Threshold discounts comparison: All-unit or incremental?

Abstract

An increasingly ubiquitous discount format in retailing that is taking over traditional price cuts is threshold discount, under which a price reduction is awarded to a purchase that meets a minimum quantity or minimum spending requirement. One of the most frequently used forms of threshold discounts is the all-unit discount, where a discount applies to all units of the promoted item that the customer purchases as long as the total basket size reaches the predetermined minimum requirement. On the other hand, under another type of threshold discounts known as incremental discount, the reduced price is granted for only the units purchased beyond the threshold; the customer still pays the full price for the first units up to the threshold. In this paper, we are interested in comparing the effectiveness of using all-unit discounts and incremental discounts in a retail setting. We consider a retailer selling to a market of heterogenous consumers, whose purchase decisions are modeled and analyzed at the individual customer level. The optimal discount terms and the seller’s profits when using the all-unit and incremental discounts are investigated under different market scenarios. We show that when the discount terms can be fully optimized, the incremental discount is always more profitable to the retailer than the all-unit discount. Furthermore, in many cases, the resulting consumer utility as well as the social welfare is also greater under the incremental discount than under other discount schemes. Our study suggests that the incremental discount, while not currently popular in retailing, is in fact a very profitable and efficient retail pricing mechanism that deserves more attention.

Appendices

Appendix 1: Extensions

Exponential Distribution of Consumer Types

Previously, we have assumed when analyzing the seller’s problem that the consumer type, $\theta$, is uniformly distributed. In this section, we numerically conduct similar analyses as before but by assuming that the consumer type is exponentially distributed with a parameter $\lambda > 0$. We observe that all key results continue to hold. That is, the seller’s profit under the optimal incremental discount is greater than that under the optimal all-unit discount, which is in turn greater than that under the optimal price cut and no discount, respectively. The social welfare under the optimal incremental discount is also the highest in all cases that we observe. The profit improvement of the incremental scheme over the other schemes also decreases in the consumer valuation ($v_0$), in the same manner as observed in Table 2. As depicted in Figure 5, the seller’s profits under different types of discounts with exponentially distributed consumer types exhibit a very similar trend as those with uniformly distributed consumer types shown in Figure 4. This suggests that our results are not dependent on a particular distribution of consumer types.

Figure 5

Seller’s profit under optimal threshold discounts when consumer types are exponentially distributed: $c = 0.2, p = 1, \lambda = 0.2.$

Endogenous Price

In the original model, we have assumed that the retail price is exogenously given to the seller. In this section, we expand our consideration to the situations when the seller can optimize the retail price as well as the discount terms. Our numerical results show that the incremental scheme is still more profitable than the all-unit scheme, the straight-off price cut, and no discount, respectively. We also compare the seller’s profit under the optimal incremental discount when the retail price is endogenous versus when the retail price is exogenous. These results are summarized in Table 3 for both when the consumer types are uniformly distributed and exponentially distributed. Notice that, quite intuitively, the benefit of the ability to set the retail price increases as the optimal price is further away from the exogenously given price. But overall, we see that the profit improvement from setting the retail price is only significant when the optimal price is substantially different from the exogenous price (at least 30 per cent different to realize the benefit of more than 5 per cent). This implies that as long as the given retail price is reasonable, relative to the consumer valuation and marginal cost, the optimal threshold discount with exogenous price would still achieve a profit close to the optimal profit with endogenous price.

Table 3 Optimal price and seller’s profit improvement under incremental discounts with endogenous price: $c = 0.2$, exogenous price $p = 1$, uniform distribution [0, 5], exponential distribution $\lambda = 0.2$

Appendix 2: Proofs

Proof of Proposition 1:

Let $q^{A*}(\theta )$ be the optimal purchase quantity of a consumer type $\theta$ when offered an all-unit discount $D^A = (r, K)$. We will show part (i), (ii), and (iii) of the proposition by showing that

$$\begin{aligned} q^{A*}(\theta ) = \left\{ \begin{array}{ll} \frac{\theta (v_0-p)}{v_0} &{} \text {if }\;\theta< \underline{\theta }^A\\ K &{} \text {if }\;\underline{\theta }^A \le \theta < \bar{\theta }^A\\ \frac{\theta (v_0-p(1-r))}{v_0} &{} \text {if }\;\theta \ge \bar{\theta }^A\\ \end{array} \right. \end{aligned}$$

where

$$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\theta } ^{A} = \left\{ {\begin{array}{*{20}l} {\frac{{v_{0} K\left[ {(v_{0} - p(1 - r)) - \sqrt {pr(2v_{0} - 2p + pr)} } \right]}}{{(v_{0} - p)^{2} }}} & {{\text{if}}\;r \le \frac{p}{{2v_{0} }}} \\ {\frac{{2v_{0} K(1 - r)}}{{2v_{0} - p}}} & {{\text{if}}\;r > \frac{p}{{2v_{0} }}} \\ \end{array} } \right.$$

and $\bar{\theta }^A = \frac{v_0K}{v_0-p(1-r)}$.

Notice that $\bar{\theta }^A > K$ for any $r \le 1$. If $r > \frac{p}{2v_0}$, then $\underline{\theta }^A = \frac{2v_0K(1-r)}{2v_0-p}< K < \bar{\theta }^A$. If $r \le \frac{p}{2v_0}$, then $\underline{\theta }^A = \frac{v_0K\left[ (v_0-p(1-r)) - \sqrt{pr(2v_0-2p+pr)}\right] }{(v_0-p)^2} \ge K$. With some algebra, one can check that $\frac{v_0K\left[ (v_0-p(1-r)) - \sqrt{pr(2v_0-2p+pr)}\right] }{(v_0-p)^2} \le \frac{v_0K}{v_0-p(1-r)}$for all $r \le \frac{p}{2v_0}$. It is also straightforward to see that both $\underline{\theta }^A$ and $\bar{\theta }^A$ defined above are monotonically increasing in K but monotonically decreasing in r.

Given the definition of $\underline{\theta }^A$ and $\bar{\theta }^A$ above, one can also check that $\frac{\theta (v_0-p)}{v_0} < K$ if $\theta < \underline{\theta }^A$ and $\frac{\theta (v_0-p(1-r))}{v_0} \ge K$ if $\theta \ge \bar{\theta }^A$.

