SMU H3 Map

Content map: SMU H3 Game Theory Map

Wrapping Up

Independent Private Values Auctions

Values are Independent Draws from the Same Distribution

Second Price Auction

Same as previous derivations
Weakly dominant to bid your value $v$

First Price Auction

Given your value $v$ , bid the expected value of the largest value among your opponents
This is assuming your opponents bid less than or equal to $v$

E[\max(v_2, v_3, ..., v_n) \space | \space v_1 > \max(v_2, v_3, ..., v_n)]

Random Variables

E[X | x \le v] = \frac{1}{F(v)} \int_a^v x f(x) dx = \frac{1}{F(v)} \int_a^v x dF(x)

Common and Known Values Auctions

The prize is worth $v$ to each bidder

First Price Auction

Bidding $v$ is a dominated strategy
Bid the biggest number less than $v$

All Pay Auction

Everyone pays their bid
No pure strategy equilibrium
Mixed strategy equilibrium exissts
For $v = 1$

F(x) = x^{\frac{1}{n}-1}, \quad \text{s.t. } x \in [0, 1]

War of Attrition (Second-Price All-Pay Auction)

Definition

Everyone pays their bid
The winner pays the second-highest bid
It is a contest which the victor is the one that stays in the contest
Choosing to remain in the contest is costly for both

Solution

Each player mixes their strategy from the interval $[0, +\infty)$

F_1(x) = 1-e^{-\alpha_2x}, \quad F_2(x) = 1-e^{-\alpha_1x}

Where $\alpha_i = \frac{1}{v_i}$
$F(x)$ is valid in the range $[0, 1]$ because as $x \rightarrow \infty$ , $e^{-\alpha_i x} \rightarrow 0$

Setup

Let $F_1$ and $F_2$ denote the mixing functions (CDFs) for P1 and P2

P1 indifference condition

Bidder 1 bids $x$ , Bidder 2 bids $y$

Case 1

If P2 bids $< x$ , P1 wins and pays $y$
Since P1 wins, P1 gets the payoff $v_1 -y$

Case 2

If P2 bids $> x$ , P2 wins and P1 pays $x$
Since P1 loses, P1 gets the payoff $-x$

E(\pi_1) = \underbrace{\int_0^x (v_1 - y) f_2(y)dy}_{\text{Win Case}} + \underbrace{\vphantom{\int_0^x}(-x)(1-F_2(x))}_{\text{Loss Case}}

Applying the indifference condition

$E(\pi_1)$ is a constant function of $x$

E(\pi_1) = \int_0^x (v_1 - y) f_2(y)dy - x(1-F_2(x))

E(\pi_1) = v_1 \int_0^x f_2(y)dy - \int_0^x y f_2(y)dy - x(1-F_2(x))

E(\pi_1) = v_1 F_2(x) - \int_0^x yf_2(y)dy - x(1-F_2(x))

To find the equilibrium, take the derivative with respect to $x$ and set it to zero

\frac{\partial}{\partial x} E(\pi_1) = v_1 f_2(x) - x f_2(x) - \left[(1-F_2(x)) + x(-f_2(x))\right] = 0

v_1 f_2(x) - x f_2(x) - 1 + F_2(x) + x f_2(x) = 0

v_1 f_2(x) = 1-F_2(x)

Since $f_2(x) = F_2'(x)$ , we can rewrite this as a differential equation

\frac{F_2'(x)}{1-F_2(x)} = \frac{1}{v_1}

Using the identity

\frac{\partial}{\partial x}\ln(1-F_2(x)) = -\frac{F_2'(x)}{1-F_2(x)}

We get

\frac{\partial}{\partial x}\ln(1-F_2(x)) = -\frac{1}{v_1} = -\alpha_1

Integration and boundary conditions

Suppose $F_2(x)$ has the domain $[a, b]$ such that $F_2(a) = 0$

\int_a^x \frac{\partial}{\partial t}\ln(1-F_2(t))dt = \int_a^x -\alpha_1 dt

\ln(1-F_2(x)) - \ln(1-F_2(a)) = -\alpha_1(x-a)

Since $F_2(a) = 0$ , $\ln(1-F_2(a)) = \ln(1) = 0$

\ln(1-F_2(x)) = -\alpha_1(x-a)

1-F_2(x) = e^{-\alpha_1(x-a)}

F_2(x) = 1-e^{-\alpha_1(x-a)}

Finding $a$ and $b$

Upper bound: as $x \rightarrow \infty$ , $F_2(x) \rightarrow 1$ , so $b = \infty$
Lower bound: for P1 to be willing to enter the contest at the minimum bid, the expected payoff at the start must be zero

