Price and capacity competition in balancing markets with energy storage

Abstract

Energy storage can absorb variability from the rising number of wind and solar power producers. Storage is different from the conventional generators that have traditionally balanced supply and demand on fast time scales due to its hard energy capacity constraints, dynamic coupling, and low marginal costs. These differences are leading system operators to propose new mechanisms for enabling storage to participate in reserve and real-time energy markets. The persistence of market power and gaming in electricity markets suggests that these changes will expose new vulnerabilities. We develop a new model of strategic behavior among storages in energy balancing markets. Our model is a two-stage game that generalizes a classic model of capacity followed by Bertrand–Edgeworth price competition by explicitly modeling storage dynamics and uncertainty in the pricing stage. By applying the model to balancing markets with storage, we are able to compare capacity and energy-based pricing schemes and to analyze the dynamic effects of the market horizon and energy losses due to leakage. Our first key finding is that capacity pricing leads to higher prices and higher capacity commitments, and that energy pricing leads to lower, randomized prices and lower capacity commitments. Second, we find that a longer market horizon and higher physical efficiencies lead to lower prices by inducing the storage to compete to have their states of charge cycled more frequently.

Appendix

Proof

(Lemma 1) We show that a mixed strategy equilibrium exists in the pricing game. Existence for a single period when B is not random is proved in Prop. 4.3 [2] using Theorem 5 of [20]. We may straightforwardly adapt their approach to the present scenario. Let $\hat{\pi }(p)=\sum _{i=1}^N\pi _i(p)$. For each firm i, let $G_i\in \mathbb {N}$. For each G with $0\le G\le G_i$ and $j\ne i$, $1\le j\le N$, let $g_{ij}^G$ be a one-to-one continuous function. Let $\hat{P}(i)$ be defined as

$$\begin{aligned} \hat{P}(i)=\left\{ p| \exists \; j\ne i,\; \exists \; G,\; 0\le G\le G_i \text { s.t. } p_j=g_{ij}^G(p_i)\right\} . \end{aligned}$$

Theorem 5 in [20] is as follows:

Theorem 2

Suppose that $\pi _i(p_i,p_{-i})$ is bounded, continuous in p except on a subset $P^*$ of $\hat{P}(i)$, and weakly lower semicontinuous in $p_i$ for all i, and that $\hat{\pi }(p)$ is upper semicontinuous in p. Then a mixed-strategy equilibrium exists.

Using a similar argument to that given in [1, 2], it can be shown that $\pi _i(p)$ is bounded, continuous in p except at a subset $P^*$, and weakly lower semicontinuous when B is deterministic. All of these properties are preserved by taking the expectation over B and thus hold for $\pi _i(p)$ when B is random. Clearly, $\hat{\pi }(p)$ is continuous as well since it is the sum of continuous functions. This establishes the existence of a mixed-strategy equilibrium.

Proof

(Lemma 2) We proceed case-wise as in the lemma.

• 1. Let $(i= \mathop {\mathrm{arg\,max}}\nolimits _{j} \;p_j)$ and assume $p_i>0$. If i is not a strict maximum, profits can be made through undercutting (i.e., reducing $p_i$ by a small $\epsilon >0$), so we assume that it is a strict maximum. Then $\mathbb {E}\left[ \sum _{t=0}^T|\mathcal {X}_i^t(p)|\right] =0$ by our assumption in the first part of the lemma. Therefore, it is profitable for firm i to reduce its price by $\epsilon >0$ such that $p_i-\epsilon \le p_j$ for some j; this guarantees firm i a portion of the demand, and hence their payoff can only improve. This implies that a profitable deviation exists if $p_i>0$. Since $p_i$ is maximal, this implies that $p_j=0$ for all j is the only candidate pure strategy equilibrium. It is straightforward to see that any unilateral deviation results in the same profit, zero, and hence $p_j=0$ for all j is the only pure strategy Nash equilibrium for this case.

• 2. Since the allocation to each firm is independent of the price vector in this case, the pure strategy equilibrium is for each firm to maximize its profits by setting the highest possible price, $p_i=R$.

The following technical result is necessary for the proof of Lemma 3.

Lemma 7

$\sum _{t=0}^T|\mathcal {X}_i^t(p)|$ is nonincreasing in $p_i$ and nondecreasing in $p_{-i}$.

