A theory of search with deadlines and uncertain recall

Abstract

We ask how the ability to recall past prices affects the dynamics of search and price formation. In the model, buyers have limited time to purchase a good and face uncertainty regarding the availability of past price quotes in the future. Sellers cannot observe a potential buyer’s remaining time until deadline nor her quote history, and hence post prices that weigh the probability of sale versus the profit once sold. We find that, in contrast to conventional wisdom, reducing the consumer’s recall ability may actually improve his expected utility because it lowers the average expected price in the market and reduces the duration of search.

Appendix A: Proofs

1.1 A.1 Proof of Lemma 1

Proof

Let $\gamma \in (0,1)$ and let $p$ satisfy $0 < F(p) < 1$.

Let $M(p,q) \equiv \max \left\{ x - \min \{p,q\} , \; d + (1 - \gamma ) W_0( \min \{p,q\}) + \gamma V_0 \right\}$, which is the integrand of Eq. 3. Thus $W_0(p) = \int _{-\infty }^{+\infty } M(p,q) dF(q)$.

If $q < p$, then $\frac{\partial M}{\partial p} = 0$, since the inferior quote $p$ is discarded.

Suppose instead that $q \ge p$. For a given $p$, the choice to accept or defer is the same for any $q \ge p$, as $q$ is discarded. If the buyer would accept $p$, then $\frac{\partial M}{\partial p} = -1$. Integrated over all $q$, this yields

$$\begin{aligned} W_0^{\prime }(p) = \int _{-\infty }^{+\infty } \frac{\partial M(p,q)}{\partial p} dF(q) = - (1 - F(p)) > -1. \end{aligned}$$

If instead the buyer would defer $p$, then $\frac{\partial M}{\partial p} = (1-\gamma ) W_0^{\prime }(p)$. Integrated over all $q$, this yields

$$\begin{aligned} W_0^{\prime }(p) = \int _{-\infty }^{+\infty } \frac{\partial M(p,q)}{\partial p} dF(q) = (1 - F(p)) (1-\gamma ) W_0^{\prime }(p). \end{aligned}$$

The first two terms on the r.h.s. are strictly between 0 and 1, so this only holds if $W_0^{\prime }(p) = 0$. Thus, in either case, $W_0^{\prime }(p) > -1$.

In light of this, consider the choice to accept versus defer an offer. The utility from accepting, $x - p$, has a derivative w.r.t. $p$ of $-1$. The utility from deferral, $d + (1 - \gamma ) W_0(p) + \gamma V_0$, has a derivative $(1 - \gamma ) W_0^{\prime }(p) > -1$. Therefore, if $p > R_0$ then $x - p < d + (1 - \gamma ) W_0(p) + \gamma V_0$ , and vice versa.

1.2 A.2 Proof of Lemma 2

Proof

Let $I(z,u) = x - z - u - (1 - \gamma ) W_0(z) - \gamma V_0$. Note that $\frac{\partial I}{\partial u} = -1$ and $\frac{\partial I}{\partial z} = -1 - (1 - \gamma ) W_0^{\prime }(z) < 0$. By implicit differentiation, $\frac{\partial z}{\partial u} = -\frac{-1}{-1 - (1 - \gamma ) W_0^{\prime }(z)} < 0$. Since $R_0$ is defined by $I(R_0,d) = 0$, $R_1$ by $I(R_1,b) = 0$, and $b > d$, this implies that $R_1 < R_0$.

1.3 A.3 Proof of Lemma 3

Proof

(of Claim 3) Suppose an atom of weight $a > 0$ existed at price $p \in (R_1,R_0]$. A seller offering $p$ would find that a positive measure of their offers is tied with another offer of $p$. In particular, this occurs with probability $(1 - \gamma ) a > 0$, since $a$ is the probability that another offer is made at price $p$, and $1 - \gamma$ is the probability that the buyer retains the first offer made.

