Appendix
Proof of Lemma 1: We first show that for any sequence of capacity investments b = (b
1, b
2...), with b
t
≥ 0:
$$ \sum^\infty_{t=1} v \cdot (1+r)\cdot b_t \cdot \gamma^t = \sum^\infty_{t=1} c \cdot k_t \cdot \gamma^t $$
where k
t
= b
t
+ β · b
t−1. Direct substitution yields:
$$ \sum^\infty_{t=1} v \cdot (1+r) \cdot b_t \cdot \gamma^{t} =v \cdot [k_1 + \gamma \left[k_2 - \beta \cdot k_1\right] + \gamma^2 [k_3 - \beta \left[ k_2 - \beta \cdot k_1\right] + \gamma^3 \left[k_4 - \beta \left[ k_3 - \beta \cdot \left[k_2 - \beta \cdot k_1\right]\right]\right] + \cdots $$
This expression is linear in each k
t
, and the coefficient on k
1 is:
$$ \begin{aligned} v\left[ 1 - \gamma \beta + \gamma^2 \cdot \beta^2 - \gamma^3 \cdot \beta^3 + \gamma^4 \cdot \beta^4 \ldots \right] =&\,v \left[ \sum^\infty_{i=0} (\gamma \cdot \beta)^{2i} - \sum^\infty_{i=0} (\gamma \cdot \beta)^{2i + 1}\right] \\ =&\,v \cdot \sum^\infty_{i=0} (\gamma \cdot \beta)^{2i} \left[1- \gamma \cdot \beta\right] \\ =&\,v \cdot \frac{1}{1 + \gamma \cdot \beta} = c \cdot \gamma. \\ \end{aligned} $$
Similarly, the coefficient on k
t
is
$$ v \cdot \frac{1} {1 + \gamma \cdot \beta} \cdot \gamma^{t-1} = c \cdot \gamma^t. $$
In terms of future capacity levels, the firm’s discounted future cash flows can therefore be expressed as:
$$ \sum^\infty_{t=1} \left[ M_{d} (k_t,\theta_t) - c \cdot k_t \right] \cdot \gamma^t. $$
This optimization problem is intertemporally separable, and the optimal \(\bar{k}^o_t\) are given by \(\bar{k}^o_t= \bar{k}^o_{1t}+ \bar{k}^o_{2t}\) , where \(\bar{k}^o_{it}\) satisfies the first order conditions:
$$ E_{\epsilon_{i}}\left[R^{'}_{i}(\bar k_{it}^o, \theta_{it},\epsilon_{it})\right] = c $$
The monotonicity condition in (2) ensures that the optimal \(\bar{k}^o_{it}\) are weakly increasing over time. Therefore the non-negativity constraints \(b_t \geq 0\) do not bind.□
Proof of Proposition 1
Using backward induction, consider the decision made by the downstream division in the last period. Independent of the current capacity stock and past decisions, its objective is to maximize:
$$ E_{\epsilon}\left[R_2 (k_{2T} , \theta_{2T} , \epsilon_{2T})\right] - c \cdot k_{2T}. $$
(24)
Let \(\bar{k}^{o}_{2T}(\theta_{2T})\) denote the maximizer of (24). Division 1 faces the constrained optimization problem:
$$ E_{\epsilon}\left[R_1 (k_{1T} , \theta_{1T} , \epsilon_{1T})\right] - c \cdot k_{1T} $$
(25)
subject to the constraint \(k_{1T} + \bar{k}^{o}_{2T}(\theta_{2T})\geq \beta \cdot b_{T-1}.\) Since Division 1’s objective function in (25) is concave, it follows that the optimal capacity level installed at date T − 1 is
$$ k_{T}^* = \max\lbrace \bar{k}^{o}_{1T}(\theta_{1T}) + \bar{k}^{o}_{2T}(\theta_{2T}), \beta \cdot b_{T-2}\rbrace, $$
where \(\bar{k}^{o}_{1T}(\theta_{1T})\) is the unconstrained maximizer of (25). In particular, the upstream division would invest b
T
= 0 if \(k_{T}^* = \beta \cdot b_{T-1}\).
In a subgame perfect equilibrium, Division 2 must select its capacity choice in period T − 1 according to \(\bar{k}^{o}_{2,T-1}(\theta_{2,T-1}),\) irrespective of past decisions. In response, Division 1 will install a capacity level:
$$ k^*_{T-1} = \max\lbrace \bar{k}^{o}_{1,T-1}(\theta_{1,T-1}) + \bar{k}^{o}_{2,T-1}(\theta_{2,T-1}), \beta \cdot b_{T-2}\rbrace. $$
Proceeding inductively, we conclude that in any period, the downstream division will rent the myopically optimal quantity \(\bar{k}^{o}_{2t}(\theta_{2t})\). In response, the upstream division cannot do better than to select the capacity level \(k^*_{t}\) in period t. The assumption that marginal revenues are increasing for each division ensures that
$$ \bar{k}^{o}_{1,t+1}(\theta_{1,t+1}) + \bar{k}^{o}_{2,t+1}(\theta_{2,t+1}) \geq \bar{k}^{o}_{1t}(\theta_{1t}) + \bar{k}^{o}_{2t}(\theta_{2t}). $$
As a consequence, the non-negativity constraint on new investments will not bind and
$$ k^*_{t} = \bar{k}^{o}_{1t}(\theta_{1t}) + \bar{k}^{o}_{2t}(\theta_{2t}) \equiv \bar{k}^{o}_{t}.\quad \square $$
Proof of Proposition 2
Let T = 2 and suppose that the random shocks ε
t
assume their expected value \(\bar{\epsilon}\) for sure. Furthermore, suppose that the divisional revenue functions are identical both cross-sectionally and intertemporally; that is, θ
it
= θ for each i ∈ {1, 2} and each t ∈ {1, 2}. Let \(k \equiv \bar{k}_t^o\) denote the efficient capacity level for each division. Thus, \(R_i^{'}(k, \theta, \bar{\epsilon})=c\). For simplicity, we also set the decay factor β equal to 1. Absent any growth in revenues and absent any decay in capacity, the optimal investment levels are b
1 = 2 · k and b
2 = 0, respectively. To show that there is no sub-game perfect equilibrium which results in efficient capacity investments for some weights u
t
, suppose the downstream division has the entire bargaining power at the negotiation stage; that is, δ = 0.Footnote 32
Step 1:
For any b
1 ≥ 2 · k, Division 2 will secure k
22 = 0 in the second period, and Division 1 will set b
2 = 0.