Observe that $\sum _{t=0}^T|\mathcal {X}_i^t(p)|$ only increases or decreases when the price-wise ordering of the firms changes, because otherwise the allocation (1)–(3) is unaffected. Since Lemma 7 holds for any realization of the imbalance sequence B, it implies that $\mathbb {E}\left[ \sum _{t=0}^T|\mathcal {X}_i^t(p)|\right]$ is nonincreasing in $p_i$ and nondecreasing in $p_{-i}$ as well.

Proof

(Lemma 7) Suppose that $p_j$ is the smallest price larger than $p_i$, and consider an increase in $p_i$ to $p_i+\epsilon$, $\epsilon <0$. If $p_i+\epsilon <p_j$, then $\sum _{t=0}^T\left| \mathcal {X}_i^t(p)\right|$ does not change. Assume that $p_i+\epsilon >p_j$. We proceed inductively.

Without loss of generality, assume $B^0\ge 0$. By construction, $|S_i^1(p)-S_i^1(p_{-i},p_i+\epsilon )|= |\mathcal {X}_i^0(p)|- |\mathcal {X}_i^0(p_{-i},p_i+\epsilon )|\ge 0$, which establishes the base case. Now assume that

$$\begin{aligned}\sum _{t=0}^k|\mathcal {X}_i^{t}(p)| - \sum _{t=0}^k\left| \mathcal {X}_i^{t}(p_{-i},p_i+\epsilon )\right| \ge |S_i^{k+1}(p)-S_i^{k+1}(p_{-i},p_i+\epsilon )|. \end{aligned}$$

Without loss of generality, assume that $B^{k+1}>0$ and thus $\mathcal {X}_i^{k+1}(p)\ge 0$ and $\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )\ge 0$. First, consider the case that $\mathcal {X}_i^{k+1}(p) -\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )\le 0$. This implies $S_i^{k+1}(p)-S_i^{k+1}(p_{-i},p_i+\epsilon )\ge 0$ because $S_i^{k+1}(p)=S_i$. By assumption, we then have

$$\begin{aligned} \sum _{t=0}^{k+1}|\mathcal {X}_i^{t}(p)| - \sum _{t=0}^{k+1}\left| \mathcal {X}_i^{t}(p_{-i},p_i+\epsilon )\right|\ge & {} \alpha _i |S_i^{k+1}(p)-S_i^{k+1}(p_{-i},p_i+\epsilon )| \\&+ \;|\mathcal {X}_i^{k+1}(p)| -|\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )| \\= & {} \alpha _i S_i^{k+1}(p)-\alpha _iS_i^{k+1}(p_{-i},p_i+\epsilon )\\&+\; \mathcal {X}_i^{k+1}(p) - \mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon ) \\= & {} |S_i^{k+2}(p)-S_i^{k+2}(p_{-i},p_i+\epsilon )|, \end{aligned}$$

where the last line is due to (4) and the fact that

$$\begin{aligned}|\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )|-|\mathcal {X}_i^{k+1}(p)|\le \alpha _i|S_i^{k+1}(p)-S_i^{k+1}(p_{-i},p_i+\epsilon )| \end{aligned}$$

due to the capacity limit.

Now assume that $\mathcal {X}_i^{k+1}(p) - \mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )\ge 0$. Then

$$\begin{aligned} \sum _{t=0}^{k+1}|\mathcal {X}_i^{t}(p)| - \sum _{t=0}^{k+1}|\mathcal {X}_i^{t}(p_{-i},p_i+\epsilon )|\ge & {} \alpha _i|S_i^{k+1}(p)-S_i^{k+1}(p_{-i},p_i+\epsilon )| \\&+\; |\mathcal {X}_i^{k+1}(p)| -|\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )| \\= & {} \alpha _i |S_i^{k+1}(p)-S_i^{k+1}(p_{-i},p_i+\epsilon )| \\&+\; |\mathcal {X}_i^{k+1}(p) -\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )| \\\ge & {} |\alpha _iS_i^{k+1}(p) + \mathcal {X}_i^{k+1}(p) \\&-\;\alpha _iS_i^{k+1}(p_{-i},p_i+\epsilon ) -\mathcal {X}_i^{k+1}(p_{-i},p_i+\epsilon )| \\= & {} |S_i^{k+2}(p)-S_i^{k+2}(p_{-i},p_i+\epsilon )|, \end{aligned}$$

