Screening

Definition

Screening is when the uninformed party takes action to learn action to learn information that is relevan to them
Examples:
- Price discrimination by firms
- Airlines splitting tickets into first class, business class, economy class
- Telecomm companies offering different mobile phone plans

Example

Willingness to pay

Software Version	Type 1	Type 2
A	180	500
B	150	200

Setup

Suppose there are $45$ students in each group (Type 1 / Type 2)
Marginal costs of producing each unit of the software is (virtually) zero
There is a fixed cost of developing, but these fixed costs do not affect pricing decisions
Price decisions are driven by revenue maximisation

Option 1: Sell only one version

Software Version	Unit Price	Quantity	Total Revenue
A	$150$	$90$	$150 \cdot 90 = 13500$
A	$200$	$45$	$200 \cdot 45 = 9000$
B	$180$	$90$	$180 \cdot 90 = 16200$
B	$500$	$45$	$500 \cdot 45 = 22500$

Option 2: (Bad) Screening

Since consumers care about surplus, i.e.

\text{Surplus} = \text{Willingness} - \text{Price}

You should NOT set the prices at the willingness to pay
Consider the following prices,

P_A = 500, \quad P_B = 150

For type 1 consumers, they will purchase software B since

\begin{aligned} \text{Surplus}_A &= 180 - 500 = -320 \\ \text{Surplus}_B &= 150 - 150 = 0 \end{aligned}

For type 2 consumers, they will purchase software B (instead of software A) since

\begin{aligned} \text{Surplus}_A &= 500 - 500 = 0 \\ \text{Surplus}_B &= 200 - 150 = 50 \end{aligned}

Option 3: Proper Screening

Considering the conundrum above, set the price of A such that the surplus derived by type 1 consumers purchasing A is more than purchasing B
Consider the following prices,

P_A = 449, \quad P_B = 150

For type 1 consumers, they will purchase software B since

\begin{aligned} \text{Surplus}_A &= 180 - 449 = -269 \\ \text{Surplus}_B &= 150 - 150 = 0 \end{aligned}

For type 2 consumers, they will purchase software A since

\begin{aligned} \text{Surplus}_A &= 500 - 449 = 51 \\ \text{Surplus}_B &= 200 - 150 = 50 \end{aligned}

Screening success (hooray!)

Auctions

Types of Auctions

Ascending Price

Increasing price
Types
- Price Sealed / Non-Price Sealed
- First Price Auction / Second Price Auction

Descending Price

Decreasing price

Environment

Common Value: Take the price provided in the open market
Private Value: Price setter since value is intrinsic

Strategic Elements

Asymmetries of information between seller and bidders, and among bidders (screening and signalling elements)
Optimal strategies depend on the type of attitude towards risk, the way values are determined and the type of auction
Revenues may or may not change with the type of auction adopted

Auctions

Second Price Sealed Auction

Bidding your value is a weakly dominant strategy if your utility depends on the surplus you earn

Strategy

If your bid is $b = v$ , your payoff is

\pi = v - v = 0

If your bid is $b < p$ , your payoff is

\pi = v - b > 0

First Price Sealed Auction

Suppose your value for the object is $5 and the other bidder’s values are $4, $3, $3, $2, and $1
These numbers are common knowledge

Strategy

Bidding your value is a weakly dominant strategy
Thus, bid the second highest price, $4

First Price Sealed Auction (Imperfect Information)

Suppose your value for the object is $5 and the other bidder’s values satisfy $b \in [0, 5]$
These numbers are not common knowledge

Strategy

Assume that you have the highest value for this object
Estimate the next highest value for the opponent

Continuous Random Variable

Definition

Random numbers that belong to a subset of the real numbers

N \sim D, \quad N \in [0, 1]

Random Variable

The probability of one outcome is negligible

P(N=k) \rightarrow 0

However, the probability that a random outcome is within a range of a certain number is valid

P(N\le 0) = p, \quad p \in [0, 1]

