimized social surplus is equal so S∗(θ) = 12 (2 + θ1 + θ2)2.

Now consider about the transfers.

t1(x) = B(x1, q∗(x))− S∗(0, x2)

where S∗(0, x2) = 12 (2 + x2)2. Similarly,

t2(x) = −C(x2, q)− S∗(x1, 0)

where S∗(x1, 0) = 12 (2 + x1)2 Hence, we have

UD(θ) =1

2[(2 + θ2)2 + (2 + θ1)2 − (2 + θ2 + θ2)2]

To see whether it is positive or not, take the partial derivative, we can see that

∂UD

∂θi= −θ−i < 0

when θ ∈ (0, 1]× (0, 1]. And both UD(0, 0) and UD(1, 1) are positive. Thus, designer’s payoff is always

positive in the model.

Remark: When dealing with questions in which some of the agents are producers, double check if θ

-58-


is increasing in cost or productivity. It is a little trick when you are solving S∗(αi, x−i). You need to

make sure that α is for lowest productivity or highest cost!

-59-


Sales of Indivisible Good

According to Myerson-Satterthwaite Impossibility Theorem (Efficient Mechanism Design III), if there is one

buyer, one seller, one unit for sale and parameters are independently and continuously distributed, then the

designer loses money in order to achieve efficiency. In this session, we consider a two buyer case and show

that designer loses money as well in this situation.

Agent 1 and 2 are buyers. Their values to the good are θi ∈ Θ = [α, β], i = 1, 2. Agent 3 is the seller.

The cost of production is c ∈ Θ = [α, β].

Given vector θ and c, efficient production should be q∗(θ, c) = 1 if and only if θmax ≥ c, where θmax =

max{θ1, θ2}. And the maximized social surplus is equal to

s(θ, c, q∗(θ, c)) = q∗(θ, c)(θmax − c)

WLOG, assume θ1 > θ2, then there are three cases to be discussed.

c > θ1 > θ2 (Case 1)

θ1 > θ2 > c (Case 2)

θ1 > c > θ2 (Case 3)

• Case 1: Obviously, q∗ = 0 in this case. So everyone, including the designer, gets zero payoff.

• Case 2: In this situation, agent 1 will get the good. So without any transfer payment, the benefit of

each agent is

B1(θ, c) = θ1

B2(θ, c) = 0

B3(θ, c) = −c

And to compute ti(θ, c), we need

S∗(θ1, θ2, c) = θ1 − c

S∗(α, θ2, c) = θ2 − c

S∗(θ1, α, c) = θ1 − c

S∗(θ1, θ2, β) = 0

Again, be careful about producer’s type space. Here, worst type means largest cost, c = β. we have

t1(θ1, θ2, c) = −θ2

t2(θ1, θc, c) = 0

t3(θ1, θ2, c) = θ1

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So the designer’s payoff equals to

UD(θ, c) = −∑

ti(θ, c) = θ2 − θ1 < 0

• Case 3: Similarly, we have

B1(θ, c) = θ1

B2(θ, c) = 0

B3(θ, c) = −c

and

S∗(θ1, θ2, c) = θ1 − c

S∗(α, θ2, c) = 0

S∗(θ1, α, c) = θ1 − c

S∗(θ1, θ2, β) = 0

So

t1(θ1, θ2, c) = −c

t2(θ1, θc, c) = 0

t3(θ1, θ2, c) = θ1 − c

The designer’s payoff in this case equals to

UD(θ, c) = −∑

ti(θ, c) = c− θ1 < 0

In conclusion, in the two-buyer case, the designer always loses money as long as the good is produced.

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Week 9

Optimal Auctions with Finite Types (COMP Spring ’13)

Buyer i, i = 1, 2, has value θi = Θ ∈ {v1, v2}. Values are independently distributed. Let ft be the probability

that buyer i has value vt. Consider the two type case where (v1, v2) = (120, 200) and (f1, f2) = (35 ,

25 ).

1. In the sealed second-price auction, what reserve price should the seller set in order to maximize expected

revenue?

Solution: In second price auction, each agent will bid true value. Hence, when reserve price p ∈[0, 120], the result should be the same. Seller’s expected revenue is

120×(

3

5

)2

+ 120× 2× 2

5

3

5+ 200×

(2

5

)2

=3320

25

If the reserved price is larger than 120, low type ill not bid anyway. If so, the optimal price is 200 and

seller’s revenue is

200×(

1− 3

5× 3

5

)=

3200

25

Thus, the optimal reserve price should be p ∈ [0, 120] - it is better to let both types enter the auction.

(b) Solve for the expected revenue maximizing selling scheme.

