Strategic advance sales, demand uncertainty and overcommitment

Abstract

We study a game in which producers can sell both before and after the realization of a random demand. Models of strategic forward trading and of advance sales to intermediaries or consumers share this structure. Demand uncertainty and committed advance sales imply that final-stage net residual demand may be so low that producers find additional sales unprofitable, or even so low that the final period price is zero, introducing some convexity into producers’ payoffs. If such an ex-post overcommitment occurs on the equilibrium path, producers reduce their advance sales, muting the pro-competitive effects found under deterministic demand. We prove existence of a unique symmetric pure-strategy equilibrium, whose nature depends on the support of the demand distribution relative to the marginal cost of production. For a narrow support, existence is assured for any distribution, while for wider support we establish a sufficient condition for existence. Our results provide a precise characterization of “minor” uncertainty, in which only the expected value of demand affects the equilibrium which is otherwise qualitatively similar to the deterministic case.

Appendix

1.1 Proof of Theorem 1

The proof proceeds as follows. We first prove that there exists a unique best response, denoted ${{\hat{x}}}_i(X_{-i})$, for each possible value of $X_{-i}$. Then, we show that the best response ${{\hat{x}}}_i(X_{-i})$ is continuous and everywhere strictly decreasing with respect to $X_{-i}$, with a slope smaller than 1 (in absolute value). Hence, a unique symmetric equilibrium to the game can be determined by solving for $x^*$ in $x^* = {{\hat{x}}}_i((n-1) x^*)$.

Consider first the case $x_i \le a_L-c - X_{-i}$. As (17) is clearly strictly decreasing in $x_i$, the expected profit is concave in $x_i$ given $X_{-i}$ on the interval $x_i\in [0, a_L-c-X_{-i}]$, and reaches a maximum either at

$$\begin{aligned} x_i^{(1)}(X_{-i}) = \frac{(n-1)(E[a] - c) - (n-1) X_{-i}}{2 n} \end{aligned}$$

(36)

if this quantity is smaller than $a_L - c - X_{-i}$, that is if

$$\begin{aligned} X_{-i} < \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}, \end{aligned}$$

(37)

or at $a_L-c-X_{-_i}$ otherwise.

Consider now the case $x_i \in [a_L-c-X_{-i},a_H-c-X_{-i}]$. From (19), at the upper bound of this interval, $x_i = a_H-c-X_{-i}$, marginal expected profit is strictly negative³², and hence, if it exists, the advance sales that maximize expected profit must be strictly smaller than $a_H-c-X_{-i}$. Differentiating (19) with respect to $x_i$ yields

$$\begin{aligned}&\frac{\partial ^2 \pi _{II}^e(x_i, X_{-i})}{\partial x_i^2} \nonumber \\&\quad = (c+X - c - 2 x_i - X_{-i}) f(c+X) + \int _{a_L}^{c+X} - 2 f(a) \hbox {d}a\nonumber \\&\qquad - \frac{(n-1) X - 2 n x_i - (n-1) X_{-i}}{(n+1)^2} f(c+X) + \int _{c+X}^{a_H} - \frac{2 n}{(n+1)^2} f(a) \hbox {d}a\nonumber \\&\quad = - x_i f(c+X) - 2 F(c+X) + \frac{(n+1) x_i f(c+X)}{(n+1)^2} - \frac{2 n (1-F(c+X))}{(n+1)^2} \nonumber \\&\quad =- \frac{n x_i f(c+X)}{n+1} - \frac{2 n}{(n+1)^2} - \frac{2 (n^2 + n +1) F(c+X)}{(n+1)^2} < 0. \end{aligned}$$

(38)

Expected profit is concave in $x_i$ given $X_{-i}$ on $x_i \in [a_L-c-X_{-i},a_H-c-X_{-i}]$, and its maximum occurs either in the interior of $[a_L-c-X_{-i},a_H-c-X_{-i}]$ or at its lower bound. We denote an interior maximum $x_i^{(2)}(X_{-i})$ which is the unique $x_i$ that equates (19) to zero.

Expected marginal profit at $x_i = a_L - c - X_{-i}$ is obviously continuous³³. As it is strictly decreasing, checking that $x_i^{(2)}(X_{-i}) > a_L-c-X_{-i}$ is equivalent to verify whether $X_{-i}$ is such that the limit of the marginal profit (19) at $x_i = a_L-c-X_{-i}$ is strictly positive. It is immediate that $x_i^{(2)}(X_{-i}) > a_L-c-X_{-i}$ is equivalent to $X_{-i} > \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}$, which is the condition on $X_{-i}$ under which the maximum of the expected profit on $[0,a_L-c-X_{-i}]$ is located at $x_i = a_L-c-X_{-i}$. Consequently, for $X_{-i} > \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}$, marginal expected profit is positive and strictly decreasing on $[0, a_L-c-X_{-i}]$, strictly decreasing on $[a_L-c-X_{-i}, a_H-c-X_{-i}]$, and strictly negative at $x_i=a_H-c-X_{-i}$. Therefore, expected profit is concave in $x_i$ for $x_i\in [0, a_H-c-X_{-i}]$ and there is a unique best response ${{\hat{x}}}_i(X_{-i})$ which maximizes the expected profit of firm i on the range $x_i \in [0,a_H-c-X_{-i}]$ when $a_H - a_L \le c$, which is given by

$$\begin{aligned} {{\hat{x}}}_i(X_{-i})=\left\{ \begin{array}{ll} x_i^{(1)}(X_{-i}) &{}\quad \text{ if } X_{-i} \le \frac{2n (a_L - c) - (n-1) (E[a] - c)}{n+1}\\ x_i^{(2)}(X_{-i}) &{}\quad \text{ if } X_{-i} > \frac{2n (a_L - c) - (n-1) (E[a] - c)}{n+1} \end{array} \right. \end{aligned}$$

(39)

We now demonstrate that each branch of the best response (39) is strictly decreasing in $X_{-i}$ with a slope smaller than 1 in absolute value. First, $x_i^{(1)}(X_{-i})$ is a Cournot-like best response and is obviously strictly decreasing with respect to $X_{-i}$, with a slope smaller than 1 in absolute value. Let us establish it is also true for $x_i^{(2)}(X_{-i})$.