If this happens, the buyer randomizes with equal weight between the two offers. If the seller had offered $p - \epsilon$, where $\epsilon > 0$ is arbitrarily small, he would have garnered $\frac{(1 - \gamma ) a}{2}$ more sales among early buyers who defer, without losing any of the other $S(p)$ sales from offering $p$. This discrete jump in expected sales (with a negligible reduction in profit per sale) means offering $p$ is strictly dominated. Thus, no firm would offer $p$ and hence it cannot be part of the support.

While we have assumed equal probability in randomizing over the two offers, other tiebreaking rules have similar effect, as in Bertrand competition.

Proof

(of Claim 4) Suppose $\hat{p} = \sup P$ and $R_1 < \hat{p} < R_0$. We will show that any $\tilde{p} \in (\hat{p},R_0]$ will yield strictly greater profits.

First, consider buyers in search state 0 with no other offer; from Lemma 1, they will accept not only $\hat{p}$ but also any $\tilde{p} \in (\hat{p},R_0]$.

Next, consider buyers in search state 0 who have retained some prior offer. Since $\hat{p} = \sup P$ and there is no atom at $\hat{p}$, their prior offer is almost surely strictly better than $\hat{p}$ or $\tilde{p}$. Thus, either new offer is almost surely rejected.

Finally, note that those who are offered $\hat{p}$ in search state 1 will always defer it and almost surely obtain a better offer in search state 0. The same would occur if had $\tilde{p}$ been offered initially.

Thus, both offers generate sales from the same fraction of the population, namely those in state 0 with no offers. But $\tilde{p}$ has a larger markup than $\hat{p}$, so offering the former generates strictly greater expected profit. Since $\tilde{p} \notin P$, this contradicts $F(p)$ as being an equilibrium price distribution.

1.4 A.4 Proof of Proposition 1

Proof

Using the properties of $F$ in Lemma 3 and the reservation price properties of $R_0$ and $R_1$, we can restate the buyer’s Bellman equations as follows. Delayed buyers with no retained offers will accept any price at or below $R_0$, which includes the full support of $F$. Therefore, his expected utility is

$$\begin{aligned} V_0 = a (x - R_1) + \int _{\underline{p}}^{R_0} (x - q) F^{\prime }(q) \mathrm{d}q. \end{aligned}$$

(25)

A delayed buyer who has retained an offer $p$ will also make a purchase in the current iteration, accepting the lower price of either $p$ or his new offer $q$:

$$\begin{aligned} W_0(p) = (1 - F(p)) (x - p) + a (x - R_1) + \int _{\underline{p}}^{p} (x - q) F^{\prime }(q) \mathrm{d}q. \end{aligned}$$

(26)

An early buyer, however, will only accept $R_1$ if it is drawn; all other offers are deferred

$$\begin{aligned} V_1 = a (x - R_1) + \int _{\underline{p}}^{R_0} \left( b + (1 - \gamma ) W_0(q) + \gamma V_0 \right) dF(q). \end{aligned}$$

(27)

Next, note that $W_0(R_0) = V_0$. By substituting this into Eq. 2, we get

$$\begin{aligned} R_0 = x - d - V_0. \end{aligned}$$

(28)

Similarly, $W_0(R_1) = x - R_1$. Substituting this into Eq. 5, we obtain

$$\begin{aligned} R_1 = x - \frac{b}{\gamma } + V_0. \end{aligned}$$

(29)

Regarding the steady state population, note that Eqs. 7 through 10 are linear in terms of $H,\,h_1,\, h_0,\, \text{ and} \ g_0(p)$, and are easily solved.

To obtain the price distribution, we turn to the equal profit conditions. Offering $R_1$ results in profit $\frac{(2-a)\delta }{1- \gamma } (R_1 - c)$. After substituting for the steady state populations, for prices between $\underline{p}$ and $R_0$, Eq. 12 profit becomes

$$\begin{aligned} \pi (p) = \left( 2(1 - F(p)) + \frac{ (1 - a) \gamma }{1 - \gamma } \right)\delta (p-c). \end{aligned}$$

(30)