Solution: In terms of selling scheme, the seller needs to determine a list of winning probabilities

and expected payments for all types of buyers - {wi, ri}. However, in auction problems, winning

probabilities cannot be arbitrarily chosen. Consider a truth-telling scheme, when two types are the

same, each agent will win with half chance; if two agents reports the same type, high type will win. If

so, the scheme is efficient. Thus, winning probability for low type is

w1 =3

5× 1

2=

3

10

and high type has winning probability of

w2 =3

5+

2

5× 1

2=

4

5

In equilibrium, low type has zero expected payoff,

r1 = w1 · v1 − U1 =3

10· 120− 0 = 36

In terms of r2, profit maximization requires LDC binding, so

U2 = w2v2 − r2 = w1v2 − r1

So r2 = 136. In summary, the best selling scheme is

{(wi, ri)}i=1,2 = {(3/10, 36), (4/5, 136)}

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(c) How might this be implemented as an auction?

Solution: There are more than one way to implement the mechanism. One is to set the auction with

only two prices

(b1, b2) = (r1/w1, r2/w2) = (120, 170)

In stead of letting buyers submit whatever they want, the revenue can be raised by restricting possible

bids within a limited choice set.

(d) Compare the buyer payoffs with those in the sealed second price auction. Is there revenue equivalence?

If so why? If not why not?

Solution: The expected revenue of seller under best selling scheme is

120×(

3

5

)2

+ 170× 2× 2

5

3

5+ 170×

(2

5

)2

=3800

25

which is larger than second price auction 332025 . So the revenue equivalence does not hold here. To

understand why it is the case, notice that the best selling scheme is similar to indirect price discrimina-

tion. Revenue maximization requires local downward constraint binding. However, if we check second

price auction, it is not. Fixed winning probability, buyers try to maximize their payoff by choosing

b, which cannot guarantee that LDC are binding. Therefore, revenue equivalence does not hold here.

However, under continuous-type model, the problem can be resolved, since w and b are continuous

functions such that LDC are binding automatically.

(e) Describe a continuous c.d.f F (θ) with p.d.f. f(θ) defined on [120,200] that is very similar to the

cumulative probability for the two point distribution above. Without going into any technical details,

discuss what you think the optimal auction will look like in this case.

Solution: Though the question does not require technical detail, let’s go through it for review purpose.

Briefly speaking, the optimal auction should be a high bid auction with an optimal reserve price p. It

is equivalent to set a benchmark type θ, below which agent cannot enter the auction. Remember that

for this sort of model, we impose regularity condition:

θ − 1− F (θ)

f(θ)

is increasing in θ. So, the reserve price is to make sure that the above term is always positive. In

equilibrium we have

U ′(θ) = w(θ) = F (θ)

where U(θ) is type θ agent’s equilibrium payoff. The revenue of seller is

R =

∫ 200

θ

w(θ)θ − U(θ)dF (θ)

=

∫ 200

θ

w(θ)θ − 1− F (θ)

f(θ)U ′(x)dF (θ)

=

∫ 200

θ

F (θ)

[θ − 1− F (θ)

f(θ)

]dF (θ)

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The above expression is almost determined. The only thing one can do is set a reserve price such that

the marginal revenue is always positive. Alternatively, we can just set p = θ such that

θ − 1− F (θ)

f(θ)= 0

Here are three examples.

• First, if θ is uniformly distributed on [120, 200], then

θ − 1− F (θ)

f(θ)= 2θ − 200 > 0

for all θ.

• If the distribution is

F (θ) =θ2

6400− 3θ

80+

9

4

Then,

θ − 1− F (θ)

f(θ)=

3θ2 − 480θ + 8000

2θ − 240→ −∞

when θ → 120. The reserve price should be around p = 20√

1383 + 80 ≈ 160.

• If the distribution is that

F (θ) =

{− 9

4 + 3160θ if θ ∈ [120, 160)

− 14 + 1

160θ if θ ∈ [160, 200]

Then we have

θ − 1− F (θ)

f(θ)=

{2θ − 520/3 if θ ∈ [120, 160)

2θ − 200/3 if θ ∈ [160, 200]

which is always positive.

The third example is similar to the two-point example. (at least compared with the previous two.)

Notice that the reason why we let both agents enter the auction is that the proportion of low type is

too large. The 3rd example is a 2-segment uniform distribution with higher density on first half. The

2nd example is somehow the opposite so we have a reserve price p > 120. The first example show us

that even if the types are uniformly distributed, letting both types enter is better than squeezing the

low type from the market.

-64-


Choosing Optimal Quantity of a Public Good

Agent i ∈ J = {1, 2} has a benefit from q units of the public good of Bi(θi, q) = θiq− q2

2 , where his type θi,

is continuously distributed on Θi = [α, β]. The cost of q units of the public good is cq where c < 2α.

(a) Solve for the efficient quantity of the public good and hence obtain an expression for the maximized

social surplus.