Call this cumulative density function $F(x) = P(N \le n)$ , we know that

F(0) = 0, \qquad F(1) = 1

$F(x)$ is an increasing function of $x$ , which is differentiable
$F'(x)$ has an important role, and deserves its own letter $f(x)$

f(x) = F'(x) = \lim_{\delta \rightarrow 0} \frac{F(x + \delta) - F(x)}{\delta}

Thus, the expected value $E(x)$ is as follows

E(x) = \int_{0}^{1} x \cdot f(x) dx \\ E(x) = \int_{0}^{1} x \cdot dF(x)

Uniform Distribution

Each outcome in a certain interval is equally likely
The uniform distribution for interval $[0, 1]$ is

F(x) = x \\ f(x) = 1 \\ E(x) = \int_0^1 x F(x) dx = \frac{1}{2}

The uniform distribution for interval $[a, b]$ is

F(x) = \frac{x-a}{b-a} \\ f(x) = \frac{1}{b-a} \\ E(x) = \int_0^1 x F(x) dx = \frac{a+b}{2}

General Distribution

For a general distribution $F$ in $[0, 1]$ , the formula for the expected value is

E(x) = \int_0^1 x f(x) dx \quad \begin{cases} u = x, \quad &du = 1 \\ v = F(x), \quad &dv = f(x) \end{cases}

E(x) = \bigg( x F(x) \bigg)_0^1 - \int_0^1 F(x) dx

E(x) = 1 - \int_0^1 F(x) dx

Equilibirum in the First-Price Auction

Setup

Suppose there are 3 bidders whose object is drawn from the uniform distribution $v \sim U(0, 1)$
Bidder with the value $s$ has the utility as follows:

\pi_x = \begin{cases} v - p \quad & \text{auction is won} \\ 0 \quad & \text{otherwise} \end{cases}

Claim

The equilibrium bidding function is

b(v) = \frac{2}{3} v

Show that the words correspond to the math
The word means $b(v)$ which is the expected value of the largest of my opponents’ $v_s$ given that they are less than my $v$
I am bidder $1$ , competitors are bidder $2$ and $3$
I want to compute the probability that the largest of the values between $v_2$ and $v_3$ is less than or equal to $x$

\max(v_2, v_3) \le v_1

This gives the equation for $F(Y_1)$ , where $Y_1 = \max(v_2, v_3)$
When is it that $\max(v_2, v_3) \le x$ , when $v_2 \le x$ and $v_3 \le x$

Illustration

Case	Constraint 1	Constraint 2	Valid?
1	$v_2 < v_3$	$\{v_2, v_3\} \le x$	✓
2	$v_2 > v_3$	$\{v_2, v_3\} \le x$	✓
3	$v_2 < v_3$	$v_2 \le x, x \ge v_3$	✗
4	$v_2 > v_3$	$v_3 \le x, x \ge v_2$	✗
5	$v_2 < v_3$	$\{v_2, v_3\} \ge x$	✗
6	$v_2 > v_3$	$\{v_2, v_3\} \ge x$	✗

Derivation

Since we are considering the probabilities for both $v_2$ and $v_3$ to be less than $x$ , we use the product rule

F(x) \cdot F(x) = F(x)^2

F'(x) = f(x) = 2x

To compute the expected value under investigation, do

E(x) = \int_0^v \frac{x \cdot f(x)}{v^2} dx = \int_0^v \frac{2x^2}{v^2} dx

E(x) = \frac{2}{v^2} \int_0^v x^2 dx = \frac{2}{v^2}\bigg( \frac{x^3}{3} \bigg)_0^v

E(x) = \frac{2v^3}{3v^2} = \frac{2}{3}v

Verification

Perform BEST RESPONSE ANALYSIS
If bidder 2 and 3 bid according to

b(v_2) = \frac{2}{3} v_2 \quad b(v_3) = \frac{2}{3} v_3 \\

Then the best bidding function for bidder 1 whose value is $v_1$ is

b(v_1) = \frac{2}{3} v_1

The expected payoff is

\pi_1 = \begin{cases} (v_1 - b) \quad &\text{if } b \ge \frac{2}{3}v_2 \text{ and } b \ge \frac{2}{3}v_3 \\ 0 \quad &\text{otherwise} \end{cases}

What is the probability that bidder 1 wins?

