Basic Pricing Theory

Abstract

This chapter provides an introduction to multi-product monopoly pricing when the variable costs are linear. Profit maximization problems with linear variable costs arise from capacity constraints, where the firm maximizes the expected profit net of the opportunity costs of the capacities used. We argue that under mild assumptions, both the optimal profit function and the expected consumer surplus are convex functions of the variable costs. Consequently, when variable costs are random, both the firm and the representative consumer benefit from prices that dynamically respond to changes in variable costs. Randomness in variable cost is often driven by randomness in demand in conjunction with capacity constraints, and this accounts for some of the benefits of dynamic pricing. We explore conditions for the existence and uniqueness of maximizers of the expected profit and analyze in detail problems with capacity constraints both when prices are set for the entire sales horizon a priori, and when prices are allowed to change during the sales horizon. The firm’s problem is discussed in Sect. 8.2, while the representative consumer’s problem is presented in Sect. 8.3. The case with finite capacity is discussed in Sect. 8.4. Details about existence and uniqueness for single product problems are discussed in Sect. 8.5. This section also includes applications to priority pricing, social planning, multiple market segments, and peak-load pricing. Multi-product pricing problems are discussed in Sect. 8.6.

Appendix

Proof of Theorem 8.1

It is clear that R(p, z) is decreasing in z and that this implies that $\mathcal {R}(z)$ is decreasing in z. To verify convexity, let α ∈ (0, 1). Then for any $z, \tilde {z}$,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{R}(\alpha z + (1-\alpha)\tilde{z}) &\displaystyle = &\displaystyle \max_{p \in X} R(p,\alpha z + (1-\alpha)\tilde{z})\\ &\displaystyle = &\displaystyle \max_{ p \in X}R(\alpha p + (1-\alpha)p, \alpha z + (1-\alpha)\tilde{z}))\\ &\displaystyle = &\displaystyle \max_{ p \in X}\left[\alpha(p-z)' + (1-\alpha)(p - \tilde{z})'\right]d(p)\\ &\displaystyle = &\displaystyle \max_{ p \in X}\left[\alpha R(p,z) + (1-\alpha)R(p,\tilde{z})\right]\\ &\displaystyle \leq &\displaystyle \alpha \max_{ p \in X}R(p,z) + (1-\alpha)\max_{ p \in X}R(p,\tilde{z})\\ &\displaystyle = &\displaystyle \alpha \mathcal{R}(z) + (1-\alpha)\mathcal{R}(\tilde{z}). \quad \end{array} \end{aligned} $$

Proof of Proposition 8.2

This follows from a direct application of the Taylor’s expansion around $\mathcal {R}(\mathbb E[Z])$.

Proof of Corollary 8.4

The proof of the first part is left as an exercise. From the concavity of g we have $g(\alpha z + (1-\alpha )\tilde {z}) \geq \alpha g(z) + (1-\alpha ) g(\tilde {z})$ for any $z, \tilde {z} \in \Re ^m$ and any α ∈ [0, 1]. Since $\mathcal {R}$ is decreasing, it follows that $\mbox{{$\mathcal {R}(g(\alpha z + (1-\alpha )\tilde {z})$}} \leq \mathcal {R}( \alpha g(z) + (1-\alpha ) g(\tilde {z}))$. From the convexity of $\mathcal {R}$, we have $\mathcal {R}( \alpha g(z) + \mbox{{$(1-\alpha ) g(\tilde {z}))$}} \leq \alpha \mathcal {R}( g(z)) + (1-\alpha )\mathcal {R}(g(\tilde {z}))$. Consequently, $\mbox{{$\mathcal {R}(g(\alpha z + (1-\alpha )\tilde {z}))$}} \leq \alpha \mathcal {R}(g(z)) + (1-\alpha )\mathcal {R}(g(\tilde {z}))$, showing that $\mathcal {R}(g(z))$ is convex in z. From Jensen’s inequality, it follows that $\mathbb E[\mathcal {R}(g(Z))] \geq \mathcal {R}(g(\mathbb E[Z]))$.  