Next, we will show for each of the three intervals of $\theta$, corresponding to part (i), (ii), and (iii) of the proposition, that $q^{A*}(\theta )$ defined above gives the maximum utility to the consumer. Suppose first that $r > \frac{p}{2v_0}$, so $\underline{\theta }^A = \frac{2v_0K(1-r)}{2v_0-p} < K$.

• (i) For $\theta < \underline{\theta }^A$, $q^{A*}(\theta ) = \frac{\theta (v_0-p)}{v_0}$.

  Since $\theta < K$, the consumer’s utility from purchasing q is given by

  $$U^{A} (\theta ,q,D^{A} ) = \left\{ {\begin{array}{*{20}l} {v_{0} q\left( {1 - \frac{q}{{2\theta }}} \right) - pq} & {{\text{if }}0 \le q < \theta } \\ {\frac{{v_{0} \theta }}{2} - pq{\text{ }}} & {{\text{if }}0 \le q < K} \\ {\frac{{v_{0} \theta }}{2} - p(1 - r)q} & {{\text{if }}q \ge K} \\ \end{array} } \right.$$

  Note from the second and third expressions that the consumer’s utility monotonically decreases in q for all $\theta \le q < K$ and $q \ge K$. Thus, $\theta< q < K$ and $q > K$ cannot be optimal. We only need to consider $0 \le q \le \theta$ and $q = K$ as candidates for the optimal purchase quantity.

  For $0\,\le\,q\,\le\,\theta$, the utility function is $U^A(\theta , q, D^A) = v_0q(1 - \frac{q}{2\theta }) - pq$, which has $\frac{dU^A(q)}{dq} = \frac{v_0(\theta -q)-\theta p}{\theta }$ and $\frac{d^2U^A(q)}{dq^2} = \frac{-v_0}{\theta } < 0$. Hence, it is concave in q with the solution to the first-order condition given by $q_0(\theta ) := \frac{\theta (v_0-p)}{v_0}$. Notice that $0 \le q_0(\theta ) < \theta$ since $0< p < v_0$ and $\theta \ge 0$. Thus, $q_0(\theta )$ is the interior maximizer of the consumer’s utility in this interval, resulting in the utility of $\frac{\theta (v_0-p)^2}{2v_0}$. For $q = K$, the utility is $\frac{v_0\theta }{2} - p(1-r)K$, which is less than $\frac{\theta (v_0-p)^2}{2v_0}$ since $\theta < \underline{\theta }^A = \frac{2v_0K(1-r)}{2v_0-p}$. Hence, the optimal quantity for $\theta < \underline{\theta }^A$ is $q^{A*}(\theta ) = \frac{\theta (v_0-p)}{v_0}$.

• (ii) For $\underline{\theta }^A \le \theta < \bar{\theta }^A$, $q^{A*}(\theta ) = K$.

  Since $\bar{\theta }^A > K$, we will consider two subintervals: (iia) $\underline{\theta }^A \le \theta < K$ and (iib) $K \le \theta < \bar{\theta }^A$.

  • (iia) For $\underline{\theta }^A \le \theta < K$, the consumer’s utility is the same as that given in part (i) since $\theta < K$. Hence, we will consider only $q_0(\theta ) := \frac{\theta (v_0-p)}{v_0}$ and $q = K$ as candidates for the optimal purchase quantity. Notice that $U^A(q_0(\theta )) = \frac{\theta (v_0-p)^2}{2v_0} \le \frac{v_0\theta }{2} - p(1-r)K = U^A(K)$ since $\theta \ge \underline{\theta }^A = \frac{2v_0K(1-r)}{2v_0-p}$. Hence, the optimal purchase quantity for this subinterval is $q^{A*}(\theta ) = K$.

  • (iib) For $K \le \theta < \bar{\theta }^A$, the consumer’s utility from purchasing q is given by

    $$U^{A} (\theta ,q,D^{A} ) = \left\{ {\begin{array}{*{20}l} {v_{0} q\left( {1 - \frac{q}{{2\theta }}} \right) - pq} & {{\text{if}}\;0 \le q{{ < }}K} \\ {v_{0} q\left( {1 - \frac{q}{{2\theta }}} \right) - p(1 - r)q} & {{\text{if}}\;K\,\le\,q\,\le\,\theta } \\ {\frac{{v_{0} \theta }}{2} - p(1 - r)q} & {{\text{if}}\;q > \theta {\text{ }}} \\ \end{array} } \right.$$

    Note from the third expression that the consumer’s utility monotonically decreases in q for all $q > \theta$. Thus, $q > \theta$ cannot be optimal. We only need to consider $0 \le q < K$ and $K \le q \le \theta$ as candidates for the optimal purchase quantity. For $0 \le q < K$, the utility function is $U^A(\theta , q, D^A) = v_0q\left( 1 - \frac{q}{2\theta }\right) - pq$. We have seen from part i) that the optimal purchase quantity in this interval is $q_0(\theta ) = \frac{\theta (v_0-p)}{v_0}$, which gives utility of $\frac{\theta (v_0-p)^2}{2v_0}$. Notice that $q_0(\theta ) < K$ since $\theta < \bar{\theta }^A = \frac{v_0K}{v_0-p(1-r)}$. For $K \le q \le \theta$, the utility is given by $U^A(\theta , q, D^A) = v_0q(1-\frac{q}{2\theta }) - p(1-r)q$, which has $\frac{dU^A(q)}{dq} = \frac{v_0(\theta -q)-\theta p(1-r)}{\theta }$ and $\frac{d^2U^A(q)}{dq^2} = \frac{-v_0}{\theta } < 0$. Hence, it is concave in q with the solution to the first-order condition given by $q_0^\prime (\theta ) := \frac{\theta (v_0-p(1-r))}{v_0}$. However, observe that $q_0^\prime (\theta ) < K$ since $\theta < \bar{\theta }^A = \frac{v_0K}{v_0-p(1-r)}$. This implies that the optimal purchase quantity for the interval $K \le q \le \theta$ is actually K, resulting in utility of $v_0K(1-\frac{K}{2\theta }) - p(1-r)K$. It remains to compare the utility at $q = K$ and $q = q_0(\theta )$. Note that $\frac{d(U^A(K) - U^A(q_0(\theta )))}{d\theta } = \frac{v_0K^2}{2\theta ^2} - \frac{(v_0-p)^2}{2v_0} > 0$ since $\theta < \bar{\theta }^A = \frac{v_0K}{v_0-p(1-r)}$. This implies $U^A(K) - U^A(q_0(\theta ))$ is minimized at the smallest $\theta$ in this interval, which is $\theta = K$. At $\theta = K$, we have $U^A(K) - U^A(q_0(\theta )) = pK(r - \frac{p}{2v_0}) > 0$ since $r > \frac{p}{2v_0}$. Thus, $U^A(K) - U^A(q_0(\theta )) > 0$ for $K \le \theta < \bar{\theta }^A$. It follows that the optimal purchase quantity for this interval of $\theta$ is K.