At the beginning of the second period, Division 1 will choose b
1 so as to maximize:
$$ R_1(b_1 -k_{22}+ b_2, \theta,\bar{\epsilon}) -c \cdot b_2, $$
subject to the constraints b
2 ≥ 0 and b
1 − k
22 + b
2 ≥ 0. We note that the charges corresponding to b
1 are sunk costs. Division 2’s second period capacity demand induces the following optimal response from Division 1:
$$ b_2(k_{22}, b_1)=\left\{ \begin{array}{ll} 0 & \hbox{ if}\quad k_{22} \leq b_1 - k\\ k_{22} + k - b_1 & \hbox{ if}\quad k_{22} \geq b_1 - k. \end{array}\right. $$
(26)
Anticipating this response, Division 2’s second-period profit is given by
$$ \Upgamma (k_{22},\theta,\bar{\epsilon})= M_{f}(b_1,\theta,\bar{\epsilon}) - R_1(b_1 -k_{22},\theta,\bar{\epsilon}) -c \cdot k_{22} $$
for any k
22 ≤ b
1 − k. Thus, we find that \(\Upgamma^{'} (k_{22}, \theta,\bar{\epsilon}) = R^{'}_1 (b_1 - k_{22},\theta,\bar{\epsilon})-c \leq R^{'}_1 (k,\theta,\bar{\epsilon})-c < 0\) for all k
22 ≤ b
1 − k. For any k
22 ≥ b
1 − k the downstream division’s payoff is
$$ \Upgamma (k_{22}, \theta,\bar{\epsilon}) = M_{f}(k +k_{22},\theta,\bar{\epsilon}) - R_1(k,\theta,\bar{\epsilon}) -c \cdot k_{22}. $$
Since by definition \(S(k_{22}+ k, \theta,\bar{\epsilon}) \leq c\), it follows that \(\Upgamma^{'} (k_{22}, \theta,\bar{\epsilon}) < S(k_{22}+k,\theta,\bar{\epsilon})- c < 0.\) This completes the proof of Step 1.
Step 2:
For any b
1 ≥ 2 · k, Division 2’s second period payoff is increasing in b
1.
From Step 1 we know that neither division will obtain additional capacity rights if b
1 ≥ 2 · k. As a consequence, Division 2’s payoff becomes
$$ \Upgamma (0, b_1,\theta,\bar{\epsilon})= M_{f}(b_1,\theta,\bar{\epsilon}) - R_1(b_1,\theta,\bar{\epsilon}). $$
This expression is increasing in b
1 because
$$ \frac{\partial} {\partial b_1}\Upgamma (0, b_1,\theta,\bar{\epsilon})= R_1^{'}(q_1^*(b_1,\theta,\bar{\epsilon}),\theta,\bar{\epsilon})- R_1^{'}(b_1,\theta,\bar{\epsilon})> 0, $$
as \(q_2^*(b_1,\theta,\bar{\epsilon})>0\). We conclude that Division 2 has an incentive to force Division 1 to acquire excess capacity in the first period, that is, to drive b
1 beyond the efficient level 2·k. Division 2 can do so unilaterally by increasing k
21. Doing so is, of course, costly in period 1. Yet, it will be an optimal strategy for the downstream division provided the performance measure weights are such that u
21 is sufficiently small relative to u
22. □
Proof of Lemma 2
Step 1: For a given capacity level k, the expected shadow price of capacity is increasing over time, that is
$$ E_{\epsilon} \left[S(k, \theta_{t+1},\epsilon_{t+1})\right] \geq E_{\epsilon} \left[S(k,\theta_{t},\epsilon_{t})\right]. $$
(27)
For any fixed pair (k, \(\epsilon\)), we claim that
$$ S(k, \theta_{t+1},\epsilon) \geq S(k,\theta_{t},\epsilon). $$
(28)
Suppose first \(0 < q_1^*(k, \theta_{t+1}, \epsilon) \leq q_1^*(k, \theta_{t},\epsilon)< k\). Since \(R_1^{'}(q,\theta_{1t},\epsilon_{1t})\) is increasing in θ1t
and θ1,t+1 ≥ θ1t
, the definition of the shadow price in (11) implies the inequality in (28). Suppose now \(q_1^*(k, \theta_{t+1}, \epsilon) \geq q_1^*(k, \theta_{t},\epsilon)\). Since the shadow price can be expressed as
$$ S(k, \theta_t, \epsilon) = R_{2}^{'}(k-q_{1}^*(k,\theta_t,\epsilon), \theta_{2t}, \epsilon_{2}), $$
θ2,t+1 ≥ θ2t
and \(R_2^{'}(q,\theta_{2t},\epsilon_{2t})\) is increasing in θ2t
, we conclude that (28) holds. The claim now follows because \(\epsilon\)
t
and \(\epsilon\)
t+1 are iid. If \( q_i^*(k, \theta_{t}, \epsilon) = 0\) , a similar argument can be made, keeping in mind that \(S(k, \theta_t, \epsilon) = R_{j}^{'}(k, \theta_{jt}, \epsilon_{j})\) if \( q_i^*(k, \theta_{t}, \epsilon) = 0\).