We then solve for $F(p)$ in this price range by equating $\pi (p) = \pi (R_0)$. Note that because this portion of the distribution is atomless and $R_0$ is the highest offered price, $F(R_0) = 1$. This yields

$$\begin{aligned} F(p) = 1 - \frac{ (1 -a) (R_0 - p) \gamma }{2 (p-c)(1 - \gamma )}, \end{aligned}$$

(31)

and taking the first derivative w.r.t. $p$ provides

$$\begin{aligned} F^{\prime }(p) = \frac{(1 - a)(R_0 - c)\gamma }{2 (p-c)^2(1 - \gamma )}. \end{aligned}$$

(32)

In addition, we must determine the value of $\underline{p}$. This is pinned down by the fact that $F(\underline{p}) = a$, since no prices between $R_1$ and $\underline{p}$ are offered. This provides the solution:

$$\begin{aligned} \underline{p} = \frac{(1- \gamma ) 2 c + \gamma R_0}{2 - \gamma }. \end{aligned}$$

(33)

With the price distribution pinned down, we can now compute expected prices in Eq. 25 to express $V_0$ in terms of $a$, $R_0$, and $R_1$. After evaluating the integral, we obtain

$$\begin{aligned} V_0 = x - c - (R_1 - c ) a - (R_0 - c) \frac{(1 - a) \gamma \ln \left( \frac{2-\gamma }{\gamma }\right)}{2(1 - \gamma )}. \end{aligned}$$

(34)

Notice that Eqs. 28, 29, and 34 provide three linear equations in terms of $R_0$, $R_1$, and $V_0$; of course, only the former two are of explicit interest. These jointly solve to obtain the proposed $R_0^*$ and $R_1^*$. After these are found, they can be written in terms of the function $\kappa (a)$, which is defined as the expected price in distribution $F(p)$, or equivalently, $\kappa (a) = x - V_0$.

Note that we obtain the proposed profit functions and price distribution by substituting $R_0^*$ and $R_1^*$ into Eqs. 30, 31, and 33. We note here that $\underline{p}^* < R_0^*$ as required:

$$\begin{aligned} c + \frac{\gamma }{2 - \gamma } (\kappa (a^*) - c - d) < \kappa (a^*) -d \quad \Longleftrightarrow \quad \frac{1 - \gamma }{2 - \gamma } (c + d - \kappa (a^*)) < 0. \end{aligned}$$

The fraction is clearly positive. If $\kappa (a^*) < c + d$, firms would earn negative profit, which violates the equilibrium conditions since charging $p = c$ earns 0 profit.

To compare $R_1^* < \underline{p}^*$, one must substitute for $\kappa (a^*)$, and the calculation depends on whether a late or a full equilibrium has occurred. If $a^* = 0$, this comparison yields

$$\begin{aligned} d \left( 1+\frac{4 (1-\gamma )^2}{(2-\gamma ) \left(2(1- \gamma ) +\gamma \ln \left(\frac{\gamma }{2-\gamma }\right)\right)} \right)< \frac{b}{\gamma }. \end{aligned}$$

The parenthetical term on the l.h.s. is always positive and is multiplied by $d < 0$, while the r.h.s. is always positive. Thus, the inequality must hold.

The comparison in the case of a full equilibrium requires more algebraic manipulation, but it always holds as well. These details are provided in Section 2.4.2 of the Technical Appendix.

If the respective profit condition holds, this solution is an equilibrium by construction. Existence is assured because $\varPi _1(a) - \varPi _0(a)$ is continuous in $a$. This falls to $-\infty$ as $a \rightarrow 1$ (which precludes $a = 1$ as an equilibrium outcome). Thus, if $\varPi _1(0) - \varPi _0(0) > 0$, there must exist an $a^* \in (0,1)$ such that $\varPi _1(a^*) = \varPi _0(a^*)$. Of course, if $\varPi _1(0) - \varPi _0(0) \le 0$, then a late equilibrium exists.

Finally, this proof by construction also demonstrates that no other equilibrium (beyond these two) can exist, as only the constructed solution will satisfy all equilibrium requirements.

1.5 A.5 Proof of Proposition 2

Proof

Suppose $\varPi _1(a^*) = \varPi _0(a^*)$ for some $a^* \in (0,1)$. Thus, Eq. 20 holds with equality.

