SMU H3 Notes Game TheoryGamesSMU H3

Signalling Entry Game with Modified Low Cost

Game theory analysis: Signalling Entry Game with Modified Low Cost.


Setup

Definition:

Signalling Entry Game with Modified Low Cost

  • Players: An incumbent firm (P1), and an entrant firm (P2).
  • Strategies:
    • Nature chooses the incumbent’s marginal cost mc1{10,15}mc_1\in\{10,15\} with probabilities α\alpha and 1α1-\alpha.
    • The incumbent chooses p1{17.5,20}p_1\in\{17.5,20\}.
    • The entrant observes the price and chooses In or Out.
  • Rules:
    • The incumbent has marginal cost 1010 or 1515 and fixed cost 00.
    • The entrant has marginal cost 1010 and fixed cost 4040.
    • If entry does not occur, the incumbent firm has the full market quantity.
p=25qp = 25-q
  • If entry occurs, firms play a Cournot duopoly.
p=25(q1+q2)p = 25-(q_1+q_2)

Game Tree

diagram

Payoff Details

Round 1

π1=pQc1Q\pi_1=pQ-c_1Q π1=(25Q)Q10Q=(15Q)Q\pi_1=(25-Q)Q-10Q=(15-Q)Q Q=7.5,p=17.5,π1=56.2556Q=7.5,\qquad p=17.5,\qquad \pi_1=56.25\approx 56 π1=(pc1)Q\pi_1=(p-c_1)Q π1=(17.515)(2517.5)=18.7519\pi_1=(17.5-15)(25-17.5)=18.75\approx 19

Using the notes’ rounded branch label:

π119\pi_1\approx 19 π1=(25Q)Q15Q=(10Q)Q\pi_1=(25-Q)Q-15Q=(10-Q)Q Q=5,p=20,π1=25Q=5,\qquad p=20,\qquad \pi_1=25

Round 2

p=25(q1+q2)p=25-(q_1+q_2)

For P1 (incumbent):

π1=pq1c1q1\pi_1=pq_1-c_1q_1 π1=q1(25q1q2c1)=q1(15q1q2)\pi_1=q_1(25-q_1-q_2-c_1)=q_1(15-q_1-q_2)

Solving for FOC:

q1E(π1)=152q1q2=0\frac{\partial}{\partial q_1}E(\pi_1)=15-2q_1-q_2=0 q1=15212q2q_1=\frac{15}{2}-\frac{1}{2}q_2

For P2 (entrant):

π2=pq2mc2q2fc2\pi_2=pq_2-mc_2q_2-fc_2 π2=q2(25q1q2mc2)fc2=q2(15q1q2)40\pi_2=q_2(25-q_1-q_2-mc_2)-fc_2=q_2(15-q_1-q_2)-40

Solving for FOC:

q2E(π2)=15q12q2=0\frac{\partial}{\partial q_2}E(\pi_2)=15-q_1-2q_2=0 q2=15212q1q_2=\frac{15}{2}-\frac{1}{2}q_1

Equating the two quantities:

q1=5,q2=5,p=15q_1=5,\qquad q_2=5,\qquad p=15 π1=25,π2=2540=15\pi_1=25,\qquad \pi_2=25-40=-15 p=25(q1+q2)p=25-(q_1+q_2)

For P1 (incumbent):

π1=pq1c1q1\pi_1=pq_1-c_1q_1 π1=q1(25q1q2c1)=q1(10q1q2)\pi_1=q_1(25-q_1-q_2-c_1)=q_1(10-q_1-q_2)

Solving for FOC:

q1E(π1)=102q1q2=0\frac{\partial}{\partial q_1}E(\pi_1)=10-2q_1-q_2=0 q1=512q2q_1=5-\frac{1}{2}q_2

For P2 (entrant):

π2=pq2mc2q2fc2\pi_2=pq_2-mc_2q_2-fc_2 π2=q2(25q1q2mc2)fc2=q2(15q1q2)40\pi_2=q_2(25-q_1-q_2-mc_2)-fc_2=q_2(15-q_1-q_2)-40

Solving for FOC:

q2E(π2)=15q12q2=0\frac{\partial}{\partial q_2}E(\pi_2)=15-q_1-2q_2=0 q2=15212q1q_2=\frac{15}{2}-\frac{1}{2}q_1

Equating the two quantities:

q1=532,q2=2037,p=50317q_1=\frac{5}{3}\approx 2,\qquad q_2=\frac{20}{3}\approx 7,\qquad p=\frac{50}{3}\approx 17 π1=2593,π2=4095\pi_1=\frac{25}{9}\approx 3,\qquad \pi_2=\frac{40}{9}\approx 5

Derivation (Best Response Analysis)

Separating Equilibrium

(p(10),p(15))(17.5,20)(p(10),p(15))\mapsto(17.5,20) p=17.5cI=10,p=20cI=15p=17.5\Rightarrow c_I=10,\qquad p=20\Rightarrow c_I=15 (x,y)(Out,In)(x,y)\mapsto(\text{Out},\text{In})