Step 2: Proceeding exactly as in the proof of Lemma 1, the firm’s expected future cash flows are
$$ \sum^\infty_{t=1} E_{\epsilon_t}\left[ M_{f} (k_t, \theta_t, \epsilon_t ) - c \cdot k_t \right] \cdot \gamma^t. $$
This problem is intertemporally separable, and the optimal k
o
t
satisfy the first order conditions:
$$ E_{\epsilon}\left[\frac{\partial} {\partial k_t} M_{f}(k^o_t, \theta_t, \epsilon_t)\right]= R_{i}^{'}(q_i^*(k^o_t, \theta_{t}, \epsilon), \theta_{it}, \epsilon_{it})=c, $$
provided \( q_i^*(k^o_t, \theta_{t}, \epsilon) > 0\). By definition,
$$ R_{i}^{'}(q_i^*(k^o_t, \theta_{t}, \epsilon), \theta_{it}, \epsilon_{it}) = E_{\epsilon}\left[S_t (k^o_t, \theta_t,\epsilon_t)\right]= c. $$
The claim therefore follows after observing that, by Step 1, the optimal capacity levels, k
o
t
are increasing over time and, as a consequence, the non-negativity constraints b
t
≥ 0 will not be binding.□
Proof of Proposition 3
The proof of Proposition 5 below shows that \(k_t^o = \bar k_t^o \equiv \bar k_{1t}^{o} + \bar k_{2t}^{o}\) when the shadow price S(k
t
, θ
t
, \(\epsilon\)
t
) is linear in \(\epsilon\)
t
. Consequently, it suffices to show that the adjusted full cost transfer pricing induces division i to secure \(\bar k_{it}^{o}\) units of capacity for each i ∈ {1, 2}. Since \(R_{i}^{'}(\cdot,\cdot,\epsilon_{it})\) and S(·, ·, \(\epsilon\)
t
) are linear in \(\epsilon\)
t
, it follows that:
$$ E_{\epsilon}\left[R_{i}^{'}(k_{it},\theta_{it},\epsilon_{it})\right] = R_{i}^{'}(k_{it},\theta_{it},\bar \epsilon_{it}) $$
and
$$ E_{\epsilon}\left[S(k_{t},\theta_{t},\epsilon_{t})\right] = S(k_{t},\theta_{t},\bar \epsilon_{t}). $$
Furthermore,
$$ S(\bar k_t^o, \theta_t, \bar \epsilon_t) =c= R_{i}^{'}(\bar k_{it}^{o},\theta_{it},\bar \epsilon_{it}). $$
As a consequence, the divisional first-order conditions:
$$ (1-\delta) \cdot R_{1}^{'} (\bar k_{1t}^{o}, \theta_{1t}, \bar \epsilon_{1t}) + \delta\cdot S(\bar k_{1t}^o + \bar k_{2t}^o, \theta_t, \bar \epsilon_t) = c $$
(29)
and
$$ \delta\cdot R_{2}^{'} (\bar k_{2t}^{o}, \theta_{2t}, \bar \epsilon_{2t}) + (1-\delta) \cdot S (\bar k_{1t}^o + \bar k_{2t}^o, \theta_t, \bar \epsilon_t ) = c $$
(30)
are met at \( (\bar k_{1t}^{o} , \bar k_{2t}^{o})\). Since, each division’s objective function is globally concave in its choice variable k
it
, the pair \( (\bar k_{1t}^{o} , \bar k_{2t}^{o})\) constitutes a Nash equilibrium. Finally, it is straightforward to check that there cannot be any other pure strategy equilibria. □
Proof of Proposition 4
Conditional on history h
t
and state information θ
t
, Division i chooses b
it
in period t to maximize:
$$ \sum_{\tau=t}^{T} u_{it} \cdot E \left[\pi_{it}(b_{it},b_{jt}|h_t, \theta_t) \right], $$
where π
it
(b
it
, b
jt
|h
t
, θ
t
) is as given in (19) and u
it
is the weight that Division i attaches to its performance measure in period t. Given that each divisions begins period 1 with zero capacity, i.e., h
1 = (0, 0), it can be readily verified that strategy \(b^* \equiv \{b_1^*(h_1,\theta_1),\ldots,b_T^*(h_T,\theta_T)\}\) with:
$$ b_{it}^*(h_t, \theta_t) = \left \{\begin{array}{ll} 0 & \hbox{for}\quad h_{it} \geq \bar k_{it}^o(\theta_t)\hbox{ and all }h_{jt} \geq 0, \\ 0 & \hbox{for} \quad h_{it} < \bar k_{it}^o(\theta_t)\hbox{ and }h_{jt} \geq h_{jt}^{+}, \\ b_{it}^{+} & \hbox{ for}\quad h_{it} < \bar k_{it}^o(\theta_t)\hbox{ and }h_{jt} \in (\bar k_{jt}^o(\theta_t), h_{jt}^{+}),\\ \bar k_{it}^o(\theta_t) - h_{it} & \hbox{ for}\quad h_{it} < \bar k_{it}^o(\theta_t)\hbox{ and }\ h_{jt} \leq \bar k_{jt}^o(\theta_t), \end{array}\right. $$
(31)
leads to efficient capacity levels in each period. To prove the proposition, we will therefore show that b