• (iii) For $\theta \ge \bar{\theta }^A$, $q^{A*}(\theta ) = \frac{\theta (v_0-p(1-r))}{v_0}$.

  Since $\theta \ge \bar{\theta }^A > K$, the consumer’s utility is given by the same expressions as in part (iib). For the same reason, we can rule out $q > \theta$ and consider only $0 \le q < K$ and $K \le q \le \theta$.

  For $0 \le q < K$, we have seen that the utility is concave and the solution to the first-order condition is $q_0(\theta ) = \frac{\theta (v_0-p)}{v_0}$. However, notice that $q_0(\theta )$ is less than K if and only if $\theta < \frac{v_0K}{v_0-p}$. This implies that for $0 \le q < K$, the utility is maximized at $q = q_0(\theta )$ if $\bar{\theta }^A = \frac{v_0K}{v_0-p(1-r)} \le \theta < \frac{v_0K}{v_0-p}$, and is maximized at $q = K$ if $\theta \ge \frac{v_0K}{v_0-p}$. For $K \le q \le \theta$, we have seen from part ii) that the utility is concave with the solution to the first-order condition of $q_0^\prime (\theta ) = \frac{\theta (v_0-p(1-r))}{v_0}$. Note that $q_0^\prime \ge K$ since $\theta \ge \bar{\theta }^A$ and $q_0^\prime \le \theta$ for any $p > 0$ and $r \le 1$. Hence, $q_0^\prime (\theta )$ is the maximizer for $K \le q \le \theta$, resulting in utility of $\frac{\theta (v_0-p(1-r))^2}{2v_0}$.

  Now, consider the utility at $q = q_0(\theta )$, $q = K$, and $q = q_0^\prime (\theta )$. It is straightforward to check that $U^A(q_0^\prime (\theta )) \ge U^A(q_0(\theta ))$ for any $r \ge 0$, and $U^A(q_0^\prime (\theta )) \ge U^A(K)$ since $\theta \ge \frac{v_0K}{v_0-p(1-r)}$. Thus, $q_0^\prime (\theta ) = \frac{\theta (v_0-p(1-r))}{v_0}$ is the optimal purchase quantity for $\theta \ge \bar{\theta }^A$.

This completes the proof for the case of $r > \frac{p}{2v_0}$. The results for the case of $r \le \frac{p}{2v_0}$ can be shown in the same manner.

Proof of Proposition 2:

Let $q^{I*}(\theta )$ be the optimal purchase quantity of a consumer type $\theta$ when offered an incremental discount $D^I = (r, K)$. We will show part (i) and (ii) of the proposition by showing that

$$\begin{aligned} q^{I*}(\theta ) = \left\{ \begin{array}{ll} \frac{\theta (v_0-p)}{v_0} &{} {\text {if }}\;\theta\,<\, \bar{\theta }^I\\ \frac{\theta (v_0-p(1-r))}{v_0} &{} {\text {if }}\;\theta\, \ge\, \bar{\theta }^I\\ \end{array} \right. \end{aligned}$$

where $\bar{\theta }^I = \frac{2v_0K}{2v_0-2p+pr}$.

Notice that $\bar{\theta }^I > K$ for any $p > 0$ and $r \le 1$. Additionally, $\bar{\theta }^I$ monotonically increases in K but monotonically decreases in r. It is also straightforward to check that if $\theta < \bar{\theta }^I$, then $\frac{\theta (v_0-p)}{v_0} < K$. If $\theta \ge \bar{\theta }^I$, then $\frac{\theta (v_0-p(1-r))}{v_0} > K$.

We will show for each of the two intervals of $\theta$, corresponding to part (i) and (ii) of the proposition, that $q^{I*}(\theta )$ defined above gives the maximum utility to the consumer.

• (i) For $\theta < \bar{\theta }^I$, $q^{I*}(\theta ) = \frac{\theta (v_0-p)}{v_0}$.

  Since $\bar{\theta }^I > K$, we will consider two subintervals: (ia) $\theta < K$ and (ib) $K \le \theta < \bar{\theta }^I$.

  • (ia) For $\theta < K$, the consumer’s utility from purchasing q is given by

    $$U^{I} (\theta ,q,D^{I} ) = \left\{ {\begin{array}{*{20}l} {v_{0} q\left( {1 - \frac{q}{{2\theta }}} \right) - pq} & {{\text{if}}\;0 \le q{{ < }}\theta } \\ {\frac{{v_{0} \theta }}{2} - pq} & {{\text{if}}\;\theta \le q{{ < }}K} \\ {\frac{{v_{0} \theta }}{2} - pK - p(1 - r)(q - K){\text{ }}} & {{\text{if }}\,q \ge K} \\ \end{array} } \right.$$

    Note from the second and third expressions that the consumer’s utility monotonically decreases in q for all $\theta \le q < K$ and $q \ge K$. Notice also that under an incremental discount, the utility function is continuous in q. Hence, we have that any $q > \theta$ cannot be optimal. We only need to consider $0 \le q \le \theta$ as candidates for the optimal purchase quantity. The utility function for $0 \le q \le \theta$ is $U^I(q) = v_0q(1 - \frac{q}{2\theta }) - pq$, which has $\frac{dU^I(q)}{dq} = \frac{v_0(\theta -q)-\theta p}{\theta }$ and $\frac{d^2U^I(q)}{dq^2} = \frac{-v_0}{\theta } < 0$. Thus, it is concave in q with the solution to the first-order condition given by $q_0(\theta ) := \frac{\theta (v_0-p)}{v_0}$. Notice that $0< q_0(\theta ) < \theta$ since $0< p < v_0$. Hence, $q_0(\theta )$ is the interior maximizer, resulting in the utility of $\frac{\theta (v_0-p)^2}{2v_0}$.

