Sharing Price Announcements

Appendix A

Proof of Lemma 1.

For any announcement $(L_{i}, H_{i})$ of firm $i$ , we define price thresholds $z_{L}$ and $y_{L}$ of the lemma, which depend on $L_{i}$ . Thresholds $z_{H}$ and $y_{H}$ depend on $H_{i}$ and are defined similarly mutatis mutandis. Threshold $y_{L}$ is defined in such a way that $L_{i}$ is binding for all $p_{- i} < y_{L}$ and non-binding for $p_{- i} > y_{L}$ . Threshold $z_{L}$ is defined in such a way that firm $i$ is restricted by $L_{i}$ for all $p_{- i} \in [z_{L}, y_{L})$ and it breaks $L_{i}$ if $p_{- i} < z_{L}$ .

If $B R (p) > L_{i}$ for any $p \in [0, \bar{p}]$ , then $L_{i}$ is never binding and we define $z_{L} = y_{L} = 0$ . If $B R (p) < L_{i}$ for any $p \in [0, \bar{p}]$ , then $L_{i}$ is always binding and we define $y_{L} = \bar{p}$ . In the remaining cases, there is a unique price $y_{L} \in [0, \bar{p}]$ such that $B R (y_{L}) = L_{i}$ so that function $y_{L} (L_{i})$ is the inverse of function $B R (p)$ .

If $y_{L} = 0$ we define $z_{L} = 0$ . If $y_{L} > 0$ , we consider $p \in [0, y_{L})$ . Due to the concavity of $π$ , the total profit decreases in $p_{i}$ on $p_{i} > L_{i}$ so that any price $p_{i} > L_{i}$ is suboptimal. Hence, either $R_{i} = L_{i}$ or $R_{i} = \tilde{B R} (p, γ) < L_{i}$ . If $\tilde{B R} (p, γ) > L_{i}$ for any $p \in [0, y_{L})$ , we define $z_{L} = 0$ . Otherwise (because $\tilde{B R} (y_{L}, γ) > L_{i}$ ), there is a unique price $z_{L} \in [0, y_{L})$ such that $\tilde{B R} (z_{L}, γ) = L_{i}$ so that function $z_{L} (L_{i}, γ)$ is the $p$ -inverse of function $\tilde{B R} (p, γ)$ .

For $p_{- i} \in [y_{L}, y_{H}]$ , firm $i$ is unrestricted so that $R (p_{- i}, L_{i}, H_{i}) = B R (p_{- i})$ . Firm $i$ is restricted by $L_{i}$ if $p \in [z_{L}, y_{L})$ so that $R (p_{- i}, L_{i}, H_{i}) = L_{i}$ . Firm $i$ breaks $L_{i}$ if $p < z_{L}$ so that $R (p_{- i}, L_{i}, H_{i}) = \tilde{B R} (p_{- i}, γ) < L_{i}$ . This ends the proof of the lemma expressions for $R (p, L_{i}, H_{i})$ .

Best response $R (p, L_{i}, H_{i})$ is non-decreasing in $p$ . By construction, $B R (y_{L}) = L_{i} = \tilde{B R} (z_{L}, γ)$ whenever $y_{L}, z_{L} \subset (0, \bar{p})$ . Hence, $R$ is continuous. Finally, $R (p, L_{i}, H_{i})$ is either constant or increases with a slope less than one. Therefore, NE always exists and is unique. Since $R$ is non-decreasing in $(L_{i}, H_{i})$ , equilibrium prices $(p_{i}^{*}, p_{- i}^{*})$ are non-decreasing in all announcements $(L_{i}, L_{- i}, H_{i}, H_{- i})$ .□

Proof of Proposition 1.

Consider the second-period best response function

p_{i} = R (p_{- i}, L_{i})

given by Lemma 1:

$R (p, L_{i}) = \{\begin{matrix} \tilde{B R} (p, γ), if p \in [0, z_{L}) \\ L_{i}, if p \in [z_{L}, y_{L}) \\ B R (p), if p \in [y_{L}, \bar{p}] \end{matrix} .$

First, we argue that for any announcement $(L_{i}, L_{- i})$ , NE prices satisfy $p_{i}^{*} \geq p^{N}$ and $p_{- i}^{*} \geq p^{N}$ . Second, we argue that for any announcements $L_{i} > p^{N}$ and $L_{- i} > p^{N}$ , NE prices satisfy $p_{i}^{*} > p^{N}$ and $p_{- i}^{*} > p^{N}$ . This implies that in any SPNE with outcome $p_{i}^{*} = p_{- i}^{*} = p^{N}$ , it must be that $L_{i}^{*} \leq p^{N}$ and $L_{- i}^{*} \leq p^{N}$ . Finally, we argue that no announcements $L_{i} \leq p^{N}$ and $L_{- i} \leq p^{N}$ can be an SPNE because each firm has a profitable deviation.

When $p < p^{N}$ , $\tilde{B R} (p, γ) > B R (p) > p$ so that $R (p, L_{i}) > p$ . Hence, reaction functions $p_{i} = R (p_{- i}, L_{i})$ and $p_{- i} = R (p_{i}, L_{- i})$ do not intersect when $p_{i} < p^{N}$ or $p_{- i} < p^{N}$ . Thus, second-period equilibrium prices satisfy $p^{*} (L_{i}, L_{- i}) \geq p^{N}$ .

When $L_{i} > p^{N}$ , best responses $p_{i} = R (p_{- i}, L_{i})$ and $p_{- i} = R (p_{i}, L_{i})$ do not intersect at $p_{i} = p_{- i} = p^{N}$ , because firm $i$ is either restricted or breaking so that $p_{i} = R (p^{N}, L_{i}^{*}) > p^{N}$ . Hence, $p_{i}^{*} > p^{N}$ . Then, in turn, $p_{- i}^{*} = R (p_{i}^{*}, L_{- i}^{*}) \geq B R (p_{i}^{*}) > p^{N}$ . Therefore, $p_{i}^{*} = p_{- i}^{*} = p^{N}$ can only occur when $L_{i} \leq p^{N}$ and $L_{- i} \leq p^{N}$ .