where the third line is due to the triangle inequality. Since

$$\begin{aligned}|S_i^{k+2}(p)-S_i^{k+2}(p_{-i},p_i+\epsilon )| \ge 0,\end{aligned}$$

we have by induction that $\sum _{t=0}^{k}|\mathcal {X}_i^{t}(p)| - \sum _{t=0}^{k}|\mathcal {X}_i^{t}(p_{-i},p_i+\epsilon )|\ge 0$ for all k, which establishes the desired result that $\sum _{t=0}^T|\mathcal {X}_i^t(p)|$ is nonincreasing in $p_i$.

$\sum _{t=0}^T|\mathcal {X}_i^t(p)|$ can be shown to be nondecreasing in $p_{-i}$ by reducing some $p_j$, $j\ne i$, by $\epsilon >0$ and repeating the above argument.

Proof

(Lemma 3) The proof is a straightforward extension of the approach of [1] to the case of random demand over multiple time periods. We first state several standard facts from game theory (see, e.g., [44]). Let $U_i$ and $L_i$ be the upper and lower boundaries of the support of $\mu _i$. By the definition of mixed-strategy equilibrium, there exists a $\pi _i^*$ and a subset $P_i\subseteq [L_i,U_i]$, $\mu _i(P_i)=1$ for which

$$\begin{aligned} \pi _i(p_i,\mu _{-i})\le & {} \pi _i^*\quad \forall p_i\in [U_i,L_i]\end{aligned}$$

(13)

$$\begin{aligned} \pi _i(p_i,\mu _{-i})= & {} \pi _i^*\quad \forall p_i\in P_i \end{aligned}$$

(14)

This means that there may be a few locations, e.g., a finite number of discrete points, inside $[U_i,L_i]$ but not $P_i$ at which $\pi _i(p_i,\mu _{-i})\le \pi _i^*$. The payoff $\pi _i(p_i,\mu _{-i})$ is continuous at $p_i$ if $\mu _{-i}$ and has no atom there ($\mu _i$ has an atom at $p_i\in [L_i,U_i]$ if $\text {Prob}(p_i)=a>0$, or, equivalently, $\mu _i(x)=a\delta (x-p_i)$, where $\delta (x)$ is the Dirac delta function). Therefore, $\pi _i(p_i,\mu _{-i})=\pi _i^*$ if $\mu _{-i}$ has no atom at $p_i$.

We now prove each point of the lemma sequentially.

• 1. First suppose that $\mu _i(p)=0$ for all i and all $p_i\in [\underline{p},\overline{p}]$, $L_j<\underline{p}<\overline{p}<U_j$ for at least one firm j. Then from increasing price from $\underline{p}$ to $p\in (\underline{p},\overline{p})$ is a profitable deviation for any firm with probability mass below $\underline{p}$ because it will raise their profits without losing their share the demand. Now suppose that only one firm i has $\mu _i(p)>0$ for $p_i\in [\underline{p},\overline{p}]$. Then the distribution

  $$\begin{aligned} \mu '_i(p)=\left\{ \begin{array}{ll} \mu _i(p) &{}\quad \text {if } p_i<\underline{p} \\ 0 &{}\quad \text {if } \underline{p}\le p_i<\overline{p}\\ \mu ( [\underline{p},\overline{p}] ) &{}\quad \text {if } p_i=\overline{p}\\ \mu _i(p) &{}\quad \text {if } p_i>\overline{p} \end{array}\right. \end{aligned}$$

  is a profitable deviation for i; essentially, all of firm i’s probability mass in $[\underline{p},\overline{p}]$ has been shifted to $\overline{p}$ without any loss of demand. Thus at least two firms i must have $\mu _i(p)>0$ for all j with $p_i\in [L_j, U_j]$.