We take the first derivative of $\varDelta (a) \equiv \varPi _1(a) - \varPi _0(a)$ w.r.t. $a$, evaluated at $a^*$. This later fact allows us to substitute for $d$ using Eq. 20. The details of this algebraic manipulation are included in Sect. 3 of the Technical Appendix; it results in

$$\begin{aligned}&\varDelta ^{\prime }(a^*) = \frac{2 \delta b}{\gamma (1 - a^*)^2} \\&\quad \cdot \left(2 (1-\gamma )(1-a^*) (3-a^* (1-\gamma )-\gamma )-\left( \gamma \left(1-a^* \right)^2 - (2-a^*)^2 \right) \ln \left( \frac{\gamma }{2-\gamma }\right) \right) \Bigg / \\&\quad \left( \left(2(1-\gamma )+\gamma \ln \left( \frac{\gamma }{2-\gamma }\right) \right) \left(-2 (1-\gamma )(1-a^*) - (2-a^*) \ln \left( \frac{\gamma }{2-\gamma }\right) \right)\right). \end{aligned}$$

The first line $\frac{2 \delta b}{\gamma (1 - a^*)^2}$ is always positive. The next term in the numerator is strictly increasing in $\gamma$. Its derivative w.r.t. $\gamma$ is

$$\begin{aligned}&- (1 - a^*)^2 \ln \left( \frac{\gamma }{2-\gamma }\right)\\&\quad +\frac{1}{\gamma (2 - \gamma )} \left(2 {a^*}^2 +2 \gamma a^* (1-a^*)(5-4 \gamma ) +4 \gamma ^2 (1-\gamma )^2 (1-a^*)^2 \right.\\&\quad +\left.4 \gamma ^3 (1-\gamma ) \left(1-2a^*+{a^*}^2\right) +(1-a^*) \left(8-18 \gamma +12 \gamma ^2\right) \right). \end{aligned}$$

Note that the logarithm will be negative since $\frac{\gamma }{2-\gamma } < 1$, so the first term is positive, as is the fractional term. Within the parentheses, one can immediately see that the first four terms are strictly positive for $\gamma \in (0,1)$. The last term is also strictly positive, since it reaches a minimum of 1 at $\gamma = 3/4$. Thus, this derivative is strictly positive for $\gamma \in (0,1)$. Thus, the numerator reaches its maximum value when $\gamma = 1$, where it evaluates to 0. Thus, the numerator is negative for any $\gamma < 1$ or $a^* \in (0,1)$.

In the denominator, the first term has a derivative w.r.t. $\gamma$ of $-\frac{2(1-\gamma )}{2 - \gamma } + \ln ( \frac{\gamma }{2-\gamma })$. The first term is obviously negative for $\gamma \in (0,1)$, as is the logarithmic term since $\frac{\gamma }{2-\gamma } < 1$. Thus, the first denominator term is strictly decreasing in $\gamma$, reaching its smallest value of 0 at $\gamma = 1$; hence, it is always positive.

The same applies to the second term in the denominator. Its derivative w.r.t. $\gamma$ is $- \frac{2-a (1-\gamma )^2-(2-\gamma ) \gamma }{(2-\gamma ) \gamma } < 0$. Note that the inequality holds even at $a = 1$ and is easier to satisfy for lower $a$. Thus, the second denominator term is decreasing in $\gamma$ and reaches its smallest value of 0 at $\gamma =1$. Thus, the term is always positive.

Thus, with a negative numerator and positive denominator, $\varDelta ^{\prime }(a^*) < 0$ for all $\gamma < 1$. $\varDelta (a)$ is a continuous function. Since it is decreasing in $a$ wherever it crosses $\varDelta (a) = 0$, it can cross at most once in $a \in [0,1]$.

Next, suppose there is an $a^* \in (0,1)$ such that $\varDelta (a^*) = 0$. Since this solution is unique and $\varDelta ^{\prime }(a^*) < 0$, then $\varDelta (0) > 0$. This precludes the existence of a late equilibrium. The reverse is also true: if $\varDelta (0) \le 0$, there could not exist another $a^* \in (0,1)$ such that $\varDelta (a^*) = 0$ without contradicting $\varDelta ^{\prime }(a^*) < 0$.