Suppose firms announce $(L_{i}, L_{- i})$ with $L_{i} \leq p^{N}$ and $L_{- i} \leq p^{N}$ . The unique NE of this subgame is $p_{i}^{*} = p_{- i}^{*} = p^{N}$ . Let firm $i$ deviate and announce $L_{i} = p^{N} + ε$ marginally above $p^{N}$ . This announcement is binding so that $p_{i}^{*} = L_{i} > p^{N}$ . Firm $- i$ remains unrestricted so that $p_{- i}^{*} = B R (L_{i}) > p^{N}$ . As a result, the deviational profit of firm $i$ is $π (L_{i}, B R (L_{i}))$ , which is increasing at $L_{i} = p^{N}$ . Thus, no SPNE with outcome $p_{i}^{*} = p_{- i}^{*} = p^{N}$ exists.□

Proof of Proposition 2.

According to Proposition 1, an equilibrium where both firms are unrestricted never exists. First, we consider an SPNE where one firm is unrestricted, then we consider an SPNE where neither of them is unrestricted.

Equilibria where one firm is unrestricted.

Let announcements $(L_{i}^{*}, L_{- i}^{*})$ followed by prices $(p_{i}^{*}, p_{- i}^{*})$ be an SPNE outcome, in which $B R (p_{- i}^{*}) < L_{i}^{*}$ and $B R (p_{i}^{*}) \geq L_{- i}^{*}$ . Then, $p_{- i}^{*} = B R (p_{i}^{*}) < y_{L}$ and firm $i$ is either restricted (when $p_{- i}^{*} \geq z_{L}$ so that $\tilde{B R} (p_{- i}^{*}, γ) \geq p_{i}^{*} = L_{i}^{*}$ ) or breaking its announcement (when $p_{- i}^{*} < z_{L}$ so that $\tilde{B R} (p_{- i}^{*}, γ) = p_{i}^{*} < L_{i}^{*}$ ). We show that in an SPNE, $p_{- i}^{*} = z_{L}$ so that $\tilde{B R} (p_{- i}^{*}, γ) = p_{i}^{*} = L_{i}^{*}$ must hold.

Suppose

p_{- i}^{*} < z_{L}

. Then, the best response of firm

i

p_{i} = \tilde{B R} (p_{- i}, γ)

is independent of

L_{i}^{*}

, and firm

i

has afv profitable deviation to reduce its announcement to

L_{i} = p_{i}^{*}

and save deviational cost

C = γ (L_{i}^{*} - p_{i}^{*})

. Hence, it must be that

p_{- i}^{*} \geq z_{L}

so that

\tilde{B R} (p_{- i}^{*}, γ) \geq L_{i}^{*}

and firm

i

is restricted. If firm

i

deviates to

L_{i}

, its profit is

π (L_{i}, B R (L_{i}))

, the leader’s profit in the Stackelberg game. Therefore,

L_{i}^{*} = p^{L}

followed by prices

(p^{L}, p^{F})

is a unique SPNE outcome if

(p^{L}, p^{F})

is the NE of the subgame, i.e., if

\tilde{B R} (B R (p^{L}), γ) \geq p^{L}

. This occurs when

γ > {\bar{γ}}^{A}

, where

{\bar{γ}}^{A}

is defined by

\tilde{B R} (B R (p^{L}), {\bar{γ}}^{A}) = p^{L}

${\bar{γ}}^{A} ≝ - \frac{\partial π}{\partial p_{i}} (p^{L}, p^{F}) .$

In an asymmetric SPNE, price announcement

L_{- i}^{*}

- i

must satisfy two conditions. First, it must be non-binding, i.e.,

L_{- i}^{*} \leq p^{F}

. Second, it must be such that firm

i

does not want to deviate and announce

L_{i} = B R (L_{- i}^{*})

, i.e.,

π (B R (L_{- i}^{*}), L_{- i}^{*}) \leq π^{L *}

. Clearly,

L_{- i}^{*} = 0

satisfies these two conditions so that these asymmetric SPNE always exist. Figure A1 (left) provides a stylized example of firm’ best response functions for the on-path pricing subgame in an asymmetric SPNE. Firm

(- i)

is unrestricted, and its second-stage best-response function

R (p_{i}, L_{- i})

coincides with

B R (p_{i})

; it is the solid black curve. Firm

i

is restricted by

L_{i}

, and its second-stage best-response function

R (p_{- i}, L_{i})

is shown by the solid grey curve.

Equilibria where neither of the firms is unrestricted.

Let announcements $(L_{i}^{*}, L_{- i}^{*})$ followed by prices $(p_{i}^{*}, p_{- i}^{*})$ be an SPNE outcome, in which $B R (p_{- i}^{*}) < L_{i}^{*}$ and $B R (p_{i}^{*}) < L_{- i}^{*}$ . Both announcements are binding, and each firm is either restricted (when $p_{- i}^{*} \geq z_{L}$ so that $\tilde{B R} (p_{- i}^{*}, γ) \geq p_{i}^{*} = L_{i}^{*}$ ) or breaking its announcement (when $p_{- i}^{*} < z_{L}$ so that $\tilde{B R} (p_{- i}^{*}, γ) = p_{i}^{*} < L_{i}^{*}$ ). We show that in an SPNE, $p_{- i}^{*} = z_{L}$ so that $\tilde{B R} (p_{- i}^{*}, γ) = p_{i}^{*} = L_{i}^{*}$ must hold for both firms.