* is a subgame perfect equilibrium strategy for arbitrary weights u
i
> 0. We first prove the following property of b
*:
Step 1:
In any given period t, suppose each division chooses b
it
to myopically maximize its current payoff π
it
(b
it
, b
jt
|h
t
, θ
t
). Then \(b_t^*(h_t, \theta_t)\) in (31) constitutes a Nash equilibrium strategy in the resulting one-stage game for any \(h_t \in {\mathbb{R}}^2_+\). □
Proof
If b
it
maximizes π
it
(b
it
, b
jt
|h
t
, θ
t
), then the following first-order condition must hold:
$$ (1- \delta_i) \cdot R_1^{'}(h_{it}+b_{it}, \theta_{it}, \bar \epsilon_{it}) + \delta_i \cdot S(h_{it}+b_{it} + h_{jt} +b_{jt}, \theta_t, \bar \epsilon_t) - c \leq 0. $$
(32)
The above conditions will hold as an equality whenever the corresponding b
it
> 0. Since each division’s payoff function is globally concave in its choice variable b
it
, the first-order conditions in (32) are necessary as well as sufficient.
Case I:
\(h_{it} \geq \bar k_{it}^o\) for each i.
In this case, the first-order conditions in (32) hold simultaneously at b
1t
= b
2t
= 0, and hence (0, 0) is a Nash equilibrium strategy.
Case II:
\(h_{it} \leq \bar k_{it}^o\) for each i.
In this case, it can be readily verified from the first-order conditions in (32) that \((\bar k_{1t}^o -h_{1t}, \bar k_{2t}^o- h_{2t})\) is a Nash equilibrium strategy.
Case III:
\(h_{it} \leq \bar k_{it}^o\) and \(h_{jt} \in (\bar k_{jt}^o, h_{jt}^{+})\).
We need to show that \(b_{it} = b_{it}^+\) and b
jt
= 0 is a Nash equilibrium strategy, where \(b_{it}^{+}\) is as given by (21). Given that \(h_{jt} < h_{jt}^{+}\) , \(b_{it}^{+}\) is an optimal response to b
jt
= 0 by definition of \(b_{jt}^+\) in (21). We now claim that \(h_{it}+b_{it}^+ < \bar k_{it}^o\). To the contrary, suppose \(h_{it}+b_{it}^+ \geq \bar k_{it}^o\). Then
$$ R_i^{'}(h_{it}+b_{it}^+, \theta_{it}, \bar \epsilon_{it}) \leq c $$
and
$$ S(h_{it}+b_{it}^{+} + h_{jt}, \theta_t, \bar \epsilon_t) < S(\bar k_{it}^o + k_{jt}^o, \theta_t, \bar \epsilon_t) = c $$
since \(h_{jt} > \bar k_{jt}^o\). But then the first-order condition in (21) cannot hold. This proves the claim that \(h_{it}+b_{it}^+ < \bar k_{it}^o\). Since \(h_{it} + b_{it}^+ < \bar k_{it}^o\) , we have \(R_i^{'}(h_{it} + b_{it}^+, \theta_{it}, \bar \epsilon_{it}) > c\). Therefore, equation (32) implies that
$$ S(h_{it}+b_{it}^{+} + h_{jt}, \theta_t, \bar \epsilon_t) < c. $$
Given the above inequality and the fact that \(R_j^{'}(h_{jt}, \theta_{jt}, \bar \epsilon_{jt}) < c\) , it follows that b
jt
= 0 is the best response to \(b_{it}^{+}\) because
$$ (1-\delta_j) \cdot R_j^{'}(h_{jt}, \theta_{jt}, \bar \epsilon_{jt}) + \delta_j \cdot S(h_{it}+ b_{it}^{+} + h_{jt}, \theta_t, \bar \epsilon_t) < c. $$
Case IV:
\(h_{it} \leq \bar k_{it}^o\) and h
jt
≥ h
+
jt
.
In this case, we need to show that (0, 0) is a Nash equilibrium. Since \(h_{jt} \geq h_{jt}^+\) , equation (20) implies that b
it
= 0 is Division i’s best response to b
jt
= 0. Furthermore, since \(h_{it} \leq \bar k _{it}^o\) , we have \(R_i^{'}(h_{it}, \theta_{it}, \bar \epsilon_{it}) \geq c\) , and therefore (32) implies that:
$$ S(h_{it}+h_{jt},\theta_t, \bar \epsilon_t) \leq c. $$
Given the above inequality and the fact that \(R_j^{'}(h_{jt}, \theta_{jt}, \bar \epsilon_{jt}) < R_j^{'}(\bar k_{jt}^{o}, \theta_{jt}, \bar \epsilon_{jt}) = c\) , it follows that
$$ (1-\delta_j) \cdot R_j^{'}(h_{jt}, \theta_{jt}, \bar \epsilon_{jt}) + \delta_j \cdot S(h_{it}+ h_{jt}, \theta_t, \bar \epsilon_t) < c. $$
Therefore, b
jt
= 0 is a best response to b
it
= 0. This completes the proof of Step 1.