• 2. First we show that no two firms may have an atom at the same location. Suppose multiple firms have an atom at $p'$. Then with positive probability all such firms set $p_i=p'$. For a given firm i, there exists an $\epsilon >0$ small enough that $\pi _i(p'-\epsilon ,\mu _{-i})>\pi _i(p',\mu _{-i})$ because they can capture a significantly larger demand via a small price decrease. This is a profitable deviation for firm i. We now show that no firm can have an atom in [L, U). Suppose firm i has an atom at $p\in [L,U)$. If $p\notin [L_j,U_j)$ for all $j\ne i$, then $p+\epsilon$, $\epsilon >0$ is a profitable deviation for firm i; this is because no other firm sets a higher price, and hence firm i loses no competitiveness by raising its own. Now assume firm i has an atom of mass $a\in (0,1]$ at $p'\in (L_j,U_j)$ for some $j\ne i$, and let $\hat{\mu }$ denote $\mu$ with the atom subtracted off. Consider the difference between firm j’s profits at $p'-\epsilon$ and $p'+\epsilon$:

  $$\begin{aligned}&\pi _j(p'-\epsilon ,\mu _{-j})-\pi _j(p'+\epsilon ,\mu _{-j})\\&\quad =(p'-\epsilon )\left[ \int _{p_{-j}}\left( \prod _{k\ne j}\hat{\mu }_k(p_k)\right) ~\mathbb {E}\left[ \sum _{t=0}^T|\mathcal {X}_j^t\left( p'-\epsilon ,p_{-j}\right) |\right] dp\right. \\&\quad \quad +\left. a\int _{p_{-i,j}}\left( \prod _{k\ne i,j}\hat{\mu }_k(p_k)\right) ~\mathbb {E}\left[ \sum _{t=0}^T|\mathcal {X}_j^t\left( p'-\epsilon ,p',p_{-i,j}\right) |\right] dp\right] \\&\quad \quad -(p'+\epsilon )\left[ \int _{p_{-j}}\left( \prod _{k\ne j}\hat{\mu }_k(p_k)\right) ~\mathbb {E}\left[ \sum _{t=0}^T|\mathcal {X}_j^t\left( p'+\epsilon ,p_{-j}\right) |\right] dp\right. \\&\quad \quad +\left. a\int _{p_{-i,j}}\left( \prod _{k\ne i,j}\hat{\mu }_k(p_k)\right) ~\mathbb {E}\left[ \sum _{t=0}^T|\mathcal {X}_j^t\left( p'+\epsilon ,p',p_{-i,j}\right) |\right] dp\right] .\\ \end{aligned}$$

  Letting $\epsilon$ tend to zero, the first and third terms cancel, while the sum of the remaining terms remains positive because $\sum _{t=0}^T|\mathcal {X}_j^t(p)|$ (by Lemma 7) and hence $\mathbb {E}[\sum _{t=0}^T|\mathcal {X}_j^t(p)|]$ are nonincreasing in $p_j$. Since firm j has no atom at $p'-\epsilon$ or $p'+\epsilon$ for some $\epsilon$, its strategy is continuous at $p'+\epsilon$ and $\pi _j(p'+\epsilon ,\mu _{-j})=\pi _j^*$. But $p'-\epsilon$ is a profitable deviation from $p'+\epsilon$ for firm j because they can capture a significantly large portion of the demand at a slightly lower price. This establishes that an atom can only exist at U.

• 3. Suppose $U<R$. First consider the case that a firm, i, has an atom at U. Then $\pi _i(R,\mu _{-i})>\pi _i(U,\mu _{-i})$ because firm i can increase its price without losing any demand, and thus a profitable deviation exists. Now suppose no firm has an atom at U. Then for any firm i with upper support U, $\pi _i(R,\mu _{-i})>\pi _i(U,\mu _{-i})$ similarly because firm i can increase its price without losing any demand. This establishes the existence of a profitable deviation.

Proof

(Theorem 1) Let $\Upsilon _i$ be the cumulative distribution of firm i’s mixed-strategy. Since, by Lemma 3, neither firm has an atom in [L, R), we have that

$$\begin{aligned} \pi _1^*= & {} x\left( \int _L^{x}\mu _2(p_2)\mathbb {E}\left[ \sum _{t=0}^T\left| \mathcal {X}_1^t(x,p_2)\right| \right] dp_2 + \int _{x}^{R}\mu _2(p_2)\,\mathbb {E}\left[ \sum _{t=0}^T\left| \mathcal {X}_1^t(x,p_2)\right| \right] dp_2 \right) \\= & {} x\left( \overline{\mathcal {X}}_1\int _L^{x}\mu _2(p_2)dp_2+\underline{\mathcal {X}}_1\int _{x}^{R}\mu _2(p_2)dp_2 \right) \\= & {} x\left( \overline{\mathcal {X}}_1\Upsilon _2(x)+\underline{\mathcal {X}}_1(1-\Upsilon _2(x)) \right) . \end{aligned}$$