Suppose

p_{- i}^{*} < z_{L}

. Then, the best response of firm

i

p_{i} = R (p_{- i}, L_{i}^{*}) = \tilde{B R} (p_{- i}, γ)

is independent of

L_{i}^{*}

, and firm

i

has a profitable deviation to reduce its announcement to

L_{i} = p_{i}^{*}

and save deviational cost

C = γ (L_{i}^{*} - p_{i}^{*})

. Hence, it must be that

p_{i}^{*} = L_{i}^{*}

and, similarly,

p_{- i}^{*} = L_{- i}^{*}

Figure A1.
Firms’ best responses and Nash equilibria in on-path subgames for asymmetric (left picture) and symmetric (right picture) SPNE. Best responses

B R (p)

are dashed curves that intersect at

p_{i} = p_{- i} = p^{N}

, generalized best responses

\tilde{B R} (p, γ)

are dotted curves, the second stage best-response functions

R (p, L)

are solid curves; grey (black) curves are of firm

i

(

- i

Figure A1.
Firms’ best responses and Nash equilibria in on-path subgames for asymmetric (left picture) and symmetric (right picture) SPNE. Best responses $B R (p)$ are dashed curves that intersect at $p_{i} = p_{- i} = p^{N}$ , generalized best responses $\tilde{B R} (p, γ)$ are dotted curves, the second stage best-response functions $R (p, L)$ are solid curves; grey (black) curves are of firm $i$ ( $- i$ ).

Suppose

\tilde{B R} (p_{- i}^{*}, γ) > p_{i}^{*} = L_{i}^{*}

and

\tilde{B R} (p_{i}^{*}, γ) \geq p_{- i}^{*} = L_{- i}^{*}

. Then, the best response of firm

i

p_{i} = R (p_{- i}, L_{i}^{*}) = L_{i}^{*}

is independent of

p_{- i}

, and firm

- i

has a profitable deviation to marginally reduce

L_{- i}^{*}

L_{- i} = L_{- i}^{*} - ε

. Both firms remain restricted. In NE, firm

i

keeps charging price

p_{i} = p_{i}^{*} = L_{i}^{*}

, and firm

- i

lowers its price to

p_{- i} = L_{- i}

. This deviation is profitable because profit of firm

- i

π (p_{- i}, p_{i}^{*})

, is decreasing in

p_{- i}

when

p_{- i} > B R (p_{i}^{*})

. Hence, it must be that

\tilde{B R} (p_{- i}^{*}, γ) = p_{i}^{*} = L_{i}^{*}

and, similarly,

\tilde{B R} (p_{i}^{*}, γ) = p_{- i}^{*} = L_{- i}^{*}

. Thus, the necessary SPNE condition is

L_{i}^{*} = \tilde{B R} (L_{- i}^{*}, γ)

that must hold for both firms. Since

\tilde{B R} (p, γ)

is increasing in

p

and has slope less than one, the equilibrium, if it exists, is unique, symmetric, and determined by

L_{i}^{*} = L_{- i}^{*} = L^{*}

where

L^{*} = \tilde{B R} (L^{*}, γ)

. Figure A1 (right) provides a stylized example of firm’ best response functions for the on-path pricing subgame in a symmetric SPNE. Both firms are restricted, their second-stage best-response functions are solid black (for

- i

) and grey (for

i

) curves.

To determine equilibrium existence conditions, let us consider firm $i$ . Its equilibrium profit is $π (L^{*}, L^{*})$ . Deviating to $L_{i} > L^{*}$ forces firm $i$ to break $L_{i}$ so that its best response is $p_{i} = \tilde{B R} (p_{- i}, γ)$ . Since the best response of firm $- i$ does not change, NE remains the same, $p_{i} = p_{- i} = L^{*}$ . This deviation only increases the cost of firm $i$ and is, therefore, not profitable. Deviating to $L_{i} < L^{*}$ can either keep firm $i$ restricted, if $L_{i} \in (B R (L^{*}), L^{*})$ , or make it unrestricted, if $L_{i} \leq B R (L^{*})$ . In both cases, firm $- i$ breaks its announcement and has best response $p_{- i} = \tilde{B R} (p_{i}, γ)$ . Profit of firm $i$ equals the generalized price-leader profit ${\tilde{π}}^{L} (p_{i}) = π (p_{i}, \tilde{B R} (p_{i}, γ))$ . Function ${\tilde{π}}^{L} (p)$ is assumed concave and is, therefore, maximized by some ${\tilde{p}}^{L} \in (0, \bar{p})$ . The deviation to $L_{i} < L^{*}$ is not profitable if an only if ${\tilde{π}}^{L} (L_{i})$ increases at $L_{i} = L^{*}$ , i.e., if ${\tilde{p}}^{L} \geq L^{*}$ . Hence, $L^{*} \leq {\tilde{p}}^{L}$ must hold in equilibrium. Therefore, $L^{*} = \tilde{B R} (L^{*}, γ)$ and $L^{*} \leq {\tilde{p}}^{L}$ are the two necessary and sufficient SPNE conditions. Condition $L^{*} \leq {\tilde{p}}^{L}$ implies $\frac{d}{d L^{*}} {\tilde{π}}^{L} (L^{*}) \geq 0$ .

To show that these conditions fail if

γ

is sufficiently large, we write

L^{*} = \tilde{B R} (L^{*}, γ)

\frac{\partial π}{\partial p_{i}} (L^{*}, L^{*}) = - γ

so that

L^{*} \leq {\tilde{p}}^{L}

can be written as follows:

$0 \leq \frac{d}{d L^{*}} {\tilde{π}}^{L} (L^{*}) = \frac{\partial π}{\partial p_{- i}} (L^{*}, L^{*}) - γ .$

This inequality necessarily fails if $γ > \max_{p \in [0, \bar{p}]} \frac{\partial π}{\partial p_{- i}} (p, p)$ . As a result, for large $γ$ , the solution $L^{*}$ (if it exists) to the equation $L^{*} = \tilde{B R} (L^{*}, γ)$ is such that $L^{*} > {\tilde{p}}^{L}$ , and each firm has a profitable deviation to $L_{i} = {\tilde{p}}^{L}$ . Hence there is some ${\bar{γ}}^{S} > 0$ such that for any $γ > {\bar{γ}}^{S}$ , no SPNE exist, including all symmetric equilibria, in which both firms make binding announcements. This ends the proof.□

Proof of Proposition 3.