P(b \ge \frac{2}{3}v_2 \text{ and } b \ge \frac{2}{3}v_3)

= P(b \ge \frac{2}{3}v_2) \cdot P(b \ge \frac{2}{3}v_3)

= P(v_2 \le \frac{3}{2}b) \cdot P(v_3 \le \frac{3}{2}b)

= F(\frac{3}{2}b) \cdot F(\frac{3}{2}b) = \bigg[F(\frac{3}{2}b)\bigg]^2 = \bigg(\frac{3}{2}b \bigg)^2

P(\text{P1 wins}) = \bigg(\frac{3}{2}b \bigg)^2

E(\pi_1) = (v-b) \cdot P(\text{P1 wins}) + 0 \cdot P(\text{P1 loses})

E(\pi_1) = (v-b) \cdot \bigg(\frac{3}{2}b \bigg)^2

\frac{\partial}{\partial b} E(\pi_1) = \frac{9}{4} \bigg[ 2b(v_1 - b) - b^2 \bigg] = 0

\frac{9}{4} b \bigg[ 2(v_1 - b) - b \bigg] = 0

b \cdot ( 2v_1 - 3b ) = 0

Does $b = 0$ maximise the expected payoff? No.

\therefore b = \frac{2}{3} v_1

All Pay Auction with Common Known Value

Setup

Prize is 1
$n$ bidders such that $n \ge 2$
Each bid is cashed but only the largest bidder gets the prize

Task

We want to compute the symmetric (each bidder adopts the same mixed strategy) mixed strategy equilibrium
Let $F(x)$ be $P(\text{bid} \le x)$
Remember that in mixed strategy equilibria, each bidder mixes indepedently of the other bidders
Let each $x$ in the mixing bunch belong to the interval $[a, b]$
Our goal is to find a formula for $F(x)$ and values for $a$ and $b$

Derivation

We use property 1 of mixed strategy equilibrium: A player must be indifferent between any two bids in the mixing bunch
A player is indifferent between all bids in $[a, b]$
We take POV of bidder 1 who bids $x$

E(x) = 1 \cdot P(\text{win}) - x

Bidder 1 wins when all other bidders bid less than $x$
Each of them bids less than $x$ with probability $F(x)$
We have $n-1$ of the bidders, so we do product rule

P(\text{win}) = F(x)^{n-1}

The expected value as calculated below must be constant for all values of $x \in [a, b]$

\therefore E(\pi_1) = F(x)^{n-1} - x = k, \space \forall x \in [a, b]

Finding $a$

F(a) = 0 \qquad E(\pi_1)_a = 0 - a = -a

If $a > 0$ and I bid $0$ , I lose for sure
Since $b \ge 0$ , $\pi = 0 > -a$ which violates PR2 of mixed strategy
Thus, $a$ must be 0

\therefore a = 0

Finding $F(x)$

With $a = 0$ , $k = 0$ , meaning the mixed strategy equilibrium is 0

E(\pi_1) = F(x)^{n-1} - x = 0

F(x)^{n-1} = x

F(x) = x^{{\frac{1}{n-1}}}

Finding $b$

F(b) = 1

F(b) = b^{\frac{1}{n-1}} = 1

\therefore b = 1

Conclusion ?

Mixed strategy involves

x \in [0, 1] \qquad F(x) = x^{\frac{1}{n-1}}

Note that any bid greater than 1 results in a guaranteed win but negative payoff (which is undesirable)
PR 2 is satisfied (hooray!)

H3 Game Theory Session 10

Screening

Definition

Example

Willingness to pay

Setup

Option 1: Sell only one version

Option 2: (Bad) Screening

Option 3: Proper Screening

Auctions

Types of Auctions

Ascending Price

Descending Price

Environment

Strategic Elements

Auctions

Second Price Sealed Auction

Strategy

First Price Sealed Auction

Strategy

First Price Sealed Auction (Imperfect Information)

Strategy

Continuous Random Variable

Definition

Random Variable

Uniform Distribution

General Distribution

Equilibirum in the First-Price Auction

Setup

Claim

Illustration

Derivation

Verification

All Pay Auction with Common Known Value

Setup

Task

Derivation

Finding aaa

Finding F(x)F(x)F(x)

Finding bbb

Conclusion ?

Finding $a$

Finding $F(x)$

Finding $b$