Strategic advance sales, demand uncertainty and overcommitment


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.

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  1. 1.

    In a paper that generalizes the proof of existence and uniqueness of the classical static Cournot equilibrium to encompass the case of degenerate equilibria, Gaudet and Salant (1991) point out, as a potential application, the importance of these equilibria when they occur in subgames.

  2. 2.

    Absent this regularity condition, multiple solutions to the first-order condition of each firm may co-exist, and the best response in pure strategy may lack continuity. In the case where the regularity condition is not satisfied and the support of demand shocks is wide enough, the game still possesses an equilibrium in mixed strategy due to the continuity of the expected payoffs (See Fudenberg and Tirole (1991))

  3. 3.

    With the exception that in the case of costless production, an increasing hazard rate distribution satisfies our assumption and hence, is sufficient to guarantee the existence of an equilibrium.

  4. 4.

    Allaz (1992) allows for uncertainty in the strategic forward contracting game, but he does not allow for the possibility that firms may decide to produce no additional output for the spot market, or that the spot price might be driven to zero.

  5. 5.

    The results in Holmberg and Willems (2015) assume that production is costless and that the hazard rate of the demand shock distribution is not too strongly decreasing. They analyze a closed-form solution using a Pareto distribution for demand uncertainty and show that expected welfare is reduced by derivative trading relative to a standard supply function equilibrium when there are only two firms or when demand uncertainty is sufficiently small. However, as the standard supply function equilibrium is more competitive than Cournot in their example, the effect on the expected welfare relative to Cournot competition is not clear.

  6. 6.

    Mitraille and Thille (2009; 2014) have multiple spot markets as well, but the forces at work are essentially the same.

  7. 7.

    Lagerlöf (2006) extends this analysis to consider a continuous distribution of demand shocks.

  8. 8.

    In these papers, additional sales after demand is revealed are not possible (with the exception of the price competition case in Klemperer and Meyer (1986)), so that there is no strategic effect of the advance sales.

  9. 9.

    See Appelbaum and Lim (1985), Perrakis and Warskett (1983), Vives (1989), Spencer and Brander (1992), and Sadanand and Sadanand (1996).

  10. 10.

    Advance sales differ from advance production as defined in the models of Saloner (1987) and Pal (1991) in that in the former, the marginal revenue firms face when going on the “spot” market is residual one, while in the later firms are made more aggressive on the “spot” market due to the production realized they sell entirely when the market opens. In advance production games, the cost structure is crucial to the outcome of the game: Equilibria can be asymmetric as a firm carrying a leader production is committed to sell it forcing its competitor to behave as a follower, but symmetric outcomes are also possible (Saloner 1987); cost differences between periods, such as discounting, then matters to equilibrium selection (Pal 1991). See also Mitraille and Moreaux (2013).

  11. 11.

    The counterparty to the forward contract receives the quantity specified in the forward contract and then resells it (if a “speculator”) or consumes it.

  12. 12.

    Linear production costs are necessary to be able to interpret the model as being consistent with both forward trading with physical delivery and competitive storage since splitting production across periods has no effect on cost. Convex production costs would give firms an additional cost-smoothing incentive to produce in the second period and would cloud the effect that we wish to analyze. In the same way, the constant marginal cost is the same in each period so that there is no inherent advantage in production costs between advance and spot sales.

  13. 13.

    A unit discount factor means that the timing of the receipt of profits does not matter to a firm. This is important to be able to interpret \(x_i\) as either advance sales made in period one (with revenue and costs realized in period one) or as forward contracts with physical delivery (with revenue and costs realized in period two).

  14. 14.

    As such, we do not consider the risk that demand would be so low that firms would regret any production at all. We also exclude degenerate distributions by imposing \(a_L<a_H\).

  15. 15.

    Observability of advance sales matters in our setting, as it does more generally in all commitment games. Indeed, the possibility of exerting some (endogenous) leadership through advance sales could be jeopardized if they are not observable. Bagwell (1995) establishes that in a (textbook) Stackelberg game, non-observability of the leader’s choice results in the Cournot outcome being the sole pure-strategy equilibrium to the game. However, van Damme and Hurkens (1997) demonstrate that a mixed strategy equilibrium close to the Stackelberg outcome exists in Bagwell’s game, and Maggi (1999) proves that private information restores the value of commitment. In the context of strategic forward trading, Ferreira (2006) shows that imperfect observability can lead to multiple equilibria, including some that are even more competitive than in Allaz and Vila (1993).

  16. 16.

    Specifically, from (9) we use

    $$\begin{aligned} p_2^*(X, a) - c =\left\{ \begin{array}{ll} -c &{} \text{ if } a \le X\\ a - X -c &{} \text{ if } X \le a \le X + c\\ (a - X - c)/(n+1) &{} \text{ if } a > X + c \end{array} \right. \end{aligned}$$
  17. 17.

    This is the result corresponding to Allaz and Vila (1993), with E[a] substituted for the demand intercept.

  18. 18.

    See Definition 2 and its discussion.

  19. 19.

    Indeed, the integral below the c.d.f. F(a) of a non-degenerate random variable is always strictly smaller than the area of the rectangle \([a_L, a_H] \times [0,1]\) to which it belongs, which is equal to \(a_H-a_L\).

  20. 20.

    A right-skewed distribution has a positive skewness and a long right-tail. The mass of this distribution is concentrated on the left of the support on which it is defined, implying that the c.d.f. F(a) increases faster to 1 than when the distribution is left-skewed. On the contrary, the mass of a left-skewed distribution lies mostly on the right of its support, and the c.d.f. increases more slowly to 1, implying that the integral below the c.d.f. is smaller.

  21. 21.