  • (ib) For $K \le \theta < \bar{\theta }^I$, the consumer’s utility from purchasing q is given by

    $$\begin{aligned} U^I(\theta , q, D^I) = \left\{ \begin{array}{ll} v_0q\left(1 - \frac{q}{2\theta }\right) - pq &{} {\text {if }}\;0\,\le\, q\, <\, K\\ v_0q\left(1 - \frac{q}{2\theta }\right) - pK - p(1-r)(q-K) &{} {\text {if }}\; K\, \le\, q\, \le \theta \\ \frac{v_0\theta }{2} - pK - p(1-r)(q-K) &{} {\text {if }}\; q\, >\, \theta \\ \end{array} \right. \end{aligned}$$

    Notice from the third expression that the consumer’s utility monotonically decreases in q for all $q > \theta$. Thus, $q > \theta$ cannot be optimal. We only need to consider $0 \le q < K$ and $K \le q \le \theta$ for the optimal purchase quantity. For $0 \le q < K$, we have seen from part (ia) that this utility is concave with the solution to the first-order condition given by $q_0(\theta ) = \frac{\theta (v_0-p)}{v_0}$. Note that $q_0(\theta ) < K$ since $\theta < \bar{\theta }^I = \frac{2v_0K}{2v_0-2p+pr}$. Hence, for $0 \le q < K$, $q_0(\theta )$ is the maximizer, resulting in the utility of $\frac{\theta (v_0-p)^2}{2v_0}$. For $K \le q \le \theta$, the utility function has $\frac{dU^I(q)}{dq} = v_0(1-\frac{q}{\theta }) - p(1-r)$ and $\frac{d^2U^I(q)}{dq^2} = \frac{-v_0}{\theta } < 0$. Thus, it is concave in q with the solution to the first-order condition given by $q_0^\prime (\theta ) := \frac{\theta (v_0-p(1-r))}{v_0}$. Observe that $q_0^\prime (\theta ) \le \theta$ for any $r \le 1$. However, $q_0^\prime (\theta ) \ge K$ if and only if $\theta \ge \frac{v_0K}{v_0-p(1-r)}$. Note that $\frac{v_0K}{v_0-p(1-r)} < \bar{\theta }^I$ for any $r > 0$. If $\theta < \frac{v_0K}{v_0-p(1-r)}$, then $q_0^\prime (\theta ) < K$, implying the utility monotonically decreases in q for $K \le q \le \theta$. Hence, $q_0(\theta ) = \frac{\theta (v_0-p)}{v_0}$ is the optimal purchase quantity. If $\frac{v_0K}{v_0-p(1-r)} \le \theta < \bar{\theta }^I$, then $q_0^\prime (\theta ) \ge K$ and $q_0^\prime (\theta )$ is the maximizer for $K \le q \le \theta$, resulting in the utility of $\frac{\theta (v_0-p(1-r))^2}{2v_0} - pKr$. However, $U^I(q_0^\prime (\theta )) < U^I(q_0(\theta ))$ since $\theta < \bar{\theta }^I = \frac{2v_0K}{2v_0-2p+pr}$. Hence, $q_0(\theta ) = \frac{\theta (v_0-p)}{v_0}$ is also optimal in this case.

• (ii) For $\theta \ge \bar{\theta }^I$, $q^{I*}(\theta ) = \frac{\theta (v_0-p(1-r))}{v_0}$. Since $\theta > K$, the consumer’s utility is the same as that given in part (ib). Hence, will consider only $0 \le q < K$ and $K \le q \le \theta$. For $0 \le q < K$, we have seen from part (ia) that $q_0(\theta ) = \frac{\theta (v_0-p)}{v_0}$ is the solution to the first-order condition. However, $q_0(\theta ) < K$ if and only if $\theta < \frac{v_0K}{v_0-p}$. If $\bar{\theta }^I \le \theta < \frac{v_0K}{v_0-p}$, then $q_0(\theta )$ is the interior maximizer, resulting in the utility of $\frac{\theta (v_0-p)^2}{2v_0}$. If $\theta \ge \frac{v_0K}{v_0-p}$, then $q_0(\theta ) \ge K$, implying that the optimal purchase quantity is in the interval of $K \le q \le \theta$. For $K \le q \le \theta$, we have seen from part (ib) that $q_0^\prime (\theta ) = \frac{\theta (v_0-p(1-r))}{v_0}$ is the solution to the first-order condition, and that $q_0^\prime (\theta ) \ge K$ if and only if $\theta \ge \frac{v_0K}{v_0-p(1-r)}$. Since $\theta \ge \bar{\theta }^I > \frac{v_0K}{v_0-p(1-r)}$, $q_0^\prime (\theta ) \ge K$ and $q_0^\prime (\theta )$ is the interior maximizer for $K \le q \le \theta$. It remains to compare the consumer’s utility at $q = q_0(\theta )$ and $q = q_0^\prime (\theta )$. With some algebra, one can check that $U^I(q_0^\prime (\theta )) \ge U^I(q_0(\theta ))$ since $\theta \ge \bar{\theta }^I$. Hence, $q_0^\prime (\theta )$ is the optimal purchase quantity for $\theta \ge \bar{\theta }^I$.

Proof of Corollary 1:

From Propositions 1 and 2, notice that $\bar{\theta }^I \ge \bar{\theta }^A$ for any $D^I = (r, K)$ and $D^A = (r, K)$ where $r \ge 0$. Employing the results from these two propositions, we have that for $\theta < \underline{\theta }^A$, $q^{A*}(\theta ) = q^{I*}(\theta ) = \frac{\theta (v_0-p)}{v_0}$. For $\underline{\theta }^A \le \theta < \bar{\theta }^A$, $q^{A*}(\theta ) = K > \frac{\theta (v_0-p)}{v_0} = q^{I*}(\theta )$. For $\bar{\theta }^A \le \theta \le \bar{\theta }^I$, $q^{A*}(\theta ) = \frac{\theta (v_0-p(1-r))}{v_0} \ge \frac{\theta (v_0-p)}{v_0} = q^{I*}(\theta )$. Lastly, for $\theta > \bar{\theta }^I$, $q^{A*}(\theta ) = \frac{\theta (v_0-p(1-r))}{v_0} = q^{I*}(\theta )$.