We consider symmetric and asymmetric SPNE separately.

Asymmetric SPNE.

In an asymmetric SPNE, $p_{i}^{*} = L_{i}^{*} > B R (p_{- i})$ and $p_{- i} = B R (L_{i}^{*})$ . In the proof of Proposition 2, a threshold ${\bar{γ}}^{A}$ has been defined in such a way that $L_{i}^{*} = p^{L}$ is not an SPNE announcement for some $γ < {\bar{γ}}^{A}$ , because $(p^{L}, B R (p^{L}))$ is not NE in the subgame following $L_{i}^{*} = p^{L}$ : firm $i$ has a profitable deviation to charge a price $p_{i} < L_{i}^{*}$ . Because any announcement $L_{i}^{*} > p^{L}$ also suffers from this deviation, only SPNE where $L_{i}^{*} < p^{L}$ may exist. Consider such an SPNE.

Let announcements $(L_{i}^{*}, L_{- i}^{*})$ followed by prices $(p_{i}^{*}, p_{- i}^{*})$ be an SPNE where $L_{i}^{*} < p^{L}$ . Announcement $L_{i}^{*}$ is restrictive so that $B R (p_{- i}^{*}) < L_{i}^{*}$ . Firm $i$ must be restricted so that $\tilde{B R} (p_{- i}^{*}, γ) \geq p_{i}^{*} = L_{i}^{*}$ . If $\tilde{B R} (p_{- i}^{*}, γ) > p_{i}^{*}$ , a marginal deviation to $L_{i} = p_{i}^{*} + ε$ is profitable because firm $i$ remains restricted and its profit $π (L_{i}, B R (L_{i})) = π^{L} (L_{i})$ is increasing in $L_{i}$ for $L_{i}^{*} < p^{L}$ . Hence, it must be that $\tilde{B R} (p_{- i}^{*}, γ) = p_{i}^{*} = L_{i}^{*}$ . Using $p_{- i}^{*} = B R (p_{i}^{*})$ , we write it as $\tilde{B R} (B R (L_{i}^{*}), γ) = L_{i}^{*}$ , Since the left-hand side is continuously increasing in $(L_{i}^{*}, γ)$ and has a slope less than one in $L_{i}^{*}$ , the solution $L_{i}^{*} = p^{A} (γ)$ is unique and is continuously increasing in $γ$ . Since $p^{A} (0) = p^{N}$ and $p^{A} ({\bar{γ}}^{A}) = p^{L}$ , $p^{A} (γ)$ exists for all $γ \in [0, {\bar{γ}}^{A}]$ . This proves part 1 of the proposition.

Symmetric SPNE.

In a symmetric SPNE, $L_{i}^{*} = L_{- i}^{*} = p^{S}$ where $p^{S} = \tilde{B R} (p^{S}, γ)$ . In the proof of Proposition 2, a threshold ${\bar{γ}}^{S}$ is defined in such a way that for any $γ > {\bar{γ}}^{S}$ , equation $\tilde{B R} (p^{S}, γ) = p^{S}$ has no solution $p^{S}$ that satisfies $p^{S} \leq {\tilde{p}}^{L}$ , where ${\tilde{p}}^{L} \in (0, \bar{p})$ is a maximizer of ${\tilde{π}}^{L} (p) = π (p, \tilde{B R} (p, γ))$ . Thus, ${\bar{γ}}^{S}$ is the largest solution to the equation $\tilde{B R} ({\tilde{p}}^{L}, γ) = {\tilde{p}}^{L}$ . Assume $γ < {\bar{γ}}^{S}$ in the rest of the proof.

Since the left-hand side of the equation $\tilde{B R} (p, γ) = p$ is continuously increasing in $(p, γ)$ and has a slope less than one in $p$ , its solution $p^{S} (γ)$ is unique and is continuously increasing in $γ$ . Since $p^{S} (0) = p^{N} < {\tilde{p}}^{L} (0) = p^{L}$ , equilibrium condition $p^{S} (γ) \leq {\tilde{p}}^{L} (γ)$ holds at $γ = 0$ . By continuity, it holds for $γ \in (0, {\underline{γ}}^{S})$ where ${\underline{γ}}^{S} \in (0, {\bar{γ}}^{S}]$ is the smallest solution to the equation $\tilde{B R} ({\tilde{p}}^{L}, γ) = {\tilde{p}}^{L}$ . By construction, the symmetric SPNE always exists for $γ \in (0, {\underline{γ}}^{S}] ⋃ {\bar{γ}}^{S}$ , never exist for $γ > {\bar{γ}}^{S}$ , and only exist for $γ \in ({\underline{γ}}^{S}, {\bar{γ}}^{S})$ if $p^{S} (γ) \leq {\tilde{p}}^{L} (γ)$ . This ends the proof of part 2.

Finally, since $\tilde{B R} (p^{N}, γ) = \tilde{B R} (B R (p^{N}), γ)$ and $B R (p) < p$ for $p > p^{N}$ , it follows that the $\tilde{B R} (p, γ) > \tilde{B R} (B R (p), γ)$ for $p > p^{N}$ .Therefore, $p^{S} (γ)$ , which is the solution to $p^{S} (γ) = \tilde{B R} (p^{S}, γ)$ , is always larger than $p^{A}$ , which is the solution to $p^{A} = \tilde{B R} (B R (p^{A}), γ)$ , i.e., $p^{S} (γ) > p^{A} (γ)$ for any $γ$ . This ends the proof.□

Proof of Lemma 2.