Pooling Equilibrium

(p(10),p(15))=(17.5,17.5)(p(10),p(15))=(17.5,17.5) P(cI=10p=17.5)=αP(c_I=10\mid p=17.5)=\alpha E(π2)In=α(2540)+(1α)(4540)E(\pi_2)_\text{In}=\alpha(25-40)+(1-\alpha)(45-40) =15α+5(1α)=520α=-15\alpha+5(1-\alpha)=5-20\alpha 520α0α145-20\alpha\leq 0 \quad\Longleftrightarrow\quad \alpha\geq\frac{1}{4} BR2(α)={Inif α<14Outif α14BR_2(\alpha)= \begin{cases} \text{In} \quad &\text{if } \alpha<\frac{1}{4} \\ \text{Out} \quad &\text{if } \alpha\geq\frac{1}{4} \end{cases}

Semi-separating Equilibrium

μ=P(cI=10p=17.5)=αα+q(1α)\mu=P(c_I=10\mid p=17.5)=\frac{\alpha}{\alpha+q(1-\alpha)} P(p=17.5cI=10)P(cI=10)=1αP(p=17.5\mid c_I=10)P(c_I=10)=1\cdot\alpha P(p=17.5)=1α+q(1α)P(p=17.5) =1\cdot\alpha+q(1-\alpha) μ=1α1α+q(1α)=αα+q(1α)\mu = \frac{1\cdot\alpha} {1\cdot\alpha+q(1-\alpha)} = \frac{\alpha}{\alpha+q(1-\alpha)} E(π2)In=μ(2540)+(1μ)(4540)E(\pi_2)_\text{In}=\mu(25-40)+(1-\mu)(45-40) μ(2540)+(1μ)(4540)=0\mu(25-40)+(1-\mu)(45-40)=0 15μ+5(1μ)=0μ=14-15\mu+5(1-\mu)=0 \quad\Longleftrightarrow\quad \mu=\frac{1}{4} αα+q(1α)=14\frac{\alpha}{\alpha+q(1-\alpha)}=\frac{1}{4}

so

4α=α+q(1α)4\alpha=\alpha+q(1-\alpha) q=3α1αq=\frac{3\alpha}{1-\alpha} 3α1α<1α<14\frac{3\alpha}{1-\alpha}<1 \quad\Longleftrightarrow\quad \alpha<\frac{1}{4} 25+3=2825+3=28 x(19+3)+(1x)(19+25)x(19+3)+(1-x)(19+25) 28=x(19+3)+(1x)(19+25)28=x(19+3)+(1-x)(19+25) 28=22x+44(1x)28=22x+44(1-x) x=811x=\frac{8}{11}

Nash Equilibrium

Result:

A pooling equilibrium exists when α14\alpha\geq\frac{1}{4} and is given by

  • Incumbent:
{cI=10,cI=15}{p=17.5,p=17.5}\{c_I=10,c_I=15\}\mapsto\{p=17.5,p=17.5\}
  • Entrant:
{p=17.5,p=20}{Out,In}\{p=17.5,p=20\}\mapsto\{\text{Out},\text{In}\}

A semi-separating equilibrium exists when α<14\alpha<\frac{1}{4} and is given by

  • Incumbent:
cI=10p=17.5c_I=10\mapsto p=17.5 cI=15{p=17.5with probability qp=20with probability 1qwhere q=3α1αc_I=15\mapsto \begin{cases} p=17.5 &\text{with probability } q \\ p=20 &\text{with probability } 1-q \end{cases} \qquad \text{where } q=\frac{3\alpha}{1-\alpha}
  • Entrant:
p=17.5{Inwith prob xOutwith prob 1xwhere x=811,p=17.5\mapsto \begin{cases} \text{In} &\text{with prob } x \\ \text{Out} &\text{with prob } 1-x \end{cases} \qquad \text{where } x=\frac{8}{11}, p=20Inp=20\mapsto\text{In}
  • Posterior beliefs of the entrant are:
P(cI=10p)={14,if p=17.50,if p=20P(c_I=10\mid p)= \begin{cases} \frac{1}{4}, & \text{if } p=17.5\\ 0, & \text{if } p=20 \end{cases}

Social Optimum

2540=15<025-40=-15<0 4540=5>045-40=5>0

Insights

Insight:

  • Raising the low-cost type from 55 to 1010 changes the equilibrium set: pooling and semi-separating can now exist.
  • Pooling works when the prior probability of the low-cost incumbent is large enough to make entry unattractive after p=17.5p=17.5.
  • Semi-separating requires two indifference conditions:
    • The entrant must be indifferent after p=17.5p=17.5.
    • The high-cost incumbent must be indifferent between p=17.5p=17.5 and p=20p=20.
  • The posterior belief μ=14\mu=\frac{1}{4} is pinned down by the entrant’s indifference condition.
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