Proof of Proposition 3:

Since we consider the situation where the consumer purchases the same quantity $q \ge \max \{K^A, K^I\}$ under both the all-unit discount $D^A = (r^A, K^A)$ and the incremental discount $D^I = (r^I, K^I)$, the consumer receives the same valuation $V(\theta , q)$ from her purchase. Hence, to see which type of discounts is preferable, we only need to compare the total payment that the consumer has to make under each type of discounts. Under the all-unit discount, the total payment is $p(1-r^A)q$. Under the incremental discount, the total payment is $pK^I + p(1-r^I)(q-K^I)$. With algebra, one can easily check that the total payment under the all-unit discount is greater than that under the incremental discount, hence the incremental discount is preferable, if and only if $r^I \ge r^A(\frac{q}{q-K^I})$.

Proof of Lemma 1:

From the closed-form expressions of the optimal threshold purchase quantity $K^{t*}(r), t \in \{A, I\},$ given in the proof of Proposition 4, it follows immediately that both $K^{A*}(r)$ and $K^{I*}(r)$ monotonically increase in r.

Proof of Proposition 4:

We will show the results for the all-unit discount first, and then the incremental discount.

All-unit discount For convenience in analyzing the seller’s profit, we will define $K_u^A(r)$ and $K_o^A(r)$ as follows. Let $K_u^A(r)$ be such that $\underline{\theta }^A(D^A = (r, K_u^A(r)) = n$. Let $K_o^A(r)$ be such that $\bar{\theta }^A(D^A = (r, K_o^A(r)) = n$. From the expressions of $\underline{\theta }^A$ and $\bar{\theta }^A$ given in the proof of Proposition 1, we can derive $K_u^A(r)$ and $K_o^A(r)$ as:

$$K_{u}^{A} (r) = \left\{ {\begin{array}{*{20}l} {\frac{{n(v_{0} - p(1 - r) + \sqrt {pr(2v_{0} - 2p + pr)} )}}{{v_{0} }}} & {{\text{if}}\;r \le \frac{p}{{2v_{0} }}} \\ {\frac{{n(2v_{0} - p)}}{{2v_{0} (1 - r)}}} & {{\text{if}}\;r > \frac{p}{{2v_{0} }}} \\ \end{array} } \right.$$

and $K_o^A(r) = \frac{n(v_0-p(1-r))}{v_0}.$

Since $\underline{\theta }^A \le \bar{\theta }^A$ and both $\underline{\theta }^A$ and $\bar{\theta }^A$ are monotonically increasing in K, it follows that $K_o^A(r) \le K_u^A(r)$. With these definitions, we have that for an all-unit discount $D^A = (r, K)$, if $K < K_o^A(r)$, then $\bar{\theta }^A < n$, implying all three terms in the seller’s profit given by equation (1) exist. If $K_o^A(r) \le K < K_u^A(r)$, then $\underline{\theta }^A < n \le \bar{\theta }^A$, implying the third term in (1) does not exist. If $K \ge K_u^A(r)$, then $\underline{\theta }^A \ge n$, implying only the first term in (1) exists, resulting in the same seller’s profit as the no-discount profit. Hence, to show that for a given r, the all-unit discount with the optimal threshold purchase quantity $K^{A*}(r)$ is more profitable than no discount, it suffices to show that $K^{A*}(r)$ is strictly less than $K_u^A(r)$. Likewise, to show that no discount is optimal, it suffices to show that $K^{A*}(r) = K_u^A(r)$. Since any $K \ge K_u^A(r)$ gives the same profit equal to the no-discount profit, we will consider only $0 \le K \le K_u^A(r)$ when characterizing $K^{A*}(r)$.

To show part i) of the proposition for the all-unit discount, we will show that for a given discount depth r under an all-unit discount, the optimal threshold purchase quantity is given by

$$K^{{A*}} (r) = \left\{ {\begin{array}{*{20}c} {K_{f}^{A} (r)} & {{\text{if }}\,r < \bar{r}^{A} } \\ {K_{u}^{A} (r)} & {{\text{if}}\;r \ge \bar{r}^{A} } \\ \end{array} } \right.$$

where

$$\bar{r}^{A} = \left\{ {\begin{array}{*{20}c} {\frac{{2(p - c)^{2} }}{{p(v_{0} + p - 2c)}}} & {{\text{if }}\,p \le 2c} \\ {\frac{{p(p - c)}}{{p^{2} + 2c(v_{0} - p)}}} & {{\text{if}}\;p\, > \,2c} \\ \end{array} } \right.$$

and

$$dK_{f}^{A} (r) = \left\{ {\begin{array}{*{20}c} {\frac{{n(p(1 - r) - c)(v_{0} - p)^{3} }}{{v_{0} [(p - c)(v_{0} - p)^{2} - 2pr(v_{0} - c)(v_{0} - p(1 - r) - \sqrt {pr(2v_{0} - 2p + pr)} )]}}} & {{\text{if}}\;r \le \frac{p}{{2v_{0} }}} \\ {\frac{{n(p(1 - r) - c)(2v_{0} - p)^{2} }}{{4v_{0} (1 - r)(pv_{0} - cv_{0} + cpr - cv_{0} r - pv_{0} r)}}} & {{\text{if}}\;r > \frac{p}{{2v_{0} }}.} \\ \end{array} } \right.$$

To show this, we will consider the two cases of: (1) $p \le 2c$ and (2) $p > 2c$.

• (1) Suppose $p \le 2c$. Then, $\bar{r}^A = \frac{2(p-c)^2}{p(v_0+p-2c)} \le \frac{p}{2v_0}$ for any $c< p < v_0$. We will characterize $K^{A*}(r)$ for the three intervals of: 1a) $r < \bar{r}^A$, 1b) $\bar{r}^A \le r \le \frac{p}{2v_0}$, and (1c) $r > \frac{p}{2v_0}$.

  • 1a) Suppose $r < \bar{r}^A$. We will show that $K^{A*}(r) = K_f^A(r)$, where $K_o^A(r)< K_f^A(r) < K_u^A(r)$.