Similar to the proof of Lemma 1, for any announcement

(L_{i}, H_{i})

, we define price thresholds

z_{L}

y_{L}

y_{H}

, and

z_{H}

, which depend on

L_{i}

and

H_{i}

, as follows:

$B R (y_{L}) = L_{i}, B R (y_{H}) = H_{i}, π (B R (z_{L}), z_{L}) - β = π (L_{i}, z_{L}), π (B R (z_{H}), z_{H}) - β = π (H_{i}, z_{H}) .$

If an equality above never holds, we set the corresponding price threshold to either 0 or to

\bar{p}

. Thresholds

y_{L}

and

y_{H}

are unique because of the monotonicity of

B R (p_{- i})

. Threshold

z_{L}

is uniquely determined by

G (z_{L}, L) = β

, where:

$G (z, A) ≝ π (B R (z), z) - π (A, z),$

because of the monotonicity of $G (z, A)$ on $z < y_{L}$ :

$\frac{\partial G}{\partial z} (z, L) = \frac{\partial π}{\partial p_{- i}} (B R (z), z) - \frac{\partial π}{\partial p_{- i}} π (L, z) < 0 .$

The inequality follows from $\frac{\partial^{2} π}{\partial p_{i} \partial p_{- i}} > 0$ and $B R (z) < L$ for $z < y_{L}$ . Similarly, threshold $z_{H}$ is uniquely determined by $G (z_{H}, H) = β$ .

When $p_{- i} \in [y_{L}, y_{H}]$ so that firm $i$ is unrestricted, $R_{i} = B R (p_{- i})$ . When it is restricted, then either $R_{i} = L_{i}$ if $p_{- i} \in [z_{L}, y_{L})$ or $R_{i} = H_{i}$ if $p_{- i} \in (y_{H}, z_{H}]$ . Otherwise, firm $i$ breaks its announcement so that $R_{i} = B R (p_{- i})$ .

Firm $i$ is restricted when either $G (p_{- i}, L) < β$ or $G (p_{- i}, H) < β$ , and it breaks the announcement when $G (p_{- i}, L) > β$ or $G (p_{- i}, H) > β$ . When $G (p_{- i}, L) = β$ or $G (p_{- i}, H) = β$ , the firm is indifferent between these two options and, therefore, has two best responses. This occurs when $p_{- i} \in z_{L}, z_{H}$ .

Best response $R_{i} = R (p, L_{i}, H_{i})$ is non-decreasing in $p$ . It has a slope less than 1 for all values of $p$ when it is continuous. Therefore, NE always exists, yet is not necessarily unique. Since $R_{i}$ is piecewise continuous with three intervals of continuity at most, the number of equilibria is finite.□

Proof of Proposition 4.

As in the proof of Proposition 2, both firms cannot be unrestricted in an SPNE, and cannot break their announcements. First, we consider an SPNE where only one firm is restricted, then we turn to an SPNE where both firms are restricted.

Equilibria where one firm is restricted.

Let announcements

(L_{i}^{*}, L_{- i}^{*})

followed by prices

(p_{i}^{*}, p_{- i}^{*})

be an SPNE outcome, in which firm

i

is restricted so that

p_{i}^{*} = L_{i}^{*}

, and firm

- i

is unrestricted so that

p_{- i}^{*} = B R (L_{i}^{*})

. Firm

i

gets profit

π (L_{i}^{*}, B R (L_{i}^{*}))

, the profit of the Stackelberg leader. Consider a subgame after firm

i

deviates and announces

L_{i}

. According to the equilibrium selection requirement 2, if

L_{i} \leq p^{N}

the firm

i

gets profit

π (p^{N}, p^{N})

, otherwise firm

i

is either restricted and gets profit

π (L_{i}, B R (L_{i}))

(L_{i}, B R (L_{i}))

is NE, or firm

i

breaks its announcement and gets profit

π (p^{N}, p^{N}) - β

in NE

(p^{N}, p^{N})

. Maximizing deviational profit

π (L_{i}, B R (L_{i}))

of firm

i

w.r.t. deviational announcement

L_{i}

we obtain that the best deviation is

L_{i} = p^{L}

. Therefore,

L_{i}^{*} = p^{L}

followed by prices

(p^{L}, p^{F})

is an SPNE outcome if

(p^{L}, p^{F})

is the NE of the subgame, i.e., if

π^{L} (p^{L}) \geq π (B R (p^{F}), p^{F}) - β

. Hence, if

β > {\bar{β}}^{A}

where:

${\bar{β}}^{A} ≝ π (B R (p^{F}), p^{F}) - π^{L} (p^{L}),$

then $L_{i}^{*} = p^{L}$ is an SPNE announcement. The two conditions of Proposition 4 that $L_{- i}^{*}$ must satisfy in an SPNE are the same as in Proposition 2. Figure A2 (left) provides a stylized example of firm’ best response functions for the on-path pricing subgame in an asymmetric SPNE. The second stage best-response function $p_{i} = R_{i} (p_{- i}, L_{i}^{*})$ is discontinuous and multi-valued at $p_{- i} = B R (L_{i}^{*})$ . As a result of this discontinuity, on-path subgame has two NE: $(L_{i}^{*}, B R (L_{i}^{*}))$ and $(p^{N}, p^{N})$ . On path, NE $(L_{i}^{*}, B R (L_{- i}^{*}))$ is played.

Equilibria where both firms are restricted.

Let announcements $(L_{i}^{*}, L_{- i}^{*})$ followed by prices $(p_{i}^{*}, p_{- i}^{*}) = (L_{i}^{*}, L_{- i}^{*})$ be an SPNE outcome, in which $B R (L_{- i}^{*}) < L_{i}^{*}$ and $B R (L_{i}^{*}) < L_{- i}^{*}$ . Both firms are restricted so that $π (L_{i}^{*}, L_{- i}^{*}) \geq π (B R (L_{- i}^{*}), L_{- i}^{*}) - β$ and $p_{i}^{*} = L_{i}^{*}$ . Similar to the proof of Proposition 2, we show that $π (L_{i}^{*}, L_{- i}^{*}) = π (B R (L_{- i}^{*}), L_{- i}^{*}) - β$ .