The slope of this best response can be studied by applying the Implicit Function Theorem to the equation $\frac{\partial \pi _{II}^e }{\partial x_i}=0$, which gives: $d x_i^{(2)}/d X_{-i} = - \frac{\partial ^2 \pi _{II}^e }{\partial x_i \partial X_{-i}}/ \frac{\partial ^2 \pi _{II}^e }{\partial x_i^2}$. The cross-partial derivative of the expected profit is given by:

$$\begin{aligned}&\frac{\partial ^2 \pi _{II}^e(x_i, X_{-i})}{\partial x_i \partial X_{-i}}\nonumber \\&\quad = (c+X - c - 2 x_i - X_{-i}) f(c+X) + \int _{a_L}^{c+X} (-1) f(a) \hbox {d}a\nonumber \\&\qquad - \frac{(n-1) X - 2 n x_i - (n-1) X_{-i}}{(n+1)^2} f(c+X) + \int _{c+X}^{a_H} - \frac{n - 1}{(n+1)^2} f(a) \hbox {d}a\nonumber \\&\quad = - x_i f(c+X) - F(c+X) + \frac{(n+1) x_i f(c+X)}{(n+1)^2} - \frac{(n-1) (1-F(c+X))}{(n+1)^2} \nonumber \\&\quad =- \frac{n x_i f(c+X)}{n+1} - \frac{n-1}{(n+1)^2} - \frac{(n^2+n+2) F(c+X)}{(n+1)^2} < 0. \end{aligned}$$

(40)

Therefore, $x_i^{(2)}(X_{-i})$ is strictly decreasing with respect to $X_{-i}$ and so is the best response ${{\hat{x}}}_i(X_{-i})$. For the slope of the best response to be less than one in absolute value we require

$$\begin{aligned}&\left| \frac{\partial ^2 \pi _{II}^e }{\partial x_i \partial X_{-i}}\right| / \left| \frac{\partial ^2 \pi _{II}^e }{\partial x_i^2}\right|< 1 \nonumber \\&\quad \Leftrightarrow \frac{n-1}{(n+1)^2} + \frac{(n^2+n+2) F(c+X)}{(n+1)^2} < \frac{2 n}{(n+1)^2} + \frac{2 (n^2 + n +1) F(c+X)}{(n+1)^2}\nonumber \\&\quad \Leftrightarrow \frac{n+1}{(n+1)^2} + \frac{(n^2 + n) F(c+X)}{(n+1)^2} >0 \end{aligned}$$

(41)

which holds.

The best response is obviously piecewise continuous. As the marginal expected profit is continuous at $X_{-i} = \frac{2n (a_L - c) - (n-1) (E[a] - c)}{n+1}$, and decreasing in $X_{-i}$ below and above that threshold, the best response (39) is also continuous at $X_{-i} = \frac{2n (a_L - c) - (n-1) (E[a] - c)}{n+1}$. Consequently, there exists a unique and symmetric equilibrium to the game, given by the individual level of advance sales $x^*$ which solves $x^* = {{\hat{x}}}_i((n-1) x^*)$. In a Type-I equilibrium, $x_I^* = x_i^{(1)}((n-1)x_I^*)$ gives (18) which must satisfy

$$\begin{aligned} (n-1)x_I^* < \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}. \end{aligned}$$

(42)

This is equivalent, after substituting for $x_I^*$, to

$$\begin{aligned} a_L > \frac{n^2-n}{n^2+1} E[a] +\frac{n +1 }{n^2+1} c \equiv {{\hat{a}}}. \end{aligned}$$

(43)

Then, for $a_L \le {\hat{a}}$ the equilibrium is of Type II, where the individual level of advance sales $x_{II}^*$ is the unique solution in x of $\frac{\partial \pi _{II}^e(x,(n-1)x)}{\partial x_i} =0$. $\square$

1.2 Proof of Corollary 1

As Theorem 1 demonstrates, the equilibrium is of type I if $a_L > \frac{n^2-n}{n^2+1} E[a] + \frac{n+1}{n^2+1} c$, where $E[a] = \int _{a_L}^{a_H} af(a) \hbox {d}a$. Integrating by parts, the expectation allows to rewrite it as a function of $a_H$ and of the integral below the c.d.f. F(a):

$$\begin{aligned} E[a] = \int _{a_L}^{a_H} af(a) \hbox {d}a = \left[ a F(a)\right] _{a_L}^{a_H} - \int _{a_L}^{a_H} F(a) \hbox {d}a = a_H - \int _{a_L}^{a_H} F(a) \hbox {d}a.\nonumber \\ \end{aligned}$$

(44)