Solution: Social surplus equals to

S∗(θ) = maxqB1(θ1, q) +B2(θ2, q)− cq = max

q(θ1q + θ2q − q2 − cq)

By FOC, the optimal quantity produced should be

q∗(θ) =1

2(θ1 + θ2 − c)

and

S∗(θ) =1

4(θ1 + θ2 − c)2

(b) Obtain an expression for agent i’s payoff under the net contribution to social surplus mechanism.

Solution: Under the mechanism, one’s utility should be

Ui(θ) = S∗(θ)− S∗(α, θ−i)

Given the social surplus S∗ derived above, the utility of each agent under the mechanism should be

U1(θ) =1

4(θ1 + θ2 − c)2 − 1

4(α+ θ2 − c)2

U2(θ) =1

4(θ1 + θ2 − c)2 − 1

4(α+ θ1 − c)2

(c) Show that the designer’s payoff is strictly decreasing in each agent’s type. Define a ≡ α − c/2 and

b ≡ β − c/2. Hence show that if a/b is sufficiently closed to 1, then the designer’s payoff is strictly

positive for all possible realization of types.

Solution: Designer’s payoff equals to

U0(θ) = S∗(θ)− U1(θ)− U2(θ) =1

4

[(α+ θ1 − c)2 + (α+ θ2 − c)2 − (θ1 + θ2 − c)2

]Take the partial derivative, we can see that

∂U0(θ)

∂θ1=

1

4[2(α+ θ1 − c)− 2(θ1 + θ2 − c)] =

1

2[α− θ2] < 0

Similar for θ2. Therefore, designer’s payoff is decreasing in agents’ types.

To see whether U0(θ) will be positive when a/b is closed to one, we can consider the worst case. That

is, when θ = (β, β), whether the designer can get positive revenue or not? Rewrite U0 in terms of a

-65-


and b.

U0(β, β) =1

4

[(α+ β − c)2 + (α+ β − c)2 − (2β − c)2

]=

1

4

[2(a+ b)2 − (2b)2

]=

1

2b2[(a/b+ 1)2 − 2

]When a/b converges to 1, obviously the term is positive. So all the realization of (θ1, θ2) will give the

designer strictly positive payoff.

(d) Is it true when a/b is small?

Solution: The answer is no. The lowest possible a could be close to zero. Then U(β, β) will be

negative.

(e) Re-examine the problem if there are three agents and the cost of production is cq where c < 3α. Define

a = α − c/3 and b = β − c/3. See what progress you can make in determining the range of values of

a/b for which the designer’s payoff is strictly positive for all possible types.

Solution: In part (c), the range of a/b is (√

2− 1, 1).

In 3-agent case, the computation is almost the same. You can easily find that

q∗(θ) =1

3(θ1 + θ2 + θ − 3− c)

and

S∗(θ) =1

6(θ1 + θ2 + θ − 3− c)2

Therefore, designer’s payoff is equal to

U0(θ) =1

6

[∑i

(θ1 + θ2 + θ3 − θi + α− c)2 − 2(θ1 + θ2 + θ2 − c)2

]

which is decreasing in θi for all i. Plug θ = (β, β, β) into the expression, we have

U(β, β, β) =1

6

[3(2β + α− c)2 − 2(3β − c)2

]=

1

6

[3(2b+ a)2 − 2(3b)2

]=

1

2b2[(2 + a/b)2 − 6

]Then the range of a/b becomes (

√6− 2, 1), which is smaller than (

√2− 1, 1). Intuitively, the designer

can make money when (all) agents are of lower type. However, when there are more agents, the

probability that agents are all of lower type becomes smaller.

-66-


Indirect Price Discrimination

Suppose that p(θ, q) = 1 − q/θ, where θ ∈ Θ = [1, 2] and F (θ) = θ − 1. THe unit cost of production is c.

The total mass of customers is 1.

(a) Explain why for any increasing {q(θ)}θ∈Θ the total payoff to the buyers can be written as follows

U =

∫ 2

1

1

2

q2

θ2(2− θ)dθ

Solution: By definition,

U =

∫ 2

1

U(θ)dF (θ)

where U(θ) is the equilibrium payoff for type θ. Integrating by part, we have

U =

∫ 2

1

U ′(θ) · 1− F (θ)

f(θ)dF (θ)

=

∫ 2

1

U ′(θ)(2− θ)dθ

By U(θ) = B(θ, q(θ))− r(θ), we have

U ′(θ) = B1(θ, q(θ)) +B2(θ, q(θ))q′(θ)− r′(θ)

in which B2(θ, q(θ))q′(θ)− r′(θ) = 0 in equilibrium. Hence,

U ′(θ) = B1(θ, q(θ)) =1

2

q(θ)2

θ2

Done.