Since the two divisions play a finite multi-stage game with observed actions, the one-stage deviation principle of Theorem 4.1 in Fudenberg and Tirole (1991) applies. To prove that b
* is a subgame perfect equilibrium strategy, it thus suffices to show that there is no division i and no strategy \(\hat b_i\) that agrees with b
* except in a single period such that \(\hat b_i\) is a better response to \(b_j^*\) than \(b_i^*\). Since the two divisions are symmetrical, to prove this result, we will show that Division 1 cannot benefit from any unilateral one-stage deviation.
Suppose the two divisions play the strategy as specified by b
* in (31) for each of the first t − 2 periods. Let \(h_{t-1}^* \equiv (\beta \cdot b_{1,t-2}^*, \beta \cdot b_{2,t-2}^*)\) denote the resulting history at the beginning of period t − 1. We note that each division will begin period t − 1 with less than \(\bar k_{i,t-1}^o\) units of capacity, i.e., \(h_{i,t-1}^* < \bar k_{i,t-1}^o\). In period t − 1, suppose Division 2 plays \(b_{2,t-1}^*(h_{t-1}, \theta_{t-1})\) , but Division 1 deviates to \(\hat b_{1,t-1}\). Let \(\hat h_t \equiv (\hat h_{1t}, \hat h_{2t})\) with
$$ \hat h_{1t} = \beta \cdot \hat b_{1,t-1}, $$
and
$$ \hat h_{2t} = \beta \cdot b_{2,t-1}^*(h_{t-1}^*, \theta_{t-1}) $$
denote the history generated by these choices. In subsequent periods, both divisions revert back to strategy b
*.
We now claim that this one-stage deviation by Division 1 affects its payoffs, and of Division 2, only in periods t − 1 and t. To prove this claim, we first note that any capacity acquired in period t − 1 is gone by the end of period t. Furthermore, strategy profile b
* has the property that \(b_{it}^{*}(h_{t}, \theta_t) \leq \bar k_{it}^{o}\) for each i and any history h
t
. Since \(\bar k_{i,t+1}^o \geq \bar k_{it}^o\) , a one-stage deviation in period t − 1 will never cause either of the two divisions to acquire more than \(\bar k_{i,t+1}^{o}\) units of capacity in period t. Consequently, following a one-stage deviation in period t − 1, continuation of strategy b
* from period t + 1 onwards will generate the same outcomes in all subsequent periods as when strategy b
* is followed during the entire game.
To prove that b
* is a subgame perfect equilibrium strategy, we will therefore show that for all u
1,t−1 ≥ 0 and all u
1t
≥ 0,
$$ \begin{aligned} & u_{1,t-1} \cdot E\left[ \pi_{1,t-1} ( \hat b_{1,t-1}, b_{2,t-1}^*(h_{t-1}^*, \theta_{t-1}) ) \right] + u_{1t} \cdot E\left[ \pi_{1t} ( b_{1t}^* ( \hat h_{t},\theta_t), b_{2t}^*(\hat h_{t}, \theta_{t}) ) \right] \\ & \leq \sum_{\tau=t-1}^{t} u_{1\tau} \cdot E\left[ \pi_{1\tau} ( b_{1\tau}^*(h_{\tau}^*,\theta_{\tau}), b_{2\tau}^*(h_{\tau}^*, \theta_{\tau}) ) \right]. \end{aligned} $$
To show that the above inequality holds for all u
1,t−1 ≥ 0 and all u
1t
≥ 0, we prove that it holds on a period-by-period basis, that is,
$$ \pi_{1,t-1} ( \hat b_{1,t-1}, b_{2,t-1}^*(h_{t-1}^*, \theta_{t-1}) ) \leq \pi_{1,t-1} ( b_{1,t-1}^*(h_{t-1}^*,\theta_{t-1}), b_{2,t-1}^*(h_{t-1}^*, \theta_{t-1})), $$
(33)
and
$$ \pi_{1t} ( b_{1t}^* ( \hat h_{t},\theta_t), b_{2t}^*(\hat h_{t}, \theta_{t}) ) \leq \pi_{1t} ( b_{1t}^*(h_{t}^*,\theta_{t}), b_{2t}^*(h_{t}^*, \theta_{t}) ). $$
(34)
for each θt−1 and θ
t
. The inequality in (33) holds, since Step 1 shows that \(b_{1,t-1}^{*}(h_{t-1}^*, \theta_{t-1})\) is a best response to \(b_{2,t-1}^{*}(h_{t-1}^*, \theta_{t-1})\) in a hypothetical single-stage game in which divisions choose their period t − 1 capacities to myopically maximize their current payoffs.
To establish inequality (34), we prove the following result:
Step 2:
\(\frac{\partial \pi_{1t}^{*}(\hat h_{1t}|\theta_t)} {\partial \hat h_{1t}} \leq 0\) for all θ
t
, where \(\pi_{1t}^{*}(\hat h_{1t}|\theta_t) \equiv \pi_{1t}( b_{1t}^*(\hat h_t, \theta_t), b_{2t}^*(\hat h_t, \theta_t))\).