1.6 A.6 Proof of Proposition 3

Proof

We examine the buyer’s decisions in the unlimited-deferrals model, which can be described as a decision tree. Each iteration begins with a draw of a new quote, followed by the buyer deciding to either accept the current best offer or defer all offers. Stationarity requires that this decision depend only on the search state (0 or 1) and the set of $n \in \mathbb N _{++}$ available offers $\{ p_i \}_{i = 1}^n$ (which includes all retained offers and his latest draw). Deferral results in incurring the utility $b$ or $d$ and realizing which of the offers are retained for the next iteration.

We define the reservation prices $R_0$ and $R_1$ as in the single-deferral model and consider the same price distribution $F$. Thus, any offers must be at or below $R_0$. Buyers in state 1 still have no prior offers, so their strategy remains as before: accept $R_1$ and defer otherwise. We will show it is a best response for a buyer in search state 0 to accept their best available offer, regardless of the number or set of available offers. If this is the case, no buyer will ever defer a second quote; thus, the steady state distribution is unchanged, as are the expected profits from offering $R_1$ or $p \in [\underline{p}, R_0]$. Thus, the same equilibrium will exist.

Consider a buyer who has a set of offers $\{ p_i \}_{i = 1}^n$ at any iteration in the decision tree. Our proposed equilibrium directs him to accept the lowest offer among these prices; we ask if he would ever strictly prefer to make a one-shot deviation by deferring for one iteration and then accepting the best available offer.

Suppose that $p_i = R_0$ for all $i$. Immediate purchase provides utility $x - R_0$. If he defers, he draws and accepts price $q$ tomorrow, since $q \le R_0$. This holds even if all $n$ deferred offers are lost. This buyer is no better off than if he had only retained one offer $R_0$ or even none at all, which is to say, $W_0\left( \{ p_i \}_{i = 1}^n \right) = W_0\left(R_0 \right) = V_0$. The definition of $R_0$ in Eq. 2 is $x - R_0 = d + (1 - \gamma ) W_0\left(R_0 \right) + \gamma V_0$. Note that since $W_0\left(R_0 \right) = V_0$, the r.h.s. can be rewritten as:

$$\begin{aligned} d + (1 - \gamma ) W_0\left(R_0 \right) + \gamma V_0&= d + V_0 \\&= d + (1 - \gamma ^n) V_0 + \gamma ^n V_0 \\&= d + (1 - \gamma ^n) W_0\left(R_0 \right) + \gamma ^n V_0. \end{aligned}$$

Thus, $x - R_0 = d + (1 - \gamma ^n) W_0\left(R_0 \right) + \gamma ^n V_0$; the buyer with $n$ retained offers of $R_0$ is indifferent between acceptance and deferral. In particular, he cannot strictly prefer to defer his $n$ offers.

Suppose instead that $p_i = p < R_0$ for all $i$. Immediate purchase provides utility $x - p$. If he defers, with probability $\gamma ^n$, he will lose all $n$ of the deferred offers and will purchase using his new offer $q$; this gives expected utility $V_0$. With probability $1 -\gamma ^n$, he will retain one of the deferred offers. With probability $1 - F(p)$, the new draw is more expensive than $p$ and so he accepts a retained offer for utility $x - p$. Otherwise, he accepts $q$. In sum, his expected utility from deferral is

$$\begin{aligned}&d + \gamma ^n V_0 + (1 -\gamma ^n) \left( (1 - F(p)) (x - p) + a (x - R_1) + \int _{\underline{p}}^{p} (x - q) F^{\prime }(q) \mathrm{d}q \right) \\&\quad = d + \gamma ^n V_0 + (1 -\gamma ^n) W_0(p). \end{aligned}$$

Recall that $x - p = d + (1 - \gamma ^n) W_0(p) + \gamma ^n V_0$ when $p = R_0$. If we take the derivative of each side for any $p$, we get $-1$ on the l.h.s. and $(1 - \gamma ^n) W_0^{\prime }(p) > -1$ on the r.h.s., where the inequality holds due to Lemma 1. These derivatives indicate that as $p$ decreases from $R_0$, the immediate purchase utility on the left increases at a strictly faster rate than the deferral utility on the right, making the former strictly preferred. Indeed, since $(1 - \gamma ^n) W_0^{\prime }(p) > -1$ for all $p$ in the support of $F$, the difference between the two options only strengthens in favor of acceptance as $p$ continues to fall.