Proof of Theorem 8.5

That $\mathcal {S}(p)$ is decreasing follows directly from the fact that S(q, p) is decreasing in p. To verify convexity, let α ∈ (0, 1). Then for any $p, \tilde {p}$

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{S}(\alpha p + (1-\alpha)\tilde{p}) &\displaystyle = &\displaystyle \max_{q \geq 0}S(\alpha p + (1-\alpha)\hat{p},q)\\ &\displaystyle = &\displaystyle \max_{q \geq 0} \left[U(q) - (\alpha p + (1-\alpha)\hat{p})'q\right]\\ &\displaystyle = &\displaystyle \max_{q \geq 0}\left[ \alpha(U(q) -p'q) + (1-\alpha)(U(q) - \tilde{p}'q)\right]\\ &\displaystyle \leq &\displaystyle \alpha \max_{q \geq 0}S(p,q) + (1-\alpha)\max_{q \geq 0}S(\tilde{p},q)\\ &\displaystyle = &\displaystyle \alpha \mathcal{S}(p) + (1-\alpha)\mathcal{S}(\tilde{p}). \end{array} \end{aligned} $$

Notice that $\mathcal {S}(p) = U(d(p)) - p'd(p)$, so $\nabla \mathcal {S}(p)= \nabla d(p)\nabla U(d(p)) - d(p) - \nabla d(p)p = - d(p)$ on account of ∇U(d(p)) = p for all $p \in \mathcal {P}$.⁵

Proof of Theorem 8.7

Let $w'q - \frac {1}{2} {\,} q'Qq$ be the quadratic approximation to an increasing concave utility function U, where w is a vector of positive components, and Q is symmetric positive definite matrix.⁶ Let B = Q ⁻¹ and write d(p) = B(w − p) over the set $\mathcal {P} = \{p: p \geq 0, B(w-p) \geq 0\}$. Then

$$\displaystyle \begin{aligned}\mathcal{S}(p) = U(d(p)) - d(p)'p = \frac{1}{2}(w-p)'B(w-p) ~~\mbox{over}~~p \in \mathcal{P},\end{aligned}$$

which is decreasing convex in $p \in \mathcal {P}$ since B is positive definite. The firm’s problem is to find p = p(z) that maximizes

$$\displaystyle \begin{aligned}R(p,z) = (p-z)'B(w-p).\end{aligned}$$

The optimizer is given by p(z) = (w + z)∕2, which is an increasing linear function of $z \in \mathcal {P}$. The composite function $\mathcal {S}(p(z))$ is therefore convex.

Proof of Theorem 8.10

Since d(p) is USC and the product of non-negative USC functions is also USC, it follows that R(p, z) is USC. The USC of d(p) implies the USC of $\bar {d}(p)$ for if $\bar {d}(p)$ is not USC at p ₀, then there exist a p ₁ > p ₀ at which $d(p_1) = \bar {d}(p_o)$ fails to be USC. As a result $\bar {R}(p,z)$ is also USC in p ∈ [z, ∞). If d(p) = 0 for all p ≥ z, then p(z) = z and $\mathcal {R}(z) = R(z,z) = 0$ and there is nothing to prove. Otherwise there exists a price $\hat {p} > z$ such that $0 < \bar {d}(\hat {p}) < \infty$, for if $\bar {d}(p) = \infty$ for all p > z, then $\bar {d}(p)$ is not o(1∕p). Let $\epsilon = \bar {R}(\hat {p},z) > 0$. We will show that there is a price $\tilde {p} > \hat {p}$ such that $\bar {R}(p,z) \leq \epsilon$ for all $p > \tilde {p}$, for if not, then for any $\tilde {p}> z$, we can find a $p > \tilde {p}$ such that $\bar {R}(p,z) > \epsilon$, or equivalently, $p\bar {d}(p) > p \epsilon /(p-z)$, contradicting the fact that $p\bar {d}(p) \rightarrow 0$ as p →∞. Given that $\bar {R}(p,z) \leq \epsilon$ for all $p \geq \tilde {p}$, we can restrict the optimization of $\bar {R}(p,z)$ without loss of optimality to the compact set $[z,\tilde {p}]$. The extreme value theorem guarantees the existence of a finite price, say $\bar {p}(z) \in [z,\tilde {p}]$, that maximizes $\bar {R}(p,z)$. We will now show that $p(z) = \bar {p}(z)$ also maximizes R(p, z) so $\mathcal {R}(z) = \bar {\mathcal {R}}(z)$. Assume for a contradiction that $\bar {p}(z)$ is not a maximizer of R(p, z). Then