In the case of a monopolist ($n=1$), a Type-I equilibrium is always chosen in which $x_I^*=0$. Then, when $n>1$, the condition governing the existence of a Type-I equilibrium rewrites as

$$\begin{aligned}&a_L> \frac{n^2-n}{n^2+1} E[a] + \frac{n+1}{n^2+1} c \Leftrightarrow a_L> \frac{n^2-n}{n^2+1} a_H \nonumber \\&\qquad - \frac{n^2-n}{n^2+1} \int _{a_L}^{a_H} F(a) \hbox {d}a + \frac{n+1}{n^2+1} c\nonumber \\&\quad \Leftrightarrow \int _{a_L}^{a_H} F(a) \hbox {d}a> a_H - \frac{n^2+1}{n^2-n} a_L + \frac{n+1}{n^2-n} c \nonumber \\&\quad \Leftrightarrow \int _{a_L}^{a_H} F(a) \hbox {d}a> a_H - a_L + \left( 1- \frac{n^2+1}{n^2-n}\right) a_L + \frac{n+1}{n^2-n} c\nonumber \\&\quad \Leftrightarrow \int _{a_L}^{a_H} F(a) \hbox {d}a > a_H - a_L - \frac{n+1}{n^2-n}(a_L - c), \end{aligned}$$

(45)

which rewrites as (20). $\square$

1.3 Proof of Theorem 2

The proof proceeds as the proof of Theorem 1. First, we prove that there exists a best response, denoted ${{\hat{x}}}_i(X_{-i})$, which is unique and continuous for each possible value of $X_{-i}$. Contrary to the case of a narrow support of demand shocks, the expected marginal profit is convex when X is large enough in the range of non-dominated advance sales. Assumption 1 then insures that this feature does not jeopardize the existence of a unique best response, by guaranteeing the expected profit is convex only when the expected marginal profit is negative, i.e., for individual quantities a firm would never play in response to $X_{-i}$. Once the best response ${{\hat{x}}}_i(X_{-i})$ is characterized, we prove that it is everywhere strictly decreasing with respect to $X_{-i}$ with a slope smaller than 1 (in absolute value). Then, a symmetric equilibrium to the game can be determined by solving for $x^*$ in $x^* = {{\hat{x}}}_i((n-1) x^*)$, which is unique.

Let us start with the characterization of the best response. As the expression of the expected profit (15) shows, we must be concerned with three different forms of the marginal profit, depending whether $x_i+X_{-i} \le a_L-c$, or whether $x_i+X_{-i} \in [a_L-c;a_L]$, or last whether $x_i+X_{-i} \in [a_L;a_H-c]$. We have already established in the proof of Theorem 1 that the expected profit reaches a maximum at $x^{(1)}(X_{-i})$ if $X_{-i} \le \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}$, and at $x^{(2)}(X_{-i})$ if $X_{-i} > \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}$, with a marginal profit strictly decreasing in $x_i$ and continuous along $x_i+X_{-i} = a_L - c$. In the wide support case we are currently considering, we must also check whether $x^{(2)}(X_{-i})$ belongs to the set of values of X on which $\pi _{II}^e(x_i,X_{-i})$ is defined, i.e., $x^{(2)}(X_{-i}) < a_L - X_{-i}$, and study the maxima of the expected profit when it is not the case.

As the marginal profit $\frac{\partial \pi _{II}^e(x_i, X_{-i})}{\partial x_i}$ is strictly decreasing in $x_i$ given $X_{-i}$, checking whether $x^{(2)}(X_{-i}) < a_L - X_{-i}$ is equivalent to verify that $X_{-i}$ is such that $\frac{\partial \pi _{II}^e(a_L - X_{-i}, X_{-i})}{\partial x_i} <0$. We have

$$\begin{aligned}&\frac{\partial \pi _{II}^e(a_L-X_{-i}, X_{-i})}{\partial x_i} = F(a_L+c) \left( E(a\mid a\le a_L+c) -c \right) \nonumber \\&\quad +\,\frac{(n-1) (1-F(a_L+c))}{(n+1)^2} \left( E(a\mid a \ge a_L+c) -c \right) \nonumber \\&\quad -\,2 \left( F(a_L+c) + \frac{n (1-F(a_L+c))}{(n+1)^2}\right) a_L \nonumber \\&\quad +\,\left( F(a_L+c) + \frac{1-F(a_L+c)}{n+1} \right) X_{-i} \end{aligned}$$

(46)

which is linear in $X_{-i}$, and hence is negative if

$$\begin{aligned} X_{-i} \le \frac{N(a_L)}{D(a_L)}, \end{aligned}$$

(47)

where

$$\begin{aligned} N(a_L)= & {} 2 \left( \frac{((n+1)^2 - n)F(a_L+c) + n}{(n+1)^2} \right) a_L\nonumber \\&-\, F(a_L + c) (E(a\mid a\le a_L+c) - c)\nonumber \\&-\,\frac{(n-1)}{(n+1)^2} (1-F(a_L+c))(E(a\mid a \ge a_L+c) - c), \end{aligned}$$

(48)

and

$$\begin{aligned} D(a_L) = \frac{n F(a_L+c)+1}{n+1}. \end{aligned}$$

(49)

Due to the fact that the expected marginal profit is continuous and decreasing in $X_{-i}$ in cases I and II, we have $\frac{N(a_L)}{D(a_L)} > \frac{2 n (a_L - c) - (n-1) (E[a]-c)}{n+1}$. If $X_{-i} > \frac{N(a_L)}{D(a_L)}$, then, as the marginal profit is decreasing for all values of $x_i$ lower than $a_L-X_{-i}$, the maximum of the expected profit on $[0,a_L -c - X_{-i}] \cup [a_L -c - X_{-i}, a_L - X_{-i}]$ is at $x_i=a_L-X_{-i}$, in which case the optimum occurs in $[a_L-X_{-i}, a_H-c -X_{-i}]$.