Finally, suppose that $\tilde{p}_1 = p < R_0$ and $\tilde{p}_i \in [p,R_0]$ for all other $n-1$ offers and compare this to the situation above where $p_i = p$ for all $i$. Note that either set of offers produces the same utility from immediate acceptance, since $\min p_i = \min \tilde{p}_i = p$. But the utility from deferral must be weakly worse under $\{ \tilde{p}_i \}_{i = 1}^n$ and strictly so if $\tilde{p}_i > p$ for some $i$. This is because with positive probability, the offer $p$ will be lost and the new offer $q$ will be worse than the other retained offers. Thus, purchase at price $p$ is still strictly preferred to deferral.

This demonstrates that buyers will never strictly prefer to make a one-shot deviation from the proposed strategy. Indeed, buyers are only indifferent if all available offers equal $R_0$, which occurs with measure zero and thus has no impact on the steady state distribution or firm pricing. For any other set of available prices, buyers cannot optimally defer one period and accept the next.

We next consider $k$-shot deviations. Our proposed strategy tells a buyer in search state 0 to always accept the best available offer. Thus, any finite-length deviation must take the form of deferring (perhaps contingent on available prices) for $k-1$ periods and then accepting the best price in the $k\text{ th}$ iteration. However, note that such a strategy includes a one-shot deviation of deferring in the $k-1\text{ th}$ iteration, which is not optimal. Thus, any finite deviation is not optimal as well.

Finally, the only infinite-length path occurs by repeatedly deferring; doing this would produce utility $-\infty$ since it involves repeatedly incurring the penalty $d$ without discounting. This strategy (or any of its subpaths) is clearly dominated by immediate acceptance of any price in $F$ and cannot be part of an optimal strategy. Therefore, the overall strategy is optimal, and the single-deferral equilibrium exists in the unlimited-deferral environment.

For the uniqueness of this equilibrium, consider the role of $F(p)$ in the preceding argument. Its functional form was completely absent from the argument, and in fact, it only relied on the fact that $R_0$ is the maximum of the support of $F$, as required under Claim 1 of Lemma 3. Thus, if an alternative distribution $\hat{F}$ was used, it would still be the case that buyers would strictly prefer to immediately accept any offer $p < R_0$.

It is true that buyers are indifferent between acceptance and deferral of an offer $R_0$. Yet Claim 3 of Lemma 3 still applies in this environment, so there are no atoms except at $R_1$. In particular, buyers almost never receive an offer of $R_0$. Thus, even if a strategy directed buyers to defer $R_0$, it would have no impact on the steady state distribution since such buyers would constitute measure zero of the population.

Therefore, the steady state distribution of buyers would be the same under any distribution $\hat{F}$ that is consistent with profit maximization. Yet, only the equilibrium $F^*$ is consistent with that steady state distribution, as shown in the proof of Proposition 1.

1.7 A.7 Proof of Proposition 4

Proof

We show that $\frac{\partial R_1}{\partial \gamma }<0$ here, but this also establishes that $\frac{\partial R_0}{\partial \gamma }<0$ and $\frac{\partial E[p]}{\partial \gamma }<0$. To see this, consider that $R_1$, $E[p]$, and $R_0$ are each expressed in terms of $\kappa (a)$. Furthermore, the equilibrium $a$ which satisfies $\phi (a) = 0$ will implicitly depend on $\gamma$, so when derivatives are taken with respect to $\gamma$, we obtain

$$\begin{aligned} \frac{\partial R_1}{\partial \gamma } \!=\! \frac{b}{\gamma ^2}+a^{\prime }(\gamma ) \kappa ^{\prime } \left(a(\gamma )\right), \; \frac{\partial E[p]}{\partial \gamma } \!=\! a^{\prime }(\gamma ) \kappa ^{\prime } \left(a(\gamma ) \right), \text{ and} \frac{\partial R_0}{\partial \gamma } \!=\! a^{\prime }(\gamma ) \kappa ^{\prime } \left(a(\gamma ) \right). \end{aligned}$$