Proof
Note that \(\hat h_{2t} \leq \bar k_{2t}^o\). Therefore, if \(\hat h_{1t} \leq \bar k_{1t}^o\) , then Division 1 will acquire \(\bar k_{1t}^o - \hat h_{1t}\) number of units in period t and hence \(\frac {\partial \pi_{1t}^{*}(\hat h_{1t}|\theta_t)} {\partial \hat h_{1t}}= 0\).
Consider now the case when \(\hat h_{1t} > \bar k_{1t}^o\). Differentiating \(\pi_{1t}^{*}(\hat h_{1t}|\theta_t)\) with respect to \(\hat h_{1t}\) yields
$$ \frac{\partial \pi_{1t}^{*}(\hat h_{1t}|\theta_t)} {\partial \hat h_{1t}} = \Updelta + \delta \cdot \left[S - R_2^{'}\right] \cdot \frac{\partial b_{2t}^{*}(\hat h_t, \theta_t)} {\partial \hat h_{1t}}, $$
(35)
where, for brevity, we have suppressed the arguments of S and R
'2
, and
$$ \Updelta \equiv (1- {\delta)} \cdot R_1^{'}(\hat h_{1t}, \theta_{1t},\bar \epsilon_{1t}) + \delta \cdot S(\hat h_{1t} + \hat h_{2t} +b_{2t}^*, \theta_t, \bar \epsilon_t) - c. $$
(36)
We note from Step 1 that \((0,b_{2t}^*)\) is a Nash equilibrium in the single-stage game in period t. Consequently, Division 1’s first-order condition in (32) applies and therefore Δ ≤ 0. If \(\hat h_{1t} \geq h_{1t}^{+}\) , then \(b_{2t}^* =0\) and hence (35) implies that \(\frac{\partial \pi_{1t}^*} {\partial \hat h_{1t}} \leq 0\) because Δ ≤ 0 and \(\frac{\partial b_{2t}^*} {\partial \hat h_{1t}}= 0\). Consider now the case \(\hat h_{1t} \in (\bar k_{1t}^o, h_{1t}^{+})\) and \(b_{2t}^* = b_{2t}^+ > 0\). Recall that \(b_{2t}^+\) is defined by equation (21), which yields
$$ \frac{\partial b_{2t}^{+}} {\partial \hat h_{1t}} = \frac{- (1-{\delta)} \cdot S^{'}} {\delta \cdot R_{2}^{''} + (1-{\delta)} \cdot S^{'}} \in (-1,0). $$
Equation (21) can be rearranged to yield
$$ S-c = \delta \cdot (S-R_2^{'}). $$
Substituting this into (35) gives
$$ \frac{\partial \pi_{1t}^{*}(\hat h_{1t}|\theta_t)} {\partial \hat h_{1t}} = \Updelta + \left[S - c\right] \cdot \frac{\partial b_{2t}^+} {\partial \hat h_{1t}}. $$
Lemma A2 follows from the above equation if S > c. Consider now the case when S ≤ c. Substituting for Δ from (36) and simplifying yields
$$ \begin{aligned} \frac{\partial \pi_{1t}^{*}(\hat h_{1t}|\theta_t)} {\partial \hat h_{1t}} = & (1-{\delta)} \cdot \left[R_1^{'}(\hat h_{1t}, \theta_{1t}, \bar \epsilon_{1t}) -S(\hat h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t) \right]\\ & + \left[S(\hat h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t) -c \right] \cdot \left[1+ \frac{\partial b_{2t}^{+}} {\partial \hat h_{1t}}\right]. \end{aligned} $$
The above expression is negative because
- (i):
-
\(S(\hat h_{1t} + \hat h_{2t} + b_2^{+} , \theta_t, \bar \epsilon_t) \leq c\) by assumption,
- (ii):
-
\( 1 + \frac{\partial b_{2t}^{+}} {\partial \hat h_{1t}} \geq 0\), and
- (iii):
-
\(R_1^{'}(\hat h_{1t}, \theta_{1t}, \bar \epsilon_{1t}) < S(\hat h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t).\)
To see why inequality (iii) holds, we recall that
$$ S(\hat h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t) = R_1^{'}(q_{1t}^{*}(h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t), \theta_{1t}, \bar \epsilon_{1t}). $$
Furthermore, since \(\hat h_{1t} > \bar k_{1t}^o,\)
$$ R_1^{'}(\hat h_{1t}, \theta_{1t}, \bar \epsilon_{1t}) < c. $$
The proof of Step 1 showed that \(\hat h_{2t}+ b_{2t}^{+} < \bar k_{2t}^{0}\) , and therefore
$$ R_2^{'}(\hat h_{2t}+b_{2t}^{+}, \theta_{2t}, \bar \epsilon_{2t}) > c. $$
The above two inequalities imply that subsequent to the realization of period t demand shock, the total capacity will be reallocated such that Division 1 transfers some of its capacity to Division 2 when the realized shock satisfies \(\epsilon_t = \bar \epsilon_t\). That is,
$$ q_{1t}^{*}(\hat h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t) < \hat h_{1t}. $$
This implies that
$$ R_1^{'}(q_{1t}^{*}(\cdot), \theta_{1t}, \bar \epsilon_{1t}) \equiv S(\hat h_{1t} + \hat h_{2t} + b_{2t}^{+}, \theta_t, \bar \epsilon_t) > R_1^{'}(\hat h_{1t}, \theta_{1t}, \bar \epsilon_{1t}), $$