E(\pi_1) = v_1F_2(a)-\int_0^a yf_2(y)dy-a(1-F_2(a)) = 0

0-\int_0^a yf_2(y)dy-a = 0

This condition is satisfied when $a = 0$

F_2(x) = 1-e^{-\alpha_1 x}

By the same logic, P1’s mixed strategy is

F_1(x) = 1-e^{-\alpha_2 x}

P2 indifference condition

The same derivation applies symmetrically, giving $F_1(x) = 1-e^{-\frac{x}{v_2}}$

Remarks

If $v_1 > v_2$ $v_{1} > v_{2}$ , then $F_2(x) < F_1(x)$ $F_{2} (x) < F_{1} (x)$ for all $x$ $x$
- The stronger bidder is more likely to bid less
- The stronger bidder, on average, bids less
- The stronger bidder wins with probability $< \frac{1}{2}$

\int_0^\infty xf_1(x)dx < \int_0^\infty xf_2(x)dx

v_2 < v_1

P(\text{P1 wins}) = \frac{v_2}{v_1+v_2} = \int_0^\infty f_1(x)F_2(x)dx

First Price Common Value Auction with Independent Values

Setup

$V = v_1 + v_2$ is the common value of the prize
$v_1, v_2$ can be any number from 0 to 100 with equal probability, i.e. $v_1, v_2 \sim U(0, 100)$
Strategy is a rule that maps any $v$ from $[0, 1]$ into a non-negative bid $b(v)$

Solution

Equilibrium where both players adopt the same rule $b(v)$ is

b(v) = v

Verify

Suppose I am P1 with the value $v_1$ and thinking of bidding $b$
Assume P2 adopts the rule $b(v_2) = v_2$
What is my expected payoff?

$0 < v_2 < b$	$b < v_2 < 100$
$\pi > 0$	$\pi < 0$

$\because F(v) = \frac{v_2-0}{100-0} = \frac{v_2}{100}, \quad \therefore f(v_2) = \frac{1}{100}$

E(\pi) = \int_0^b(v_1+v_2-b)f(v_2)dv_2

E(\pi) = \frac{1}{100} \int_0^b(v_1+v_2-b)dv_2

E(\pi) = \frac{1}{100}\bigg[ v_2(v_1-b) \bigg]^b_0 + \frac{1}{100}\bigg( \frac{v_2^2}{2} \bigg)^b_0

E(\pi) = \frac{1}{200}\bigg[ 2b(v_1 - b) + b^2 \bigg] = \frac{1}{200}\space b( 2v_1 - b )

E(\pi) = -\frac{1}{200}\space b( b - 2v_1 )

The value of $b$ that maximises this function is the midpoint between $b = 0$ and $b = 2v_1$
Thus, the best bid is $b(v_1) = v_1$

Second Price Common Value Auction

Solution

b(v_1) = 2v_1

Verify

Suppose I am P1 with the value $v_1$ and thinking of bidding $b$
Assume P2 adopts the rule $b(v_2) = 2v_2$

\pi = v_1 + v_2 - 2v_2

What is my expected payoff?

$0 < v_2 < \frac{b}{2}$	$\frac{b}{2} < v_2 < 100$
$\pi > 0$	$\pi < 0$

f(v_2) = \frac{1}{100}

E(\pi) = \int_0^\frac{b}{2} (v_1-v_2)f(v_2)dv_2

E(\pi) = \frac{1}{100}\int_0^\frac{b}{2}(v_1-v_2)dv_2

E(\pi) = \frac{1}{100} \bigg[ v_1v_2 - \frac{v_2^2}{2} \bigg]_0^\frac{b}{2}

E(\pi) = \frac{1}{100} \bigg( \frac{v_1b}{2} - \frac{b^2}{8} \bigg)

E(\pi) = \frac{b}{100} \bigg( \frac{v_1}{2} - \frac{b}{8} \bigg)

E(\pi) = \frac{1}{800} \space b( 4v_1 - b )

The value of $b$ that maximises this function is the midpoint between $b = 0$ and $b = 4v_1$
Thus, the best bid is $b(v_1) = 2v_1$

H3 Game Theory Session 11

SMU H3 Map

Wrapping Up

Independent Private Values Auctions

Second Price Auction

First Price Auction

Random Variables

Common and Known Values Auctions

First Price Auction

All Pay Auction

War of Attrition (Second-Price All-Pay Auction)

Definition

Solution

Setup

P1 indifference condition

Case 1

Case 2

Applying the indifference condition

Integration and boundary conditions

Finding aaa and bbb

P2 indifference condition

Remarks

First Price Common Value Auction with Independent Values

Setup

Solution

Verify

Second Price Common Value Auction

Solution

Verify

Finding $a$ and $b$