    For $0 \le K \le K_o^A(r)$, we can derive from equation (1) with $\theta \sim U[0, n]$ using the expressions of $\underline{\theta }^A$ and $\bar{\theta }^A$ from the proof of Proposition 1 that

    $$\begin{aligned} \frac{d\Pi ^A(K)}{dK} = \frac{prv_0K(v_0-c)(v_0-p(1-r)-\sqrt{pr(2v_0-2p+pr)})^2}{n(v_0-p)^3(v_0-p(1-r))}. \end{aligned}$$

    (3)

    Notice that $\frac{d\Pi ^A(K)}{dK} > 0$ for all $0< r \le \frac{p}{2v_0}, c< p < v_0,$ and $K \ge 0$. This implies that the seller’s profit monotonically increases in K for $0 \le K \le K_o^A(r)$. Hence, $K^{A*}(r) \ge K_o^A(r)$. For $K_o^A(r) \le K \le K_u^A(r)$, we derive

    $$\begin{aligned} \frac{d^2\Pi ^A(K)}{dK^2}= & {} \frac{v_0J(r)}{n(v_0-p)^3} \\ {\text {where }}\, J(r):= & {} -(p-c)(v_0-p)^2 + 2pr(v_0-c)(v_0-p(1-r) - \sqrt{pr(2v_0-2p+pr)}) \\ \frac{dJ(r)}{dr}= & {} \frac{2p(v_0-c)\left(\sqrt{pr(2v_0-2p+pr)} - pr\right)\left(v_0-p(1-r)-\sqrt{pr(2v_0-2p+pr)}\right)}{\sqrt{pr(2v_0-2p+pr)}} \\ \left. \frac{d\Pi ^A(K)}{dK}\right| _{K_u^A(r)}= & {} \frac{(v_0-c)pr - (p-c)\sqrt{pr(2v_0-2p+pr)}}{v_0-p} \end{aligned}$$

    (4)

    First, we will show that the seller’s profit is concave in K by showing $J(r) < 0$ for all $r < \bar{r}^A$. Notice that $\frac{dJ(r)}{dr} > 0$ for all $r \in (0, 1]$. With some algebra, one can check that $\left. \frac{dJ(r)}{dr}\right| _{\bar{r}^A} < 0$, implying the concavity of the seller’s profit in K. The solution to the first-order condition of the seller’s profit with respect to K is

    $$K_f^A(r) := \frac{n(p(1-r)-c)(v_0-p)^3}{v_0\left[ (p-c)(v_0-p)^2-2pr(v_0-c)(v_0-p(1-r)-\sqrt{pr(2v_0-2p+pr)})\right] }.$$

    Note that $\left. \frac{d\Pi ^A(K)}{dK}\right| _{K_u^A(r)} < 0$ for all $r < \bar{r}^A$. Together what we have shown earlier that the seller’s profit is increasing in K at $K = K_o^A(r)$, it follows that $K_o^A(r)< K_f^A(r) < K_u^A(r)$ and $K_f^A(r)$ is the interior maximizer of the seller’s profit function.

  • (1b) Suppose $\bar{r}^A \le r \le \frac{p}{2v_0}$. We will show that $K^{A*}(r) = K_u^A(r).$

    Notice that we can employ equations (3) and (4) from part 1a) as those equations are the same for all $r \le \frac{p}{2v_0}$. For $0 \le K \le K_o^A(r)$, we have seen from (3) that the seller’s profit monotonically increases in K for $0 \le K \le K_o^A(r)$ when $r \le \frac{p}{2v_0}$. Hence, $K^{A*}(r) \ge K_o^A(r)$.

    For $K_o^A(r) \le K \le K_u^A(r)$, notice from (4) that $\frac{d^2\Pi ^A(K)}{dK^2}$ is independent of K, implying that the seller’s profit function is either convex or concave in K and can switch the sign of its slope at most once. Note also that $\left. \frac{d\Pi ^A(K)}{dK}\right| _{K_u^A(r)} \ge 0$ for $r \ge \bar{r}^A$. Together with what we learned earlier that the seller’s profit increases in K at $K = K_o^A(r)$, we know that the seller’s profit must monotonically increase in K for $K_o^A(r) \le K \le K_u^A(r)$ as well. Hence, $K_u^A(r)$ is the boundary maximizer of the seller’s profit.

  • (1c) Suppose $r > \frac{p}{2v_0}$. We will show that $K^{A*}(r) = K_u^A(r).$

    First, note that for $\frac{p-c}{p} < r \le 1$, the seller’s profit from selling to consumers who purchase at least K is negative. Hence, it is straightforward to see that $K^{A*}(r) = K_u^A(r)$, so that the seller can get the no-discount profit.

    It remains to show $K^{A*}(r) = K_u^A(r)$ for $\frac{p}{2v_0} < r \le \frac{p-c}{p}$. For $0 \le K \le K_o^A(r)$, we can derive

    $$\frac{d\Pi ^A(K)}{dK} = \frac{v_0K\left[ p^2(p(1-r)-c) + 4cp^2r(1-r)^2 + 4v_0r(p(1-r)-cr)(v_0-2p+pr)\right] }{n(2v_0-p)^2(v_0-p(1-r))},$$

    which is positive for $\frac{p}{2v_0} < r \le \frac{p-c}{p}$. Hence, $K^{A*}(r) \ge K_o^A(r).$

    For $K_o^A(r) \le K \le K_u^A(r)$, we derive

    $$\begin{aligned} \frac{d^2\Pi ^A(K)}{dK^2}= & {} \frac{4v_0(1-r)(-pv_0+cv_0-cpr+cv_0r+pv_0r)}{n(2v_0-p)^2} \\ \left. \frac{d\Pi ^A(K)}{dK}\right| _{K_u^A(r)}= & {} \frac{2cr(v_0-p)-p(p(1-r)-c)}{2v_0-p}. \end{aligned}$$

    Notice that $\frac{d^2\Pi ^A(K)}{dK^2}$ is independent of K, implying the seller’s profit can switch the sign of its slope at most once. One can also show that $\left. \frac{d\Pi ^A(K)}{dK}\right| _{K_u^A(r)} > 0$ for $r > \frac{p}{2v_0}$. Together with $\left. \frac{d\Pi ^A(K)}{dK}\right| _{K_o^A(r)} > 0$, it follows that the seller’s profit monotonically increases in K for $K_o^A(r) \le K \le K_u^A(r)$ as well. Hence, $K_u^A(r)$ is the boundary maximizer of the seller’s profit.