$$\displaystyle \begin{aligned}(\bar{p}(z) - z) \bar{d}(\bar{p}(z)) = \bar{\mathcal{R}}(z) \geq \mathcal{R}(z) > (\bar{p}(z) - z) d(\bar{p}(z))\end{aligned}$$

implies that $d(\bar {p}(z)) < \bar {d}(\bar {p}(z)) = \sup _{p \geq \bar {p}(z)}d(p)$. Then there exists a $p' > \bar {p}(z)$ such that $d(p') = \bar {d}(p(z))$, but then $\bar {R}(p',z) > \bar {R}(\bar {p}(z),z) = \bar {\mathcal {R}}(z)$ contradicting the optimality of $\bar {p}(z)$.

Proof of Theorem 8.11

First, we show Part a. If h(p) is continuous and increasing in p, then f(p) is continuous and strictly decreasing in p ≥ z. Equivalently, (p − z)h(p) is continuous and strictly increasing in p. Now f(z) = 1 > 0 implies that p(z) > z, while f(z + 1∕h(z)) = 1 − h(z + 1∕h(z))∕h(z) ≤ 0 on account of h(z + 1∕h(z)) ≥ h(z) > 0 implies that p(z) ≤ z + 1∕h(z). Because (p − z)h(p) is continuous and strictly increasing in p, there exist a unique p(z) satisfying $p(z) = \mbox{{$\sup \{p: f(p) \geq 0\}$}}$ that is bounded below by z and above by z + 1∕h(z). Suppose that z′ > z, then (p(z) − z′)h(p(z)) < 1, so p(z′) > p(z) showing that p(z) is strictly increasing in z. To show that Δ(z) = p(z) − z is decreasing in z, let p′ = z′ + Δ(z) and notice that (p′− z′)h(p′) = Δ(z)h(p′) ≥ Δ(z)h(p(z)) = 1, so p(z′) = z′ + q(z′) ≤ p′ = z′ + Δ(z) implying that Δ(z′) ≤ Δ(z). For the exponential demand function d(p) = λe ^−p∕θ, we have h(z) = 1∕θ and p(z) = z + θ = z + 1∕h(z), so the upper bound is attained.

Next, we show Part b. If ph(p) is continuous and strictly increasing in p and $\tilde {z}h(\tilde {z}) > 1$, then f(p) is continuous in p > z and the equation f(p) = 0 can be written as ph(p) = p∕(p − z) with the left hand side increasing in p and the right hand side strictly decreasing to one for p > z. Since zh(z) < ∞ it follows that p(z) > z. Notice that $z/(1-\tilde {z}h(\tilde {z}))$ is the root of $\tilde {z}h(\tilde {z}) = p/(p-z)$. Since $ph(p) \geq \tilde {z}h(\tilde {z}) \geq p/(p-z)$ for all $p \geq z/(1-\tilde {z}h(\tilde {z}))$, it follows that p(z) is unique and bounded above by $z/(1-1/\tilde {z}h(\tilde {z}))$. Suppose that z′ > z, then p(z′) > z′, so if z′≥ p(z) it follows immediately that p(z′) ≥ p(z). Suppose now that z < z′ < p(z), then at p = p(z) we have ph(p) < p∕(p − z′) implying that p(z′) > p(z). For d(p) = λp ^−b, with b > 1, we have ph(p) = b for all p, and $p(z) = bz/(b-1) = z/(1-1/b) = z/(1-1/\tilde {z}h(\tilde {z}))$, so the upper bound is attained. Let m(z) := 1∕h(z). Then, using the implicit function theorem on f(p, z) = 0, we can find the first and second derivatives of p(z) in terms of m(z). It is easy to see that the first derivative is given by p′(z) = (1 − m′(p(z))⁻¹, so the second derivative is given by