For $x_i \in [a_L-X_{-i}, a_H-c -X_{-i}]$, the expected marginal profit is given by (21), which takes into account the probability of a zero price and can be expressed as

$$\begin{aligned}&\frac{\partial \pi _{III}^e(x_i, X_{-i})}{\partial x_i} \nonumber \\&\quad = - c F(X) + (F(c+X) - F(X)) \left( E(a \mid X \le a \le c+X) - c - 2 x_i - X_{-i}\right) \nonumber \\&\qquad +\, (1- F(c+X)) \frac{(n-1) (E(a \mid a \ge c+X) - c) - 2 n x_i - (n-1) X_{-i}}{(n+1)^2}.\nonumber \\ \end{aligned}$$

(50)

First, expected marginal profit is continuous at $x_i = a_L-X_{-i}$:

$$\begin{aligned}&\frac{\partial \pi _{III}^e(a_L - X_{-i}, X_{-i})}{\partial x_i}\nonumber \\&\quad = F(c+a_L) \left( E(a \mid a_L \le a \le c+a_L) - c - 2 (a_L - X_{-i}) - X_{-i}\right) \nonumber \\&\qquad +\, (1- F(c+a_L)) \frac{(n-1) (E(a \mid a \ge c+a_L) - c) - 2 n (a_L-X_{-i}) - (n-1) X_{-i}}{(n+1)^2}\nonumber \\&\quad = F(c+a_L) \left( E(a \mid a_L \le a \le c+a_L) - c\right) \nonumber \\&\qquad +\, (1- F(c+a_L)) \frac{(n-1) (E(a \mid a \ge c+a_L) - c)}{(n-1)^2}\nonumber \\&\qquad - \,2 \left( F(a_L+c) + \frac{n}{(n+1)^2}(1- F(c+a_L)) \right) a_L \nonumber \\&\qquad - \,\left( F(a_L+c) + \frac{1- F(c+a_L)}{n+1} \right) X_{-i}\nonumber \\&\quad = \frac{\partial \pi _{II}^e(a_L - X_{-i}, X_{-i})}{\partial x_i} \end{aligned}$$

(51)

However, expected marginal profit is not always decreasing in the range $x_i \in [a_L-X_{-i},a_H-c -X_{-i}]$. Indeed, the derivative of expected marginal profit, (22), changes sign at most once in the range $x_i \in [a_L-X_{-i}, a_H-c-X_{-i}]$ under Assumption 1, from negative to positive values. Therefore under Assumption 1, two situations must be distinguished: the case (a) where, given $X_{-i}$, expected marginal profit is strictly increasing in $x_i$ for $x_i \in [a_L-X_{-i}, a_H - c - X_{-i}]$, from the case (b) where, given $X_{-i}$, there exists ${{\bar{x}}}(X_{-i}) \in (a_L-X_{-i},a_H-c-X_{-i})$ such that marginal expected profit is decreasing for $x_i \le {{\bar{x}}}(X_{-i})$ and increasing for $x_i > {{\bar{x}}}(X_{-i})$. Before analyzing these two cases, note that at the upper bound, $x_i = a_H - c - X_{-i}$, the total amount of advance sales X is equal to $a_H - c$, so that $c+X = a_H$, and the expected marginal profit is equal to

$$\begin{aligned}&\frac{\partial \pi _{III}^e(a_H-c-X_{-i}, X_{-i})}{\partial x_i} = - c F(a_H-c) \nonumber \\&\quad +\, (1-F(a_H-c)) \left( E(a \mid a_H-c \le a \le a_H) - 2 a_H + c + X_{-i}\right) , \end{aligned}$$

(52)

which is strictly negative for non-dominated advance sales $x_i$. Indeed, as $E(a \mid a_H-c \le a \le a_H) \le a_H$,

$$\begin{aligned} E(a \mid a_H-c \le a \le a_H) - 2 a_H + c + X_{-i} \le -a_H + c + X_{-i} \le 0 \end{aligned}$$

(53)

as we are considering non-dominated levels of advance sales, $X_{-i} \le a_H - c$. Note also that when the support is unbounded, $a_H = +\infty$, then the limit of the marginal expected profit above is equal to $-c <0$; hence our analysis also applies to the case where the support of demand shocks is unbounded from above. Consequently in case (a), marginal expected profit is strictly increasing in $x_i$ to a negative value at $x_i=a_H-c-X_{-i}$; therefore, marginal expected profit is strictly negative for all values of $x_i \in [a_L-X_{-i},a_H-c-X_{-i}]$. On the other hand in case (b), negative marginal expected profit at $x_i = a_H-c-X_{-i}$ implies that it is also negative at ${{\bar{x}}}(X_{-i})$ and so is negative whenever it is increasing. We can now complete the determination of the global optimum of the expected profit.

In the case where $X_{-i} \le \frac{N(a_L)}{D(a_L)}$ analyzed above, marginal expected profit is negative at $x_i=a_L-X_{-i}$. Since it is continuous, then either it is strictly increasing for $x_i \in [a_L-X_{-i},a_H-c-X_{-i}]$ but always remains negative (case (a)), or is decreasing and then increasing again for $x_i \in [a_L-X_{-i},a_H-c-X_{-i}]$, but remains again always negative. In both cases, the global optimum of the expected profit is $x^{(2)}(X_{-i})$.