• (2) Suppose $p > 2c$. Then, $\bar{r}^A = \frac{n(2v_0-p)}{2v_0(1-r)} > \frac{p}{2v_0}$ for any $2c< p < v_0$. We can show in a similar manner as in part (1) that: (2a) For $r \le \frac{p}{2v_0},$

  $$K^{A*}(r) = K_f^A(r) = \frac{n(p(1-r)-c)(v_0-p)^3}{v_0[(p-c)(v_0-p)^2-2pr(v_0-c)(v_0-p(1-r)-\sqrt{pr(2v_0-2p+pr)})]};$$

  (2b) For $\frac{p}{2v_0}< r < \bar{r}^A,$

  $$K^{A*}(r) = K_f^A(r) = \frac{n(p(1-r)-c)(2v_0-p)^2}{4v_0(1-r)(pv_0-cv_0+cpr-cv_0r-pv_0r)};$$

  and (2c) For $r \ge \bar{r}^A$, $K^{A*}(r) = K_u^A(r)$.

Incremental discount: For convenience in analyzing the seller’s profit, let $K_o^I(r)$ be such that $\bar{\theta }^I(D^I = (r, K_o^I(r))) = n$. From the expression of $\bar{\theta }^I$ given in the proof of Proposition 2, we can derive $K_o^I(r) := \frac{n(2v_0-2p+pr)}{2v_0}$.

With this definition, we have that for an incremental discount $D^I = (r, K)$, if $K < K_o^I(r)$, then $\bar{\theta }^I < n$, implying both terms in the seller’s profit given by equation (2) exist. Otherwise, if $K \ge K_o^I(r)$, then $\bar{\theta }^I \ge n$, implying only the first term in (2) exists, resulting in the same seller’s profit as the no-discount profit. Hence, to show that for a given r, the incremental discount with the optimal threshold purchase quantity $K^{I*}(r)$ is more profitable than no discount, it suffices to show that $K^{I*}(r)$ is strictly less than $K_o^I(r)$. Likewise, to show that no discount is optimal, it suffices to show that $K^{I*}(r) = K_o^I(r)$. Since any $K \ge K_o^I(r)$ gives the same profit equal to the no-discount profit, we will consider only $0 \le K \le K_o^I(r)$ when characterizing $K^{I*}(r)$.

To show part (i) of the proposition for the incremental discount, we will show that for a given discount depth r under an incremental discount, the optimal threshold purchase quantity is given by

$$K^{{I*}} (r) = \left\{ {\begin{array}{*{20}c} {\frac{{n(2v_{0} - 2p + pr)^{2} }}{{4v_{0} (v_{0} - c)}}} & {{\text{if }}\,r < \bar{r}^{I} } \\ {K_{o}^{I} (r)} & {{\text{if}}\;r \ge \bar{r}^{I} } \\ \end{array} } \right.$$

where $\bar{r}^I = \frac{2(p-c)}{p}$.

From the seller’s profit function given by (2) with $\theta \sim U[0, n]$, we derive

$$\begin{aligned} \frac{d^2\Pi ^I(K)}{dK^2}= & {} \frac{-4v_0pr(v_0-c)}{n(2v_0-2p+pr)^2}\\ \left. \frac{d\Pi ^I(K)}{dK}\right| _{K = 0}= & {} pr \\ \left. \frac{d\Pi ^I(K)}{dK}\right| _{K_o^I(r)}= & {} \frac{pr(2c-2p+pr)}{2v_0-2p+pr} \end{aligned}$$

Notice that the seller’s profit is strictly concave in K since $\frac{d^2\Pi ^I(K)}{dK^2} < 0$ for all $r > 0$ and $v_0 > c$. Note also that $\left. \frac{d\Pi ^I(K)}{dK}\right| _{K = 0} > 0$ for all $r > 0$, and $\left. \frac{d\Pi ^I(K)}{dK}\right| _{K_o^I(r)} < 0$ if and only if $r < \bar{r}^I$. This implies that the solution to the first-order condition of the seller’s profit with respect to K, given by $\frac{n(2v_0-2p+pr)^2}{4v_0(v_0-c)}$, is the interior maximizer if $r < \bar{r}^I$. Otherwise, if $r \ge \bar{r}^I$, the seller’s profit monotonically increases in K, implying $K_o^I(r)$ is the boundary maximizer.

Lastly, it is straightforward to see that part (ii), (iii), and (iv) of the proposition follow directly from the expressions of $\bar{r}^A$ and $\bar{r}^I$ given above.

Proof of Proposition 5:

From the proof of Proposition 4, we have that $K_o^I(r) \le K_o^A(r) \le K_u^A(r)$ for any $0 \le r \le 1$. To show the results in this proposition, we will compare the seller’s profit for a given $r \le \bar{r}^A$ and a given K under the all-unit and incremental schemes for the four regions of K: (1) $0 \le K < K_o^I(r)$, (2) $K_o^I(r) \le K < K_o^A(r)$, (3) $K_o^A(r) \le K < K_u^A(r)$, and (4) $K \ge K_u^A(r)$. Let $\Delta (r, K) := \Pi ^A(D^A = (r, K)) - \Pi ^I(D^I = (r, K))$ denote the profit difference between the seller’s profit from offering the all-unit discount and the incremental discount.

• (1) Suppose $0 \le K < K_o^I(r)$. Recall from the proof of Proposition 4 that $\frac{d^2\Pi ^I(K)}{dK^2} < 0$ for all $r > 0$ and $c< p < v_0$, and we can derive $\frac{d^2\Pi ^A(K)}{dK^2} > 0$ for $r < \bar{r}^A$. Hence, it follows that $\frac{d^2\Delta (r, K)}{dK^2} > 0$. Additionally, we can derive

  $$\begin{aligned} \left. \frac{d\Delta (r, K)}{dK}\right| _{K=0}= & {} -pr \\ \left. \frac{d\Delta (r, K)}{dK}\right| _{K=K_o^I(r)}= & {} \frac{d\Pi ^A(K)}{dK} \\ \left. \Delta (r, K)\right| _{K=0}= & {} 0. \end{aligned}$$

  This shows that $\Delta (r, K)$ is convex in K in this region; it decreases from zero at $K = 0$ and increases at $K = K_o^I(r)$. This implies that $\Delta (r, K)$ is negative, equivalently $\Pi ^A(D^A) < \Pi ^I(D^I)$, for at least some K close to 0.

• (2) Suppose $K_o^I(r) \le K < K_o^A(r)$. We learn from Proposition 4 that in this region, the seller’s profit under the incremental discount is the no-discount profit since $K \ge K_o^I(r)$. We have that $\frac{d\Delta (r, K)}{dK} = \frac{d\Pi ^A(D^A)}{dK} > 0$ and $\frac{d^2\Delta (r, K)}{dK^2} = \frac{d^2\Pi ^A(D^A)}{dK^2}> 0$ for all $r < \bar{r}^A$. This implies that $\Delta (r, K)$ is convexly increasing in K in this region.