Note that $\mathbb {E}[\sum _{t=0}^T|\mathcal {X}_1^t(x,p_2)|]=\overline{\mathcal {X}}_1$ within the limits of the first integral and $\mathbb {E}[\sum _{t=0}^T|\mathcal {X}_1^t(x,p_2)|]=\underline{\mathcal {X}}_1$ within the limits of the second integral. Solving for $\Upsilon _2(x)$ over [L, R), we have

$$\begin{aligned} \Upsilon _2(x)= & {} \frac{\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1} -\frac{\pi _1^*}{\left( \underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1\right) x}. \end{aligned}$$

By the same argument, we similarly have

$$\begin{aligned} \Upsilon _1(x)= & {} \frac{\underline{\mathcal {X}}_2}{\underline{\mathcal {X}}_2-\overline{\mathcal {X}}_2} -\frac{\pi _2^*}{\left( \underline{\mathcal {X}}_2-\overline{\mathcal {X}}_2\right) x}, \end{aligned}$$

also over [L, R).

From Lemma 3, only one firm can have an atom at R. If $S_1=S_2$, they are interchangeable. Assume now that $S_1>S_2$ and that $\underline{\mathcal {X}}_1\ge \underline{\mathcal {X}}_2$ and $\overline{\mathcal {X}}_1\ge \overline{\mathcal {X}}_2$. Suppose for the sake of contradiction that firm two, which has smaller capacity, has an atom at R. Then, since firm one cannot have an atom at R,

$$\begin{aligned} \pi _2^*= & {} R\overline{\mathcal {X}}_2\\ \pi _1^*= & {} \frac{R\overline{\mathcal {X}}_2\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_2}. \end{aligned}$$

Subbing $\overline{\pi }(C,R,S)$ into the above expression for $\Upsilon _2(R)$, we have

$$\begin{aligned} \Upsilon _2(R)= & {} \frac{\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1} -\frac{\overline{\mathcal {X}}_2\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_2\left( \underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1\right) }\\> & {} \frac{\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1} -\frac{\overline{\mathcal {X}}_2\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_1\left( \underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1\right) } \quad (\text {because} \; \underline{\mathcal {X}}_1>\underline{\mathcal {X}}_2)\\= & {} \frac{\underline{\mathcal {X}}_1-\overline{\mathcal {X}}_2}{\underline{\mathcal {X}}_1-\overline{\mathcal {X}}_1} \\> & {} 1 \quad (\text {because} \; \overline{\mathcal {X}}_1>\overline{\mathcal {X}}_2). \end{aligned}$$

This contradicts the assumption that $\mu _2$ has an atom at R. Therefore, the equilibrium payoffs are

$$\begin{aligned} \pi _1^*= & {} R\overline{\mathcal {X}}_1\\ \pi _2^*= & {} \frac{R\overline{\mathcal {X}}_1\underline{\mathcal {X}}_2}{\underline{\mathcal {X}}_1}. \end{aligned}$$

Now setting $\Upsilon _2(L)=0$, we have that $L=R\overline{\mathcal {X}}_1/\underline{\mathcal {X}}_1$. The atom at R in the large firm’s strategy can be shown to be

$$\begin{aligned} \frac{\underline{\mathcal {X}}_2\overline{\mathcal {X}}_1-\overline{\mathcal {X}}_2\underline{\mathcal {X}}_1}{\underline{\mathcal {X}}_1\left( \underline{\mathcal {X}}_2-\overline{\mathcal {X}}_2\right) }. \end{aligned}$$

The mixed strategies, can be obtained by substituting the above into $\Upsilon _i(x)$ and differentiating.