Suppose, on the contrary, that

π (L_{i}^{*}, L_{- i}^{*}) > π (B R (L_{- i}^{*}), L_{- i}^{*}) - β

, as is shown on Figure A2 (middle). The subgame has two NE:

(L_{i}^{*}, L_{- i}^{*})

and

(p^{N}, p^{N})

; NE

(L_{i}^{*}, L_{- i}^{*})

is on path. Inequality

π (L_{i}^{*}, L_{- i}^{*}) > π (B R (L_{- i}^{*}), L_{- i}^{*}) - β

is equivalent to

z_{L} < L_{- i}^{*}

where

p = z_{L}

is the point of discontinuity of

R_{i} (p, L_{i}^{*})

defined by

π (L_{i}^{*}, z_{L}) = π (B R (z_{L}), z_{L}) - β

. We can now see that firm

- i

has a profitable deviation

L_{- i} = \hat{L} \in (z_{L}, L_{- i}^{*})

. Consider a subgame following announcements

(L_{i}^{*}, \hat{L}) .

The second stage best-response function

p_{- i} = R (p_{i}, \hat{L})

of firm

- i

is shown by a bold dotted black curve. There are two NE in this subgame:

(L_{i}^{*}, \hat{L})

and

(p^{N}, p^{N})

; these are the intersections of this dotted black curve

p_{- i} = R (p_{i}, \hat{L})

and the solid gray curve

p_{i} = R (p_{- i}, L_{i}^{*})

According to the equilibrium selection requirement 2, NE of this subgame is

(L_{i}^{*}, \hat{L})

. Profit of firm

- i

in this deviation is

π (\hat{L}, L_{i}^{*}) > π (L_{- i}^{*}, L_{i}^{*})

because (1)

π (p_{- i}, L_{i}^{*})

decreases with

p_{- i}

p_{- i} > B R (L_{i}^{*})

, and (2)

\hat{L} \in (B R (L_{i}^{*}), L_{- i}^{*})

as it can be seen on Figure A2 (middle). Thus,

π (L_{i}^{*}, L_{- i}^{*}) = π (B R (L_{- i}^{*}), L_{- i}^{*}) - β

must hold for both firms. Figure A2 (right) presents this case. The on-path subgame has two NE:

(L_{i}^{*}, L_{- i}^{*})

and

(p^{N}, p^{N})

. On path, NE

(L_{i}^{*}, L_{- i}^{*})

is played.

Using function

G (z, A)

, which is defined in the proof of Lemma 2:

$G (z, A) ≝ π (B R (z), z) - π (A, z),$

we write the necessary SPNE conditions $π (L_{i}^{*}, L_{- i}^{*}) = π (B R (L_{i}^{*}), L_{- i}^{*}) - β$ as $G (L_{i}^{*}, L_{- i}^{*}) = β$ and $G (L_{- i}^{*}, L_{i}^{*}) = β$ .

In order to see that the equilibrium is necessarily symmetric, we note that for the relevant range of the arguments, i.e., for

A = L

and

z < y_{L}

so that

B R (z) < L

G (z, A)

is decreasing in

z

(this is shown in the proof of Lemma 2) and is increasing in

A

. Hence, the equilibrium equation

G (L_{i}^{*}, L_{- i}^{*}) = β

defines

L_{i}^{*}

as a continuously increasing function of

L_{- i}^{*}

, i.e.,

L_{i}^{*} = F (L_{- i}^{*}, β)

. Similarly,

L_{- i}^{*} = F (L_{i}^{*}, β)

. In equilibrium, both conditions

L_{i}^{*} = F (L_{- i}^{*}, β)

and

L_{- i}^{*} = F (L_{i}^{*}, β)

must hold. Suppose that there is an asymmetric SPNE

(L_{i}^{*}, L_{- i}^{*}) = (L_{1}, L_{2})

with

L_{1} > L_{2}

. Then, both conditions

L_{1} = F (L_{2}, β)

and

L_{2} = F (L_{1}, β)

must hold for

L_{1} > L_{2}

, which contradicts the monotonicity of

F

. Thus, the equilibrium, if it exists, is necessarily symmetric,

L_{i}^{*} = L_{- i}^{*} = L^{*}

, and is determined by condition

G (L^{*}, L^{*}) = β

$π (B R (L^{*}), L^{*}) - π (L^{*}, L^{*}) = β .$

Figure A2.
Firms’ best responses and Nash equilibria in on-path subgames for asymmetric (left picture) and symmetric (right picture) SPNE. The middle picture illustrates why

π (L_{i}^{*}, L_{- i}^{*}) > π (B R (L_{- i}^{*}), L_{- i}^{*}) - β

cannot happen on path in a symmetric SPNE. Best responses

B R (p)

are dashed curves that intersect at

p_{i} = p_{- i} = p^{N}

, the second stage best-response functions

R (p, L)

are thick solid curves,

z_{L}

is the point of discontinuity of

R (p_{- i}, L_{i})

, grey (black) curves are of firm

i

(

- i

Figure A2.
Firms’ best responses and Nash equilibria in on-path subgames for asymmetric (left picture) and symmetric (right picture) SPNE. The middle picture illustrates why $π (L_{i}^{*}, L_{- i}^{*}) > π (B R (L_{- i}^{*}), L_{- i}^{*}) - β$ cannot happen on path in a symmetric SPNE. Best responses $B R (p)$ are dashed curves that intersect at $p_{i} = p_{- i} = p^{N}$ , the second stage best-response functions $R (p, L)$ are thick solid curves, $z_{L}$ is the point of discontinuity of $R (p_{- i}, L_{i})$ , grey (black) curves are of firm $i$ ( $- i$ ).