• (3) Suppose $K_o^A(r) \le K < K_u^A(r)$. In this region, the seller’s profit under the incremental discount is the no-discount profit since $K \ge K_o^I(r)$. From Proposition 4, the seller’s profit under the all-unit discount is concave in K and it has an interior solution $K_f^A(r)$ such that $K_o^A(r)< K_f^A(r) < K_u^A(r)$ since $r < \bar{r}^A$. This implies that $\Delta (r, K)$ is concave in K and it takes some positive values for some K in this region since $\Pi ^A(D^A = (r, K_f^A(r)) > \Pi ^N = \Pi ^I(D^I)$.

• (4) Suppose $K \ge K_u^A(r)$. In this region, the seller’s profit under the incremental discount and that under the all-unit discount are the same, which is equal to the no-discount profit, since $K \ge K_u^A(r) \ge K_o^I(r)$.

From part (1), (2), and (3), the existence of $\underline{K}(r)$ as described in the proposition is guaranteed. In fact, $\underline{K}(r)$ is the smallest K where $\Delta (r, K)$ first takes a positive value, which could occur in either region (1), (2), or (3). Part (4) shows the existence of $\bar{K}(r)$, which is in fact $K_u^A(r)$.

Proof of Theorem 1:

To show this theorem, we will first show that (1) the optimal all-unit discount is $D^{A*} = (r^{A*} = \frac{(p-c)^2}{2p(v_0-c)}, K^{A*} = \frac{n(2v_0-p-c)}{2v_0})$ and (2) the optimal incremental discount is $D^{I*} = \left(r^{I*} = \frac{2(p-c)}{3p}, K^{I*} = \frac{n(3v_0-2p-c)^2}{9v_0(v_0-c)}\right)$.

• (1) Optimal all-unit discount From Proposition 4, we can infer that the optimal all-unit discount $D^{A*} = (r^{A*}, K^{A*})$ must be such that $r^{A*} < \bar{r}^A$ and $K^{A*} = K_f^A(r^{A*})$. Since we already know $K^{A*}$ as a function of $r^{A*}$, it remains to determine only $r^{A*}$. For this, we will consider two cases: $r^{A*} \le \min \{\frac{p}{2v_0}, \bar{r}^A\}$ and $\frac{p}{2v_0}< r^{A*} < \bar{r}^A$.

  For $r \le \min \{\frac{p}{2v_0}, \bar{r}^A\}$, the seller’s profit with the optimal K, denoted by $\Pi ^A(r, K_f^A(r))$, has a unique feasible solution to the first-order condition with respect to r, given by $r_f^A := \frac{(p-c)^2}{2p(v_0-c)} < \frac{p}{2v_0}$, resulting in $K_f^A(r_f^A) = \frac{n(2v_0-p-c)}{2v_0})$. One can check that $(r_f^A, K_f^A(r_f^A))$ satisfies $\underline{\theta }(D^A = (r_f^A, K_f^A(r_f^A)))< n < \bar{\theta }(D^A = (r_f^A, K_f^A(r_f^A)))$, and results in $\left. \frac{d^2\Pi ^A(r, K_f^A(r))}{dr^2}\right| _{r = r_f^A} = \frac{-np^2(v_0-c)^2}{v_0(p-c)(2v_0-p-c)} < 0$ for any $c< p < v_0$. By the Envelope Theorem, this implies that $(r_f^A, K_f^A(r_f^A))$ is the unique maximizer of the seller’s profit under the all-unit discount for $r \le \min \{\frac{p}{2v_0}, \bar{r}^A\}.$

  For $\frac{p}{2v_0}< r^{A*} < \bar{r}^A$, the seller’s profit with the optimal K, denoted by $\Pi ^A(r, K_f^A(r))$, has a unique feasible solution to the first-order condition with respect to r, given by $r_f^A := \frac{p(p-c)}{p^2+2c(v_0-p)}$, resulting in $K_f^A(r_f^A) = \frac{n(p^2+2c(v_0-p))}{2cv_0}$. However, notice that $r_f^A = \bar{r}^A$, implying the seller’s profit cannot be greater than the no-discount profit, as shown in Proposition 4.

  Hence, the optimal all-unit discount is $D^{A*} = (r^{A*} = \frac{(p-c)^2}{2p(v_0-c)}, K^{A*} = \frac{n(2v_0-p-c)}{2v_0})$.

• (2) Optimal incremental discount From Proposition 4, we can infer that the optimal incremental discount $D^{I*} = (r^{I*}, K^{I*})$ must be such that $r^{I*} < \bar{r}^I$ and $K^{I*} = K_f^I(r^{I*})$. Since we already know $K^{I*}$ as a function of $r^{I*}$, it remains to determine only $r^{I*}$. The seller’s profit with the optimal K, denoted by $\Pi ^I(r, K_f^I(r))$, has a unique feasible solution to the first-order condition with respect to r, given by $r_f^I := \frac{2(p-c)}{3p} < \bar{r}^I = \frac{2(p-c)}{p}$, resulting in $K_f^I(r_f^I) = \frac{n(3v_0-2p-c)^2}{9v_0(v_0-c)}$. This solution satisfies $\left. \frac{d^2\Pi ^I\left( r, K_f^I(r)\right) }{dr^2}\right| _{r = r_f^I} = \frac{-np^2(p-c)}{2v_0(v_0-c)} < 0$ for all $c< p < v_0$. Thus, the optimal incremental discount is $D^{I*} = \left( r^{I*} = \frac{2(p-c)}{3p}, K^{I*} = \frac{n(3v_0-2p-c)^2}{9v_0(v_0-c)}\right)$.

From part (1), we can derive the seller’s profit under the optimal all-unit discount as $\Pi ^{A*} := \frac{n(p-c)(2v_0-p-c)^2}{8v_0(v_0-c)}$. From part (2), we can derive the seller’s profit under the optimal incremental discount as $\Pi ^{I*} := \frac{n(p-c)\left( 27v_0(v_0-p-c)+8p^2+11cp+8c^2\right) }{54v_0c}.$ Thus, we obtain $\Pi ^{I*} - \Pi ^{A*} = \frac{5n(p-c)^3}{216v_0(v_0-c)}$, which is positive for any $c< p < v_0$ and $n > 0.$