$$\displaystyle \begin{aligned}p''(z) = \frac{m''(p(z)p'(z))}{(1-m'(p(z)))^2} \leq 0,\end{aligned}$$

since m″(z) ≤ 0 and p′(z) > 0.

Finally, we show Part c. Clearly $\tilde {f}(p) \leq f(p)$ so $\tilde {p}(z) \leq p(z)$.

Proof of Proposition 8.13

Since the sum of USC is USC it follows that d _S(p) is USC. Moreover $\bar {d}_m(p) = o(1/p)$ for all $m \in \mathcal {M}$ implies that $\bar {d}_S(p) = o(1/p)$. As a result d _S(p) satisfies the conditions of Theorem 8.10 so there exists a finite price p _S(z), increasing in z, such that $\mathcal {R}_S(z) = R_S(p_S(z),z)$ is decreasing convex in z.

Proof of Proposition 8.16

It is easy to see that p _m(z) > z is the root of p∕(p − z) = ph _m(p). Since the left hand side is decreasing in p and ph _m(p) is increasing in p, it follows that there is a unique root p > z. This observation implies that f _m(p) > 0 on p < p _m(z) and f _m(p) < 0 on p > p _m(z). Let f _S(p) = 1 − (p − z)h _S(p) where h _S(p) is the hazard rate of d _S(p). Since f _S(p) is a convex combination of f _m(p) = 1 − (p − z)h _m(p) with weights θ _m(p) = d _m(p)∕d _S(p), it follows that f _S(p) > 0 for all p <min_{m ∈ S} p _m(z) because over that interval f _m(p) > 0 for all m ∈ S. Also f _S(p) < 0 for all p >max_{m ∈ S} p _m(z) because over that interval f _m(p) < 0 for all m ∈ S. Since the derivative of R _S(p, z) is proportional to f _S(p) it follows that R _S(p, z) is increasing over p <min_{m ∈ S} p _m(z) and decreasing over p >max_{m ∈ S} p _m(z). Moreover, since R _S(p, z) is continuous over the closed and bounded interval [min_{m ∈ S} p _m(z), max_{m ∈ S} p _m(z)], appealing to the EVT yields the existence of a global maximizer p _S(z) of R _S(p, z).

Proof of Theorem 8.18

Clearly

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{Q_J(z)}{Q_M(z)} &\displaystyle \geq &\displaystyle \frac{Q^h_J(z)}{Q_M(z)} = \frac{\sum_{j=1}^J \sum_{m \in M_j}R_m(q_j,z)}{Q_M(z)}\\ &\displaystyle = &\displaystyle \sum_{j=1}^J \sum_{m \in M_j}e(q_j,p_m(z),z)\frac{\mathcal{R}_m(z)}{Q_M(z)}\\ &\displaystyle \geq &\displaystyle \gamma_J(z) \frac{\sum_{m \in \mathcal{M}} \mathcal{R}_m(z)}{Q_M(z)}\\ &\displaystyle = &\displaystyle \gamma_J(z). \end{array} \end{aligned} $$

Proof of Theorem 8.21

We have already shown that inequalities in (8.13). To show (8.14), notice that the super-modularity of R(p, z) in p for fixed z, allows us to select p _i(z|p _−i) so that it is increasing in p _−i. Consequently, p _i(z|p _−i) ≤ p _i(z _i|∞) = p _i(z _i) for all i. In particular, p _i(z) = p _i(z|p _−i(z)) ≤ p _i(z _i) for all i ∈ N.