This condition necessarily fails for $β > \max_{p \in [0, \bar{p}]} (π (B R (p), p) - π (p, p))$ . Hence, there is some ${\bar{β}}^{S} > 0$ such that for any $β > {\bar{β}}^{S}$ , no SPNE exist, including all symmetric equilibria, in which both firms make binding announcements. This ends the proof.□

Proof of Proposition 5.

We consider symmetric and asymmetric SPNE separately.

Asymmetric SPNE.

In an asymmetric SPNE, $p_{i}^{*} = L_{i}^{*} > B R (p_{- i})$ and $p_{- i} = B R (L_{i}^{*})$ . In the proof of Proposition 4, a threshold ${\bar{β}}^{A}$ has been defined in such a way that $L_{i}^{*} = p^{L}$ is not an SPNE announcement for some $β < {\bar{β}}^{A}$ . Consider an SPNE where $L_{i}^{*} < p^{L}$ .

Let announcements

(L_{i}^{*}, 0)

followed by prices

(p_{i}^{*}, p_{- i}^{*})

be an SPNE where

L_{i}^{*} < p^{L}

and

p_{- i}^{*} = B R (p_{i}^{*})

. Firm

i

is restricted by

L_{i}^{*}

so that

B R (p_{- i}^{*}) < L_{i}^{*} = p_{i}^{*}

and:

$π (L_{i}^{*}, B R (L_{i}^{*})) \geq π (B R (B R (L_{i}^{*})), B R (L_{i}^{*})) - β .$

Profit of firm $i$ in this SPNE is $π^{L} (L_{i}^{*}) = π (L_{i}^{*}, B R (L_{i}^{*}))$ , which is increasing in $L_{i}^{*}$ for $L_{i}^{*} < p^{L}$ . This implies that if $L_{i}^{*} < p^{A}$ then firm $i$ has a profitable deviation $L_{i}^{*} = p^{A}$ (similar to the proof of Proposition 4, we use the equilibrium selection requirement 2 for making this argument). Hence, $L_{i}^{*} = p^{A}$ must hold.

To show that

p^{A} (β)

is well defined and increasing, we consider a function

x (p, β)

x (p, β) \geq B R (p)

, implicitly defined by:

$π (x, p) = π (B R (p), p) - β .$

Function

x (p, β)

is well defined on

β < {\bar{β}}^{A}

and is increasing:

$\frac{\partial x}{\partial β} = - {(\frac{\partial π}{\partial p_{i}} (x, p))}^{- 1} > 0,$

because $x (p, β) > B R (p) > p^{N}$ for $β > 0$ and $p > p^{N}$ so that $\frac{\partial π}{\partial p_{i}} (x, p) < 0$ , and:

$\frac{\partial x}{\partial p} = \frac{\frac{\partial π}{\partial p_{- i}} (B R (p), p) - \frac{\partial π}{\partial p_{- i}} (x, p)}{\frac{\partial π}{\partial p_{i}} (x, p)} = \frac{\int_{B R (p)}^{x} \frac{\partial^{2} π (t, p)}{\partial p_{i} \partial p_{- i}} d t}{\int_{B R (p)}^{x} (- \frac{\partial^{2} π (t, p)}{\partial p_{i}^{2}}) d t} \in (ε, 1 - ε),$

because of the assumption $(\frac{\partial^{2} π}{\partial p_{i} \partial p_{- i}}) / (- \frac{\partial^{2} π}{\partial p_{i}^{2}}) \in (ε, 1 - ε)$ .

An SPNE announcement $p^{A} (β)$ is a solution to $p^{A} = x (B R (p^{A}), β)$ , i.e., a fixed point of $x (B R (p), β)$ , and $p^{A} ({\bar{β}}^{A}) = p^{L}$ by construction. Since $x (p^{N}, β) > p^{N}$ , for any $β \in (0, {\bar{β}}^{A})$ there is a unique solution $p^{A} (β) \in (p^{N}, p^{L})$ that is increasing in $β$ . This proves part 1 of the proposition.

Symmetric SPNE.

In a symmetric SPNE,

L_{i}^{*} = L_{- i}^{*} = p^{S}

, where

p^{S}

satisfies:

$π (p^{S}, p^{S}) = π (B R (p^{S}), p^{S}) - β .$

In the proof of Proposition 4, a threshold ${\bar{β}}^{S}$ has been defined in such a way that for any $β > {\bar{β}}^{S}$ , this equation does not define an SPNE announcement $p^{S}$ . Assume $β \in (0, {\bar{β}}^{S})$ in the rest of the proof.

The symmetric SPNE announcement $p^{S} (β)$ is the unique solution of $p^{S} = x (p^{S}, β)$ and satisfies $p^{S} > p^{N}$ . The asymmetric SPNE announcement $p^{A} (β)$ is the unique solution of $p^{A} = x (B R (p^{A}), β)$ , and it also satisfies $p^{A} > p^{N}$ . Since $B R (p) < p$ for $p > p^{N}$ , it follows that $x (B R (p), β) < x (p, β)$ . As a result, $p^{A} (β)$ , the fixed point of $x (B R (p), β)$ , is always smaller than $p^{S} (β)$ , the fixed point of $x (p, β)$ : $p^{A} (β) < p^{S} (β)$ . This proves part 3 of the proposition.