Proof

(Lemma 5) The second derivative of $\overline{\pi }(S_i,S_{-i})-\gamma _iS_i$ with respect to $S_i$ is $-Rf(S_{i}+S_{-i})$, and hence it is strictly concave because f is positive. Since $\underline{\pi }(S_{-i},S_i)$ is also differentiable, it suffices for quasiconcavity to show that its first derivative with respect to $S_i$ is initially positive and crosses zero exactly once [11]. Let $\overline{S}_i=\int _0^{S_i}Bf(B)dB+S_i(1-F(S_i))$. The first derivative is given by

$$\begin{aligned} \frac{d\underline{\pi }(S_{-i},S_i)}{dS_i}= & {} \frac{1}{\overline{S}_{-i}}\left[ (1-F(S_i))\overline{\pi }(S_{-i},S_i) + R\left( F(S_{i})-F(S_i+S_{-i})\right) \overline{S}_{i}\right] \nonumber \\ \end{aligned}$$

(15)

where $\overline{S}_i=\int _0^{\infty }\min (S_i,D)f(D)dD$. Dividing through by $\overline{S}_i$, define

$$\begin{aligned} M(S_i)= & {} \frac{(1-F(S_i))\overline{\pi }(S_{-i},S_i)}{\overline{S}_i}\\ N(S_i)= & {} R\left( F(S_i)-F(S_i+S_{-i})\right) \end{aligned}$$

$\underline{\pi }(S_{-i},S_i)$ is equal to zero at $S_i=0$ and $S_i=\infty$ but is not constantly zero. Therefore, by the mean value theorem, (15) is zero at least once in between, which implies that $M(S_i) +N(S_i)=0$ at any such point.

Observe that $M(S_i)$ is always positive and $N(S_i)$ always negative. One may straightforwardly differentiate to see that the magnitude of $(1-F(S_i))$ shrinks faster than that of $(F(S_i)-F(S_i+S_{-i}))$ as $S_i$ increases; since the fraction $\frac{\overline{\pi }(S_{-i},S_i)}{\overline{S}_i}$ also approaches zero with $S_i$, we may conclude that the magnitude of $M(S_i)$ shrinks faster than that of $N(S_i)$. Since the magnitudes are identical at any point $S_i'$ where $M(S_i') +N(S_i')=0$, then $M(S_i'') +N(S_i'')<0$ for any $S_i''>S_i'$. Therefore (15) is zero at only one finite value, henceforth denoted $S_i'$, establishing the quasiconcavity of $\underline{\pi }(S_{-i},S_i)$ with respect to $S_i$.

Because the magnitude of $M(S_i)$ shrinks more rapidly than that of $N(S_i)$, $\frac{d\underline{\pi }}{dS_2}(S_{-i},S_i)$ is strictly decreasing in $S_i$ prior to crossing zero. Since $-\gamma _i$ is negative and constant, $\frac{d\underline{\pi }(S_{-i},S_i)}{dS_i}-\gamma _i$ is also strictly decreasing in $S_i$ before crossing zero and then remains negative, implying quasiconcavity of $\underline{\pi }(S_{-i},S_i)-\gamma _{-i}S_{-i}$ with respect to $S_i$.

Proof

(Theorem 6) We prove that $\hat{S}^i$ are the only possible pure-strategy equilibria by showing that an equilibrium can only exist where $\psi _i$ has zero slope in $S_i$. This is equivalent to showing that no ‘ridges’ exist in $\psi _i$. A sufficient condition is

$$\begin{aligned} \left. \frac{d\underline{\pi }\left( S_{-i},S_i\right) }{dS_i}\right| _{S_i=S_{-i}}\le \left. \frac{d\overline{\pi }\left( S_i,S_{-i}\right) }{dS_i}\right| _{S_i=S_{-i}}. \end{aligned}$$

(16)

We have that

$$\begin{aligned} \left. \frac{d\underline{\pi }\left( S_{-i},S_i\right) }{dS_i}\right| _{S_{-i}=S_i}= & {} \left( 1-F\left( S_i\right) \right) \frac{\overline{\pi }\left( S_i,S_i\right) }{\overline{S}_i}+R\left( F\left( S_i\right) -F\left( 2S_i\right) \right) \\\le & {} R\left( 1-F\left( S_i\right) \right) +R\left( F\left( S_i\right) -F\left( 2S_i\right) \right) \\= & {} R\left( 1-F\left( 2S_i\right) \right) \\= & {} \left. \frac{d\overline{\pi }\left( S_i,S_{-i}\right) }{dS_i}\right| _{S_{-i}=S_i}, \end{aligned}$$

establishing the claim.