Proof of Theorem 8.23

Maximizing R(p, z) = (p − z)′d(p) with respect to p is equivalent to minimizing $\frac {1}{2}p'Sp - (a + B'z)'p + a'z$ which is quadratic function. A sufficient condition for this function to be convex is that S is positive definitive. It is known that S is positive definitive, if and only if B is, see Johnson (1970). If B is positive definitive then S is invertible and since S is symmetric, so it is inverse S ⁻¹. If B is positive definitive then the maximizer of R(p, z) is given by (8.15). A sufficient condition for p(0) = S ⁻¹ a ≥ 0 is for S ⁻¹ ≥ 0, since a > 0. However, this is true because S is an s-matrix, i.e. a real symmetric, positive definitive matrix with non-positive off-diagonal elements. It is known that an s-matrix has a non-negative inverse implying that S ⁻¹ ≥ 0, and consequently that p(0) = S ⁻¹ a ≥ 0. Since p(z) is non-decreasing in z by Theorem 8.1, it follows that p(z) ≥ p(0) ≥ 0 for all z ≥ 0 such that d(z) ≥ 0.

By adding and subtracting Bz to the expression in parenthesis on the right hand side of (8.15) we can write p(z) − z = S ⁻¹ d(z), where d(z) is the demand at p = z. It is also possible to write d(p(z)) = a−Bp(z) = a−B(p(z)+z−z) = a−Bz−B(p(z)−z) = (I−BS ⁻¹)d(z) and then use the fact that I − BS ⁻¹ = B′S ⁻¹ to obtain d(p(z)) = B′S ⁻¹ d(z). This allows us to write $\mathcal {R}(z) = (p(z)-z)'d(p(z)) = d(z)'S^{-1}B'S^{-1}d(z) = d(z)'S^{-1}B'S^{-1}d(z)$ resulting in (8.16).

Proof of Theorem 8.24

The first-order conditions are of the form

$$\displaystyle \begin{aligned}\frac{\partial R(p,z)}{\partial p_i} = d_i(p)[ 1 + \beta R(p,z)/\lambda - \beta (p_i - z_i)] = 0~~~ \forall {\,} i \in N.\end{aligned}$$

For every subset F ⊆ N, let p ^F(z) be the solution to the first-order conditions obtained by setting the expression in brackets equal to zero for all i ∈ F and by setting d _i(p) = 0 for all i∉F. Then,

$$\displaystyle \begin{aligned}p_i = z_i + 1/\beta + R(p,z)/\lambda~~~\forall {\,} i \in F~~~\mbox{and}~~~~p_i = \infty~~~\forall {\,} i \notin F.\end{aligned}$$

For each F ⊆ N, there exists a constant, θ _F = R(p ^F(z), z)∕λ, given by the root of the Lambert type equation

$$\displaystyle \begin{aligned} \beta \theta e^{\beta \theta} = \sum_{j \in F}e^{\alpha_j - \beta z_j - 1}, \end{aligned} $$

such that $p^F_i(z) = z_i + 1/\beta + \theta _F$ for all i ∈ F, and R(p ^F(z), z) = λθ _F, so θ _F represents the optimal profit per customer when we are allowed to offer only products in F. Since the root θ _F is increasing in F, it follows that among all the 2ⁿ solutions to the first-order conditions, the one with the highest profit corresponds to F = N. Thus, at optimality, we have

$$\displaystyle \begin{aligned}p_i(z) = z_i + \frac{1}{\beta} + \theta,\end{aligned}$$

where θ is the root of the Lambert equation for F = N. Moreover, $\mathcal {R}(z) = \lambda \theta$ is the optimal profit.