(b) Hence show that maximized profit (holding the allocation rule fixed) is

U0 =

∫ 2

1

[(1− c)q − 1

2

q2

θ− 1

2

q2

θ2(2− θ)dθ

Solution: By definition, seller’s payoff equals to

U0 =

∫ 2

1

r(θ)− cq(θ)dF (θ)

where r(θ) = B(θ, q(θ))− U(θ). Therefore, seller’s payoff becomes

U0 =

∫ 2

1

B(θ, q(θ))− U(θ)dθ

According to the result in part (a), we have

U0 =

∫ 2

1

[(1− c)q − 1

2

q2

θ− 1

2

q2

θ2(2− θ)

]dθ

-67-


(c) Maximize the integrand point-wise and confirm that the allocation q∗(θ, c) is incentive compatible.

Remark: In this question, we need only to confirm that the q∗(θ, c) is increasing. There is a proposition

in the lecture notes. (for discrete type model)

Proposition 3. Necessary and sufficient conditions for a mechanism to be incentive compatible and

revenue maximizing are that

• q(θt) is increasing.

• Participating constraint satisfied for lowest type.

• LDC is binding for all other types.

Extending to continuous model is easy. Just change LDC binding to MR = MC where

MR = p(θ, q)− 1− F (θ)

f(θ)

∂p(θ, q)

∂θ

Part (a) and (b) are based on this. So what’s left is to check monotonicity.

Solution: Point-wise maximization gives us

(1− c)− q

θ− q

θ2(2− θ) = 0

Therefore, optimal quantities

q∗(θ, c) =θ2

2(1− c)

which is indeed increasing in θ.

(d) Suppose that the objective is to maximize the expected total gains to buyers and seller: U0+U . What’s

the optimal allocation?

Solution: Now we are consider about efficient allocation.

U0 + U =

∫ 2

1

B(θ, q(θ))− cq(θ)dθ

which implies that

q∗∗(θ, c) = θ(1− c)

(e) How could this be achieved?

Solution: Implementation of efficient allocation is easy here. Just set p = c and let consumers choose

quantity. Actually, the mechanism is equivalent to V-C-G (why?)

-68-


Fixed Supply and Marginal Revenue

(a) Consider the model of previous question. For a given allocation rule explain why maximum revenue is

R =

∫ 2

1

[q − 1

2

q2

θ− 1

2

q2

θ2(2− θ)

]dθ

Solution: It has been discussed in previous question.

Henceforth assume that the monopoly has a fixed supply Q. Thus a feasible allocation must satisfy the

constraint ∫ 2

1

q(θ)f(θ)dθ ≤ Q

(b) Form the Lagrangian and show that it can be written as follows

L =

∫ 2

1

[q(1− λ)− 1

2

q2

θ− 1

2

q2

θ2(2− θ)

]dθ + λQ

Solution: The lagrangian is

L = R+ λ

(Q−

∫ 2

1

q(θ)f(θ)dθ

)By simple algebra, we can get the expression we need.

(c) Given the assumptions show that the solution to the problem q∗(θ, λ) is increasing for all q∗(θ, λ) > 0

Solution: Point-wise maximization gives us

(1− λ)− q

θ− q

θ2(2− θ) = 0

So we have q∗(θ, λ) = θ2(1−λ)2 , which is obviously increasing in θ.

(d) Explain why λ is the marginal revenue of the firm, i.e. λ = dRdQ .

Solution: Recall that the Lagrangian is

L = R+ λ

(Q−

∫ 2

1

q(θ)f(θ)dθ

)Hence, we have

∂L∂Q

= λ

Hence, by Envelope Theorem, we have the desired result, λ = dRdQ .

Notice: Envelope Theorem implies that, for a program

maxxf(x, φ)

s.t. g(x, φ) ≤ 0

-69-


the objective function satisfiesdF

dφ=∂L∂φ

where F (φ) = f(x∗(φ), φ) For this model, F is R here. φ is the paremeter of the program. In this

question, it is Q.

(e) Substitute the solution into the constraint to obtain a relationship between Q and λ, that is, λ = λ(Q).

Solution: By the result in part (c), we have q∗(θ, λ) = θ2(1−λ)2 . Integrate q∗ w.r.t. dθ, let it equal to

Q, we have

Q =1− λ

2

∫ 2

1

θ2dθ

=7

6(1− λ)

which implies that λ(Q) = 1− 67Q.

(f) What is total revenue?

Solution: Integrate λ(Q), we have

R(Q) = Q− 3

7Q2

There should be a constant term. But it has to be zero since R(0) = 0.

(g) If TC(Q) = γ1Q+ γ2Q2, what is the profit maximizing output?

Solution: Prevous parts assume Q is a constant. Here, the problem is

maxQ

Π(Q) = Q− 3

7Q2 − (γ1Q+ γ2Q

2)

The answer is

Q∗ =1− γ1

2γ2 + 6/7

-70-

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