and therefore inequality (iii) holds. This completes the proof of Step 2 and Proposition 4. □
Proof of Proposition 5
We first note that the efficient capacity level in the dedicated capacity setting, \(\bar k_t^o\) , can be alternatively defined by the following equation:
$$ S(\bar k_t^o, \theta_t, \bar \epsilon_t) =c. $$
(37)
This holds because (i)\(S(\bar k_t^o, \theta_t, {\bar \epsilon}_t) \equiv R_{1}^{'}(q_1^*(\bar k_t^o, \theta_{t},\bar \epsilon_{t}), \theta_{1t},\bar \epsilon_{1t})\) , and (ii) \(q_1^*(\bar k_t^o, \theta_t, \bar \epsilon_t) = \bar k_{1t}^o\) , since \(\bar k_t^o = \bar k_{1t}^o +\bar k_{2t}^o\) and, given assumption A1, \(\bar k_{it}^o\) satisfies:
$$ E_{\epsilon}\left[R_i^{\prime}(\bar k_{it}^o, \theta_{it}, \epsilon_{it})\right] = R_i^{\prime}(\bar k_{it}^o, \theta_{it}, \bar \epsilon_{it}) = c. $$
The efficient capacity level in the fungible capacity setting is given by
$$ E_{\epsilon}\left[S(k_t^o, \theta_t, \epsilon_t)\right] =c. $$
(38)
When S(·) is linear in \(\epsilon\)
t
, \(E_{\epsilon}\left[S(k_t^o, \theta_t, \epsilon_t)\right] = S(k_t^o, \theta_t, \bar \epsilon_t)\). Equations (37) and (38) therefore imply that \(k_t^o = \bar k_t^o\).
If S(·) is concave in \(\epsilon\)
t
, the application of Jensen’s inequality yields:
$$ E_{\epsilon}\left[S(\bar k_t^o, \theta_t, \epsilon_t)\right] < S(\bar k_t^o, \theta_t, \bar \epsilon_t) =c. $$
(39)
The result \(k_t^o < \bar k_t^o\) then follows because S(k, θ
t
, \(\epsilon\)
t
) is decreasing in k. A similar argument proves that \(k_t^o > \bar k_t^o\) when S(·) is convex. □
Proof of Proposition 6
We first note that a pure strategy Nash equilibrium \((k^{*}_{1t}, k^{*}_{2t})\) always exists. To see this, let f
it
(k
jt
) denote the best response by Division i if it conjectures that Division j chooses k
jt
. For convenience, we drop the subscript t. Let \(\bar{k}_i=f_i(0)\). The functions f
i
(·), defined on [0, ∞), are continuous and monotonically decreasing, because each party’s payoff is concave in its own capacity level. Consider now the function
$$ \Updelta (k_2) \equiv f_1(k_2)- f_2^{-1}(k_2), $$
(40)
defined on the interval \(\left[0,\bar{k}_2\right]\). We have: Δ(0) ≤ 0 because \(f_2(\bar{k}_1)\geq 1\) and f
2(·) is (weakly) decreasing. Furthermore \(\Updelta(\bar{k}_2) \geq 0\) because \(f_1(\bar{k}_2)\geq 0\) , yet \(f_2(0)=\bar{k}_2\). Since Δ (·) is continuous, the Intermediate Value Theorem ensures the existence of a value \(k_2^*\) such that \(\Updelta (k_2^*)=0 \). Letting \(k_1^*= f_1(k_2^*)\) , we conclude that \((k_1^*, k_2^*) \) is a Nash equilibrium.
Suppose now the shadow price S(k
t
, θ
t
, \(\epsilon\)
t
) is a concave function of \(\epsilon\)
t
, and hence \(k_t^o < \bar k_{t}^o\). Suppose also, the contrapositive of the claim, that \((k^{*}_{1t}, k^{*}_{2t})\) induces under-investment in the sense that \(k^{*}_t \equiv k^{*}_{1t}+ k^{*}_{2t} \leq k_t^o\). Since S(k, ·, ·) is a decreasing function, this implies:
$$ E_{\epsilon}\left[S(k_{t}^*,\theta_{t},\epsilon_{t})\right] \geq E_{\epsilon}\left[S(k_{t}^o,\theta_{t},\epsilon_{t})\right] = c. $$
Suppose first that \((k^{*}_{1t}, k^{*}_{2t})>0\). The first-order conditions for a Nash equilibrium then yield:
$$ (1-\delta) \cdot R_{1}^{'} (k^{*}_{1t},\theta_{1t}, \bar \epsilon_{1t}) + \delta\cdot E_{\epsilon}\left[S(k_{t}^{*},\theta_{t},\epsilon_{t})\right] = c $$
(41)
$$ \delta \cdot R_{2}^{'} (k^{*}_{2t}, \theta_{2t}, \bar \epsilon_{2t}) + (1-\delta) \cdot E_{\epsilon}\left[S(k_{t}^{*},\theta_{t},\epsilon_{t})\right] = c. $$
(42)
In order for these conditions to be met, the following inequalities would have to hold:
$$ R_{1}^{'} (k^{*}_{1t}, \theta_{1t},\bar \epsilon_{1t}) \leq c =R_{1}^{'} (\bar k_{1t}^o, \theta_{1t}, \bar \epsilon_{1t}) $$
and
$$ R_{2}^{'} (k^{*}_{2t},\theta_{2t}, \bar \epsilon_{2t}) \leq c =R_{2}^{'} (\bar k_{2t}^o, \theta_{2t}, \bar \epsilon_{2t}). $$