On the contrary when $X_{-i} > \frac{N(a_L)}{D(a_L)}$, expected marginal profit is by continuity positive at the right of $x_i=a_L-X_{-i}$. As it is negative at $x_i = a_H - c - X_{-i}$, it must decrease as long as $x_i$ is small enough in $[a_L-X_{-i},a_H-c-X_{-i}]$, to become negative and then increase again to $x_i=a_H-c-X_{-i}$, where it is negative. Consequently, a unique maximum to the expected profit must exist in $[a_L-X_{-i},a_H-c-X_{-i}]$, denoted $x_i^{(3)}(X_{-i})$ which is the solution in $x_i$ of

$$\begin{aligned} 0= & {} -c F(x_i+X_{-i}) \nonumber \\&+\, (F(c+x_i+X_{-i}) - F(x_i+X_{-i})) \nonumber \\&\left( E(a \mid x_i+X_{-i} \le a \le c+x_i+X_{-i}) - c - 2 x_i - X_{-i}\right) \nonumber \\&+\, (1- F(c+x_i+X_{-i})) \frac{(n-1) (E(a \mid a \ge c+x_i+X_{-i}) - c) - 2 n x_i - (n-1) X_{-i}}{(n+1)^2}.\nonumber \\ \end{aligned}$$

(54)

To summarize, we have demonstrated that the best response exists and is given by

$$\begin{aligned} {{\hat{x}}}_i(X_{-i})=\left\{ \begin{array}{ll} x_i^{(1)}(X_{-i}) &{} \text{ if } X_{-i} \le \frac{(2n (a_L - c) - (n-1) (E[a] - c))}{n+1}\\ x_i^{(2)}(X_{-i}) &{} \text{ if } \frac{(2n (a_L - c) - (n-1) (E[a] - c))}{n+1} < X_{-i} \le \frac{N(a_L)}{D(a_L)}\\ x_i^{(3)}(X_{-i}) &{} \text{ if } X_{-i} > \frac{N(a_L)}{D(a_L)}. \end{array} \right. \end{aligned}$$

(55)

The second step of the proof consists in demonstrating that the best response defined above is continuous and strictly decreasing. First, it is continuous at $X_{-i}=\frac{(2n (a_L - c) - (n-1) (E[a] - c))}{n+1}$ as we demonstrated in the proof of Theorem 1. The same reasoning allows to prove that the best response is also continuous at $X_{-i} = \frac{N(a_L)}{D(a_L)} \equiv {\tilde{X}}$. First note that at $\left( x^{(2)}({\tilde{X}}), {\tilde{X}}\right)$, $x^{(2)}({\tilde{X}})+ {\tilde{X}} = a_L$ and $\frac{\partial \pi _{II}^e(x^{(2)}({\tilde{X}}),{\tilde{X}})}{\partial x_i} =0$. Then as the expected marginal profit is strictly decreasing in $x_i$ at the best response, if $\frac{\partial \pi _{III}^e(x^{(2)}({\tilde{X}}),{\tilde{X}})}{\partial x_i} =0$, $x^{(3)}({\tilde{X}}) = x^{(2)}({\tilde{X}})$ and the best response is continuous at ${\tilde{X}}$. Using its definition, we can compute

$$\begin{aligned}&\frac{\partial \pi _{III}^e(x_i^{(2)}({\tilde{X}}),{\tilde{X}})}{\partial x_i} = - c F(x^{(2)}({\tilde{X}})+ {\tilde{X}}) \nonumber \\&\qquad + \int _{x^{(2)}({\tilde{X}})+ {\tilde{X}}}^{c+x^{(2)}({\tilde{X}})+ {\tilde{X}}} (a - c - 2 x^{(2)}({\tilde{X}}) - {\tilde{X}}) \hbox {d}F(a)\nonumber \\&\qquad + \int _{c+x^{(2)}({\tilde{X}})+ {\tilde{X}}}^{a_H} \frac{(n-1)(a - c) - 2 n x^{(2)}({\tilde{X}}) -(n-1) {\tilde{X}}}{(n+1)^2} \hbox {d}F(a)\nonumber \\&\quad = - c F(a_L) + \int _{a_L}^{c+a_L} (a - c - 2 x^{(2)}({\tilde{X}}) - {\tilde{X}}) \hbox {d}F(a)\nonumber \\&\qquad + \int _{c+a_L}^{a_H} \frac{(n-1)(a - c) - 2 n x^{(2)}({\tilde{X}}) -(n-1) {\tilde{X}}}{(n+1)^2} \hbox {d}F(a)\nonumber \\&\quad = \frac{\partial \pi _{II}^e(x_i^{(2)}({\tilde{X}}),{\tilde{X}})}{\partial x_i} = 0. \end{aligned}$$

(56)

Therefore under Assumption 1, the best response must be continuous at $X_{-i}=\frac{(2n (a_L - c) - (n-1) (E[a] - c))}{n+1}$ and at $X_{-i}={\tilde{X}}$.

We can now determine the slope of ${{\hat{x}}}^i(X_{-i})$. For the first two cases in (55), it is strictly decreasing as already demonstrated in the proof of Theorem 1. For the third case, applying the Implicit Function Theorem to $x_i^{(3)}(X_{-i})$, we have:

$$\begin{aligned} \frac{d x_i^{(3)}}{d X_{-i}} = - \frac{\frac{\partial ^2 \pi _{III}^e(x_i, X_{-i})}{\partial x_i \partial X_{-i}}}{\frac{\partial ^2 \pi _{III}^e(x_i, X_{-i})}{\partial (x_i)^2}} \end{aligned}$$

(57)

with the numerator given by (22) and the denominator by (23). We have already established that the denominator is negative at the best response: Indeed, $x^{(3)}(X_{-i})$ must belong to the set of values $S_1(X_{-i})$ in which the second derivative of the expected profit with respect to $x_i$ is negative.