    Indeed \(\frac{n+1}{n^2-n}\) is strictly decreasing in n, from 3 / 2 (for \(n=2\)) to 0 (for \(n=+\infty \)).

  22. 22.

    To respect the assumption that \(a_L\ge c\) we must have \(\delta \le 1-c\) and to have a narrow support requires \(\delta \le c/2\), so we limit attention to \(\delta \in (0, \min [c/2, 1-c])\). Setting \(c=2/3\), \(\delta \in (0, 1/3)\) satisfies these restrictions.

  23. 23.

    See also Lagerlöf (2006, 2007) in the context of static Cournot games with demand uncertainty.

  24. 24.

    Indeed \(1-F(X) \ge 1 - F(c+X)\) (with strict inequality for \(c>0\)), and \((n^2 + n +1)/(n+1)^2 < 1\). Hence \((1-F(X)) > \frac{n^2 + n + 1}{(n+1)^2} (1-F(X+c))\) and \(B(X) < 0\).

  25. 25.

    The argument is similar to footnote 24, with the exception that \(\frac{n^2+n+2}{(n+1)^2} < 1\) for \(n> 1\).

  26. 26.

    Mas-Colell et al. (1995) proposition 8.D.3. (p. 253) establishes that normal form games with continuous and quasi-concave payoffs have pure-strategy Nash equilibrium, and apply it to static models of oligopoly (section 12.C).

  27. 27.

    Bagnoli and Bergstrom (2005) provide a detailed literature review of articles that use an increasing hazard rate or log-concavity assumption on the distribution of random preferences of buyers in order to establish the existence of pure-strategy equilibria. They show that log-concavity and an increasing hazard rate are closely related: Log-concavity of a p.d.f. implies log-concavity of a c.d.f., which implies that the hazard rate of the distribution is increasing; however a distribution with an increasing hazard rate is not automatically log-concave.

  28. 28.

    In this case, (26) is of the same sign as \(\frac{x_i}{n+1} - \frac{2 n}{(n+1)^2 h(x_i+X_{-i})}\) and (27) is of the same sign as \(\frac{x_i}{n+1} - \frac{n-1}{(n+1)^2 h(x_i+X_{-i})}\). Both of these expressions are monotonically increasing in \(x_i\) for a given \(X_{-i}\) and so satisfy Assumption 1. See also Holmberg and Willems (2015).

  29. 29.

    The parameters used are \(c=0.2\) and \(n=3\), while, to respect \(a_L=1-\delta \ge c\) we consider now \(\delta \le 1 - c = 0.8\). Under these assumptions on the model parameters, the equilibrium is of Type I for \(\delta < 0.32\), of Type II for \(0.32<\delta <0.65\) and of Type III for \(\delta \ge 0.65\).

  30. 30.

    We have examined examples using other uni-modal distributions, such as the Truncated Normal and the uni-modal Beta, with Assumption 1 satisfied in each case.

  31. 31.

    The p.d.f. of a truncated distribution on \([a_L;a_H]\) with mean \(\mu _t\) and standard deviation \(\sigma _t\) is given by \(\phi _t(a) = \frac{\frac{1}{\sqrt{2\pi } \sigma _t} \exp \{-(a-\mu _t)^2/2 \sigma _t^2\}}{\Phi ((a_H - \mu _t)/\sigma _t) - \Phi ((a_L-\mu _t)/\sigma _t)}\), where \(\Phi (a)\) is the c.d.f. of the standard Normal distribution. Truncation preserves log-concavity (see Bagnoli and Bergstrom 2005.), but mixture does not in the particular case of Normal distributions when means differ (available upon request). The p.d.f. of the mixture of two truncated Normal distributions on \([a_L,a_H]\) is \(f(a) = \alpha \phi _1(a) + (1-\alpha ) \phi _2(a)\), where \(\alpha \in [0,1]\).

  32. 32.

    The first term in (19) is strictly negative, while the second term is zero.

  33. 33.

    Indeed, \(X=a_L-c\) in this case and the first term in (19) is zero, while the second is integrated between \(a_L\) and \(a_H\) which results in (17).


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The authors thank for their useful comments René Kirkegaard, Kurt Annen, Asha Sadanand, James Amegashie and seminar participants at the University of Guelph; Sergei Izmalkov, Rabah Amir, and participants of the Oligoworkshop 2016 (Université de Paris - Dauphine); Philippe Choné, Laurent Linnemer, Thibaud Vergé, and seminar participants at CREST-LEI; Simon Cowan and participants of the RES 2017 conference (University of Bristol); Benoît Sévi and participants to the Commodity Markets Winter Workshop (Université de Nantes - Audencia); Jacques Crémer and participants to the CEA 2018 conference (McGill University)).

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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

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

if this quantity is smaller than \(a_L - c - X_{-i}\), that is if

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

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 negativeFootnote 32, 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

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

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 continuousFootnote 33. 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

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

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:

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

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

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

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

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

This is equivalent, after substituting for \(x_I^*\), to

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

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 \(\frac{\partial \pi _{II}^e(x,(n-1)x)}{\partial x_i} =0\). \(\square \)

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):

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

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

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

which rewrites as (20). \(\square \)

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

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

which is linear in \(X_{-i}\), and hence is negative if

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


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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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 \(\frac{\partial \pi _{III}^e(x,(n-1)x)}{\partial x_i} =0\). \(\square \)

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

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

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 \(a_L < {\tilde{a}}'\). \(\square \)

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

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

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 :

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

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

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

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. \(\square \)

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Mitraille, S., Thille, H. Strategic advance sales, demand uncertainty and overcommitment. Econ Theory 69, 789–828 (2020).

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  • Oligopoly
  • Advance sales
  • Uncertainty
  • Overcommitment

JEL Classification

  • C72
  • D43
  • L13