That, however, would contradict the hypothesis:
$$ k^{*}_{1t} + k^{*}_{2t} \leq k_{t}^o < \bar k_{t}^o = \bar k_{1t}^o + \bar k_{2t}^o. $$
If the Nash equilibrium \((k^{*}_{1t}, k^{*}_{2t})\) , involves a boundary value, say \(k^{*}_{1t} =0\) , the first-order condition in (41) is replaced by:
$$ (1-\delta) \cdot R_{1}^{'} (0,\theta_{1t}, \bar \epsilon_{1t}) + \delta\cdot E_{\epsilon}\left[S(k_{t}^*,\theta_{t},\epsilon_{t})\right] \leq c. $$
But that is impossible as well since \( E_{\epsilon}\left[S(k_{t}^*,\theta_{t},\epsilon_{t})\right] = c\) and \(R_{1}^{'} (0,\theta_{1t}, \bar \epsilon_{1t})>c\). A parallel argument shows that there will be under-investment when S(·, ·, \(\epsilon\)
t
) is convex in \(\epsilon\)
t
. □
Proof of Proposition 7
We first claim that, if the two divisions reach an upfront agreement under which Division 2 receives \(k_t^o - \bar k_{1t}^o\) units of capacity for some lump-sum transfer payment of TP
t
, Division 1 will choose the efficient capacity level \(k_t^o\). To prove this, note that, given \(k_{2t} = k_t^o-\bar k_{1t}^o\) , Division 1 will choose k
1t
to maximize:
$$ (1-{\delta)} \cdot E_{\epsilon} \left[R_1(k_{1t}, \theta_{1t},\epsilon_{1t})\right] + \delta \cdot E_{\epsilon} \left [M_f(k_t^{o}+k_{1t}-\bar k_{1t}^o, \theta_t,\epsilon_t) - R_2(k_t^{o}-\bar k_{1t}^o, \theta_{2t}, \epsilon_{2t}) \right] - c \cdot (k_t^{o}+ k_{1t} -\bar k_{1t}^o). $$
(43)
We note that TP
t
is a sunk payment and hence irrelevant to Division 1’s capacity decision. The above maximization problem’s first-order condition, which is necessary as well as sufficient, yields
$$ E_{\epsilon} \left[(1-{\delta)} \cdot R_1^{\prime}(k_{1t}, \theta_{1t},\epsilon_{1t}) + \delta \cdot S(k_{t}^o + k_{1t}-\bar k_{1t}^o, \theta_t,\epsilon_t)\right] = c, $$
which shows that Division 1 will indeed choose \(k_{1t} = \bar k_{1t}^o\) , and hence \(k_t = k_t^o\).
To complete the proof, we need to show that there exists a transfer payment TP
t
such that the ex ante contract \((k_t^o-\bar k_{1t}^o, TP_t)\) will be preferred by both divisions to the default point of no agreement. If the two divisions fail to reach an agreement, Division 1 will choose its capacity level unilaterally, and Division 2 will receive no capacity rights (i.e., k
2t
= 0). Let \(\hat k_t\) denote Division 1’s optimal choice of capacity under this “default” scenario. Division 1’s expected payoff under the default scenario is then given by
$$ \hat \pi_{1t} =E_{\epsilon} \left[(1-\delta) \cdot R_1(\hat k_t,\theta_{1t},\epsilon_{1t}) + \delta\cdot M_f(\hat k_t,\theta_t,\epsilon_t) \right] - c \cdot \hat k_t, $$
(44)
while Division 2’s default payoff is
$$ \hat \pi_{2t} = (1-{\delta)} \cdot E_{\epsilon} \left[M_f(\hat k_{t},\theta_t,\epsilon_t) - R_1(\hat k_t, \theta_{1t}, \epsilon_{1t}) \right]. $$
By agreeing to transfer \(k_t^o - \bar k_{1t}^o\) units of capacity rights to Division 2, the two divisions can increase their ex ante joint surplus by
$$ \Updelta M \equiv E_{\epsilon} \left[M_f(k_{t}^{o},\theta_{t},\epsilon_{t}) - c \cdot k_t^o\right] - E_{\epsilon} \left[M_f(\hat k_{t},\theta_{t}, \epsilon_{t}) -c \cdot \hat k_t \right]. $$
The two divisions can then split this additional surplus between them in proportion to their relative bargaining power. The transfer price that implements this is given by
$$ E_{\epsilon}\left[(1-{\delta)} \cdot R_1(\bar k_{1t}^o,\theta_{1t},\epsilon_{1t}) + \delta \cdot \left[M_f(k_t^o,\theta_t,\epsilon_t) - R_2(k_t^o-\bar k_{1t}^o, \theta_{2t},\epsilon_{2t})\right] \right] + TP_t = \hat \pi_{1t} + \Updelta \cdot \delta M. $$
Division 2’s expected payoff with this choice of transfer payment will be equal to \(\hat \pi_{2t} + (1-{\delta)} \cdot \Updelta M\). Therefore, both divisions will prefer the upfront contract \((k_t^o-\bar k_{1t}^o, TP_t)\) to the default point of no agreement.□