To establish the sign of the cross-partial derivative, note that the marginal expected profit at $X_{-i}= N(a_L)/D(a_L)$ must be positive, otherwise the best response would not be $x_i^{(3)}(X_{-i})$. To sign the denominator, consider levels of advance sales such that $x_i + X_{-i} = a_H-c$. Clearly, from (50), the expected marginal profit is negative at that level of advance sales. From (23), we have

$$\begin{aligned} \frac{\partial ^2 \pi _{III}^e(x_i,X_{-i})}{\partial x_i\partial X_{-i}}\Big |_{X=a_H-c} = x_i f(a_H-c) - \frac{n}{n+1} f(a_H) - (1-F(a_H-c)),\nonumber \\ \end{aligned}$$

(58)

the sign of which can be negative or positive. If the cross-partial derivative is negative at $x_i+X_{-i}=a_H-c$, then Assumption 1 implies that $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}} < 0$ for any $x_i \in [a_L-X_{-i}, a_H-c-X_{-i}]$; the best response $x_i^{(3)}(X_{-i})$ is clearly decreasing in that case.

If the cross-partial derivative is positive at $x_i+X_{-i}=a_H-c$, then we demonstrate now that Assumption 1 implies that $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}}$ must be negative at the best response $x_i^{(3)}(X_{-i})$. Indeed, under Assumption 1, $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}} >0$ for $x_i+X_{-i} \in [X^0, a_H-c]$, with $X^0 \ge a_L$, while $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}} <0$ for $x_i + X_{-i} \in [a_L, X^0]$. Therefore, since $\frac{\partial \pi _{III}^e}{\partial x_i} <0$ at $x_i+X_{-i} = a_H-c$, Assumption 1 implies that the expected marginal profit $\frac{\partial \pi _{III}^e}{\partial x_i}$ is also negative wherever it is increasing in $X_{-i}$, i.e., wherever $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}} >0$. But for the expected marginal profit $\frac{\partial \pi _{III}^e}{\partial x_i}$ to be equal to 0 at $x_i^{(3)}(X_{-i})$, it must be the case that $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}} <0$ as, if $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}}$ were strictly positive, $\frac{\partial \pi _{III}^e}{\partial x_i}$ would be strictly negative. Therefore, the best response $x_i^{(3)}(X_{-i})$ belongs to $[a_L-X_{-i}, X^0-X_{-i}]$ and at the best response, $\frac{\partial ^2 \pi _{III}^e}{\partial x_i\partial X_{-i}} <0$. Then $x_i^{(3)}(X_{-i})$ is decreasing, and so is ${{\hat{x}}}_i(X_{-i})$.

We demonstrate now that the slope of the best response is (piecewise) lower than 1, in absolute value. As the proof of Theorem 1 established, this property holds for its branches $x^{(1)}$ and $x^{(2)}$. We simply need to prove it for $x^{(3)}$. As the own and cross second-order derivatives of the expected profit are both negative under Assumption 1 at the best response, $-A(X) x_i - 2 B(X) >0$ and $-A(X) x_i - {{\tilde{B}}}(X) >0$, and $\frac{d x_i^{(3)}}{d X_{-i}} <1$ in absolute value if

$$\begin{aligned}&\frac{|A(X) x_i + {{\tilde{B}}}(X)|}{|A(X) x_i + 2 B(X)|} < 1 \nonumber \\&\quad \Leftrightarrow {{\tilde{B}}}(X) - 2 B(X)> 0 \nonumber \\&\quad \Leftrightarrow - (1-F(X)) + \frac{n^2+n+2}{(n+1)^2} (1-F(c+X))\nonumber \\&\qquad + \,2 (1-F(X)) - 2 \frac{n^2 + n + 1}{(n+1)^2} (1-F(X+c))>0 \nonumber \\&\quad \Leftrightarrow (1-F(X)) - \frac{n}{n+1} (1-F(X+c)) > 0 \end{aligned}$$

(59)

which holds as $1-F(X) > 1- F(X+c)$ and $\frac{n}{n+1} <1$. Consequently, $x^{(3)}(X_{-i})$ has a slope smaller than 1 in absolute value.

To conclude, under Assumption 1, the best response ${{\hat{x}}}_i(X_{-i})$ is continuous and decreasing. Hence, there exists a unique $x^*$ solution to $x^* = {{\hat{x}}}_i((n-1) x^*)$, which can be on one of the three branches of the best response depending on the model parameters. It suffices to study the intersection between the best response ${{\hat{x}}}_i(X_{-i})$ and the line $x_i=X_{-i}/(n-1)$ in the graph $(X_{-i}, x_i)$, that is the solutions in $x^*$ to $x^* = {{\hat{x}}}_i((n-1) x^*)$. As established in the proof of Theorem 1, the equilibrium individual level of advance sales is given by $x_I^*$ if $a_L > {{\hat{a}}}$, where ${{\hat{a}}}$ is defined in (43). When this condition is not satisfied, the unique equilibrium level of advanced sales is given by either $x_{II}^*$, the solution in x of $\frac{\partial \pi _{II}^e(x,(n-1)x)}{\partial x_i} =0$, or by $x_{III}^*$ the solution in x of $\frac{\partial \pi _{III}^e(x,(n-1)x)}{\partial x_i} =0$. $\square$

1.4 Proof of Corollary 2

To demonstrate the corollary, it suffices to demonstrate that $(n-1) x_{II}^*$ and $\frac{N(a_L)}{D(a_L)}$ (defined in (47)) intersect as $a_L$ changes, and that $(n-1) x_{II}^* < \frac{N(a_L)}{D(a_L)}$ for large values of $a_L$, case in which the equilibrium advance sales are $x_{II}^*$, while $(n-1) x_{II}^* > \frac{N(a_L)}{D(a_L)}$ for small values of $a_L$, case in which the equilibrium advance sales are $x_{III}^*$. As $a_L$ varies between c and $a_H-c$, let us compare $(n-1) x_{II}^*$ and $\frac{N(a_L)}{D(a_L)}$ for these two extreme values of $a_L$.