In a symmetric SPNE, firms’ profit is

π (p^{S}, p^{S})

. It can be seen in Figure A2 (right), that any deviation to

L_{i} > p^{S}

and a marginal deviation to

L_{i} < p^{S}

result in a subgame with the unique ‘normal’ NE

(p^{N}, p^{N})

. Indeed, when

L_{i}

increases, the point of discontinuity of the second-stage best-response function

p_{i} = R (p_{- i}, L_{i})

(solid grey curve) shifts to the left. When

L_{i}

marginally decreases, this point of discontinuity shifts to the right, but the horizontal part of the graph shifts downwards. In both cases,

(p^{N}, p^{N})

is the unique NE after these deviations. Since

p^{S} > p^{N}

$π (p^{S}, p^{S}) = π (B R (p^{S}), p^{S}) - β > π (B R (p^{N}), p^{N}) - β = π (p^{N}, p^{N}) - β .$

Therefore, neither of these deviations are profitable. Only infra-marginal deviations to $L_{i} < p^{S}$ that lead to subgames with NE $(L_{i}, B R (L_{i}))$ can be profitable. Deviational profit of firm $i$ is $π (L_{i}, B R (L_{i})) = {\tilde{π}}^{L} (L_{i}, 0)$ which increases on $L_{i} \in [p^{N}, p^{L}]$ . Hence, the most profitable deviation is either $L_{i} = p^{L}$ if $β \geq {\bar{β}}^{A}$ or $L_{i} = p^{A}$ if $β < {\bar{β}}^{A}$ .

This implies that if

β \geq {\bar{β}}^{A}

, then

p^{S}

is an SPNE announcement if and only if:

$π (p^{S}, p^{S}) \geq π (p^{L}, B R (p^{L})) = π (p^{L}, p^{F}),$

and if $β \leq {\bar{β}}^{A}$ , then $p^{S}$ is an SPNE announcement if and only if:

$π (p^{S}, p^{S}) \geq π (p^{A}, B R (p^{A})) .$

To show that a symmetric SPNE exists for small values of

β

, we write:

$(p^{S}, p^{S}) = π (B R (p^{S}), p^{S}) - β, π (p^{A}, B R (p^{A})) = π (B R (B R (p^{A})), B R (p^{A})) - β,$

so that

$π (p^{S}, p^{S}) - π (p^{A}, B R (p^{A})) = π (B R (p^{S}), p^{S}) - π (B R (B R (p^{A})), B R (p^{A})) = π^{F} (p^{S}) - π^{F} (B R (p^{A})) > 0,$

where $π^{F} (p) = π (B R (p), p)$ , because $π^{F} (p)$ is increasing and $p^{S} > p^{A} > B R (p^{A})$ . Hence, the deviation to $L_{i} = p^{A}$ is not profitable for all $β \leq {\bar{β}}^{A}$ .

This implies that there exists a ${\underline{β}}^{S} \in ({\bar{β}}^{A}, {\bar{β}}^{S})$ such that for all $β < {\underline{β}}^{S}$ , $p^{S}$ exists and is an SPNE announcement. This proves part 2 of the proposition.

To prove part 4, we consider functions

p^{A} (β)

and

p^{S} (β)

, and the defining them equations

G_{A} (p) = 0

and

G_{S} (p) = 0

. At

β = 0

, they result is

p^{A} (0) = p^{S} (0) = p^{N}

, and in derivatives:

$\frac{d G_{A}}{d p} (p^{N}) = \frac{d G_{S}}{d p} (p^{N}) = - \frac{\partial π}{\partial p_{i}} π (p^{N}, p^{N}) = 0,$

due to $\frac{\partial π}{\partial p_{i}} (B R (p), p) = 0$ (this is the first-order condition that defines $B R (p)$ ) and $\frac{\partial π}{\partial p_{i}} (p^{N}, p^{N}) = 0$ (this is the definition of $p^{N}$ ). Since $p^{A} (β)$ and $p^{S} (β)$ are the inverses of $G_{A} (p)$ and $G_{S} (p)$ correspondingly, their derivatives at $β = 0$ are infinite. Derivatives of the asymmetric equilibrium profits at $β = 0$ are also infinite:

$\begin{array}{l} \frac{d π_{i}}{d β} = \frac{d}{d β} π (p^{A}, B R (p^{A})) = \frac{\partial π}{\partial p_{- i}} (p^{N}, p^{N}) {B R}^{'} (p^{N}) \frac{d p^{A}}{d β}, \\ \frac{d π_{- i}}{d β} = \frac{d}{d β} π (B R (p^{A}), p^{A}) = \frac{\partial π}{\partial p_{- i}} (p^{N}, p^{N}) \frac{d p^{A}}{d β} . \end{array}$

Derivatives of the symmetric equilibrium profits at

β = 0

are infinite as well:

$\frac{d π_{i}}{d β} = \frac{d π_{- i}}{d β} = \frac{d}{d β} π (p^{S}, p^{S}) = \frac{\partial π}{\partial p_{- i}} (p^{N}, p^{N}) \frac{d p^{S}}{d β} .$

This ends the proof.□

Cost Thresholds.

For the numerical example used in this paper, cost thresholds are given by:

$\begin{matrix} {\bar{γ}}^{A} & = \frac{(2 b + 3 d) d^{2}}{4 (2 {(b + d)}^{2} - d^{2}) (b + d)}, \\ {\underline{γ}}^{S} & = {\bar{γ}}^{S} = \frac{d^{2}}{2 b (2 b + 3 d) + d^{2}} > {\bar{γ}}^{A}, \\ {\bar{β}}^{A} & = \frac{d^{4} {(2 b + 3 d)}^{2}}{64 {(b + d)}^{3} {(2 {(b + d)}^{2} - d^{2})}^{2}}, \\ {\underline{β}}^{S} = {\bar{β}}^{S} & = {(\frac{d}{4 (b + d) b} (\sqrt{(b + d)} + \sqrt{(b + d) - \frac{b d^{2}}{2 (2 {(b + d)}^{2} - d^{2})}}))}^{2} > {\bar{β}}^{A} . \end{matrix}$

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Maarten Janssen www.mdpi.com

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Sharing Price Announcements

Appendix A

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Appendix A

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