By definition, $x_{II}^*$ cannot be negative even if $a_L=c$: indeed the marginal profit (19) evaluated at (0, 0) is a weighted average of differences between the demand intercept and the marginal cost c, which is strictly positive. Let us study the value of $N(a_L)/D(a_L)$ at the smallest possible value for $a_L$ we assume, $a_L=c$. We have

$$\begin{aligned} N(c)= & {} 2 \left( \frac{((n+1)^2 - n)F(2c) + n}{(n+1)^2} \right) c - F(2c) (E(a\mid a\le 2c) - c) \nonumber \\&-\,\frac{(n-1)}{(n+1)^2} (1-F(2c))(E(a\mid a \ge 2c) - c)\nonumber \\= & {} 3 F(2c) c - F(2c) (E(a\mid a\le 2c)\nonumber \\&+ \,\frac{(3n-1)}{(n+1)^2} (1-F(2c)) c - \frac{(n-1)}{(n+1)^2} (1-F(2c)) E(a\mid a \ge 2c) \end{aligned}$$

(60)

which is negative if c is sufficiently small. Hence $(n-1) x_{II}^* > \frac{N(a_L)}{D(a_L)}$ for $a_L = c$ and c sufficiently small and the equilibrium is of Type III.

Similarly, $N(a_H-c)/D(a_H-c) = 2 (a_H - c) - \int _{a_H -c}^{a_H} a\hbox {d}F(a) + c > a_H-c$, which is a strictly dominated level of advance sales for all firms, and in particular larger than $(n-1) x^*_{II}$ and $(n-1) x^*_{III}$. Hence $(n-1) x_{II}^* < \frac{N(a_L)}{D(a_L)}$ for $a_L = a_H-c$ and the equilibrium is of Type II.

As $x^*_{II}$, $N(a_L)$ and $D(a_L)$ are continuous in $a_L$, this is sufficient to ensure that $(n-1) x_{II}^*$ and $\frac{N(a_L)}{D(a_L)}$ cross at least once for a particular value of $a_L$ if c is sufficiently small, and that the equilibrium advance sales are $x_{II}^*$ for $a_L$ large enough while it is $x_{III}^*$ for $a_L$ small enough. Let us denote the upper threshold value ${\tilde{a}}''$ at which both curves cross, and ${\tilde{a}}'$ the lower threshold value (both can be identical to each other for many distributions F(a), but we cannot rule out the theoretical possibility that ${\tilde{a}}' < {\tilde{a}}''$ with multiple thresholds). Then from the argument above, $(n-1) x_{II}^* < \frac{N(a_L)}{D(a_L)}$ for $a_L > {\tilde{a}}''$ and $(n-1) x_{II}^* > \frac{N(a_L)}{D(a_L)}$ for $a_L < {\tilde{a}}'$. $\square$

1.5 Proof of Corollary 3

To demonstrate Corollary 3, we analyze separately the sign of each component of the second-order derivatives of the expected profit, A(X), B(X) and ${{\tilde{B}}}(X)$. Since A(X) depends directly on f(X) and $f(c+X)$, it is possible to determine the sign of A(X) depending on $x_i$, and we denote $x_i^0$ the value at which A(X) changes sign. B(X) and ${{\tilde{B}}}(X)$ are negative no matter f(X) and $f(c+X)$, but the signs of their derivatives with respect to X do depend on f(X) and $f(c+X)$. We denote $x_i^1$ and $x_i^2$ the values of $x_i$ such that the signs of the derivatives of B(X) and ${{\tilde{B}}}(X)$ change. With a double-peaked distribution p.d.f., we demonstrate that f(a) is such that A(X) is positive for $x_i$ close to $a_L-X_{-i}$ (whenever this quantity is positive, else close to 0), while B(X) and ${\tilde{B}}(X)$, whose derivatives are very close to A(X), are negative but increasing in $x_i$. Then, it is possible to find a sufficient condition under which second-order derivatives are positive and then negative, which breaks Assumption 1.

Suppose the two peaks of the distribution are located at the extremes of the support, $M_1=a_L$ and $M_2=a_H$. Then, f(a) decreases from $a_L$ to a minimum (denoted m) and then increases to $a_H$. With such a p.d.f., $A(X) = f(X) - \frac{n}{n+1} f(c+X) >0$ for $x_i \in [\max (a_L - X_{-i},0), x_i^0]$ where $x_i^0$ solves $f(x_i^0+ X_{-i}) = \frac{n}{n+1} f(c+x_i^0+ X_{-i})$ for any given $X_{-i}$. Note that $x_i^0$ need not be unique for all n. Indeed $f(c+X)$ reaches its minimum at $m-c$, and is shifted to the left of f(X). Then, $\frac{n}{n+1} < 1$ and tends to 1 as n increases to $+\infty$. Hence, $\frac{n}{n+1} f(c+X)$ stands clearly to the south-west of f(X) and is flatter than f(X) when n is close to 1. When n increases, the curvature of $\frac{n}{n+1} f(c+X)$ becomes more similar to the curvature of f(X), which, given c, implies uniqueness of $x_i^0$, as the increasing branch of $\frac{n}{n+1} f(c+X)$ intersects the decreasing branch of f(X). Therefore when n is large enough, given $c>0$, $x_i^0 \in (m-c-X_{-i}, m-X_{-i})$ and is unique; moreover when $x_i > x_i^0$ given $X_{-i}$, A(X) is negative.

Whereas B(X) and ${{\tilde{B}}}(X)$ are still negative, their derivatives are given by

$$\begin{aligned} B'(X) = f(X) - \frac{n^2+n+1}{(n+1)^2} f(c+X) \quad \text{ and } \quad \tilde{B}'(X) = f(X) - \frac{n^2+n+2}{(n+1)^2} f(c+X).\nonumber \\ \end{aligned}$$

(61)

As $\frac{n^2+n+1}{(n+1)^2}< \frac{n^2+n+2}{(n+1)^2} < 1$, using the same analysis than for A(X) above, these derivatives are positive for $x_i$ smaller than $x_i^1$ and $x_i^2$, respectively, which are, when n is large enough, the unique solutions of $f(x_i^1 + X_{-i}) = \frac{n^2+n+1}{(n+1)^2} f(c+x_i^1 + X_{-i})$ and $f(x_i^2 + X_{-i}) = \frac{n^2+n+2}{(n+1)^2} f(c+x_i^2 + X_{-i})$ given $X_{-i}$, respectively. Else the derivatives $B'(X)$ and ${{\tilde{B}}}'(X)$ are negative when $x_i > x_i^1$ and $x_i > x_i^2$, respectively. Consequently when n is large enough, B(X) and ${{\tilde{B}}}(X)$ are negative and reach their maximum at $x_i=x_1^i$ and $x_i^2$, respectively, given $X_{-i}$. Moreover as $\frac{n}{n+1}< \frac{n^2+n+1}{(n+1)^2} < \frac{n^2+n+2}{(n+1)^2}$, and as $f(c+X)$ is double-peaked and reaches its minimum at $x_i=m-c-X_{-i}$, then when these intersections are unique we have $x_i^0> x_i^1 > x_i^2$.

When $x_i^0$, $x_i^1$ and $x_i^2$ are unique, clearly if $x_i \in [x_i^0, a_H-c-X_{-i}]$, $A(X) x_i + 2 B(X) < 0$ and $A(X) x_i + {{\tilde{B}}}(X) < 0$: Both second-order derivatives are negative when $x_i$ is larger than $x_i^0$. But if at $X=a_L$, $2 B(a_L)$ and ${{\tilde{B}}}(a_L)$ are close enough to 0, and if at the same time $A(a_L)$ is large enough, the second-order and cross-partial derivatives of the expected profit are positive for some $X_{-i}$, from $x_i=\max (a_L -X_{-i},0)$ to some value of $x_i < x_i^0$. The conditions leading to this case, which clearly violates Assumption 1, can be described more precisely using the expressions of A(X), B(X) and ${{\tilde{B}}}(X)$ evaluated at $X=a_L$, replacing $F(a_L)=0$, and imposing $X_{-i}=0$: $A(a_L) a_L + 2 B(a_L) > 0$ and $A(a_L) a_L + {{\tilde{B}}}(a_L) > 0$ are, respectively, equivalent to :

$$\begin{aligned}&\left( f(a_L) - \frac{n}{n+1} f(c+a_L)\right) a_L - 2 \left( \frac{n}{(n+1)^2} + \frac{n^2+n+1}{(n+1)^2} F(c+a_L)\right)> 0 \nonumber \\&\left( f(a_L) - \frac{n}{n+1} f(c+a_L)\right) a_L - \frac{n-1}{(n+1)^2} - \frac{n^2+n+2}{(n+1)^2} F(c+a_L)> 0 \end{aligned}$$

(62)

As $2 \left( \frac{n}{(n+1)^2} + \frac{n^2+n+1}{(n+1)^2} F(c+a_L)\right) > \frac{n-1}{(n+1)^2} + \frac{n^2+n+2}{(n+1)^2} F(c+a_L)$, it suffices to verify the first inequality to verify both. Therefore when

$$\begin{aligned} \left( f(a_L) - \frac{n}{n+1} f(c+a_L)\right) a_L - 2 \left( \frac{n}{(n+1)^2} + \frac{n^2+n+1}{(n+1)^2} F(c+a_L)\right) > 0\nonumber \\ \end{aligned}$$

(63)

then $A(a_L) a_L + 2 B(a_L) > 0$ and $A(a_L) a_L + {{\tilde{B}}}(a_L) > 0$. This condition simplifies to $(n+1) f(a_L) a_L - 2 n >0$ when $c=0$. As A(X) is increasing for $x_i < x_i^0$, and as B(X) and ${{\tilde{B}}}(X)$ are also increasing for $x_i < x_i^1$ and $x_i < x_i^2$, Assumption 1 is clearly violated: Second-order derivatives change sign from positive to negative values, which is the opposite of the condition we require. We established this condition for n large enough and for $X_{-i}=0$, but by continuity this counter-example holds for larger values of $X_{-i}$.

When peaks are closer to each other, $a_L< M_1 \le m \le M_2 < a_H$, the logic illustrated above is reverted: A(X) is negative for $x_i$ small enough and hence B(X) and ${{\tilde{B}}}(X)$ are negative and decreasing, which goes in the direction of satisfying Assumption 1. To conclude, double-peaked probability distribution functions fail to satisfy Assumption 1 if peaks $M_1$ and $M_2$ are distant enough from each other, if n is sufficiently large given c, and if a further condition on f(a), n and c, described in (63), is verified. $\square$