Hence, if $\frac{b}{\gamma ^2}+a^{\prime }(\gamma ) \kappa ^{\prime }(a(\gamma ))<0$, then each of the derivatives will be negative.

Thus, we consider the derivative of $R_1$. To compute this, we use the implicit derivative for $a^{\prime }(\gamma )$, obtaining:

$$\begin{aligned} \frac{\partial R_1}{\partial \gamma } = \frac{N}{D}, \end{aligned}$$

where

$$\begin{aligned} N&\equiv b \left(4 (1-\gamma ) \left(\gamma (a^2 (3-2 a) - 1) - (3-2 a) a - \gamma ^2 (2-3 a) (1-a)^2 + \gamma ^3 (1-a)^3 \right) \right.\\&+ 2 (2 - \gamma ) \left(\gamma - a (2-a) - a (5-2 a (3-a)) \gamma -\gamma ^2 (1- a)^2 (3-2 a) \right) \ln \left(\frac{\gamma }{2-\gamma }\right) \\&\left.+\gamma (2 - \gamma ) (2 - a) (1- a)^2 \ln \left(\frac{\gamma }{2-\gamma }\right)^2\right), \end{aligned}$$

and

$$\begin{aligned} D&\equiv \gamma ^2 (2 - \gamma ) \left(- 2 (1 - \gamma ) (1- a) - (2-a) \ln \left(\frac{\gamma }{2-\gamma }\right)\right) \\&\times \left(2 (1 \!-\! \gamma ) (1\!-\! a) (3\!-\!a (1\!-\!\gamma )\!-\!\gamma )\!-\! \left(\gamma (1 - a)^2 - (2-a)^2 \right) \ln \left(\frac{\gamma }{2 - \gamma }\right) \right). \end{aligned}$$

We verify that $\frac{N}{D} < 0$ for any $0 < \gamma < 1$ and $0 < a < 1$, which necessarily includes the equilibrium $a^*$. Regarding the elements of $D$, it is obvious that $\gamma ^2 (2 - \gamma ) > 0$. The other two elements each appeared in $\varDelta ^{\prime }(a)$ in the proof of Proposition 2. There, we showed that the last element on the first line is positive, while the term on the second line is negative. Thus, we find that $D < 0$.

We will now show that $N > 0$. To do so, we will show that $N$ is increasing in $a$ and that $N$ is positive at its minimum. Consider the first derivative $N^{\prime }$ w.r.t. $a$

$$\begin{aligned} N^{\prime }&= 4 (1 - \gamma ) \left(4 a- 3 + 6 \gamma a (1- a) + \gamma ^2 (1- a) (7- 9 a)- 3 \gamma ^3 (1-a)^2\right) \\&-(2-\gamma ) \ln \left(\frac{\gamma }{2-\gamma }\right) \left(4(1-a)+2 \gamma (5 - 6 a (2-a)) -4 \gamma ^2 (1-a) (4-3 a) \right.\\&+\left.\gamma (1-a) (5-3 a) \ln \left(\frac{\gamma }{2-\gamma }\right)\right). \end{aligned}$$

We need to show that $N^{\prime } > 0$, but this is a complicated expression. Hence, we will show that $N^{\prime }$ is decreasing in $a$ and that its minimum value is positive. The derivative of $N^{\prime }$ w.r.t. $a$ is

$$\begin{aligned} N^{\prime \prime } =2 (2 - \gamma ) \left(2(1 - \gamma ) + \ln \left(\frac{\gamma }{2-\gamma }\right) \right) T(a), \end{aligned}$$

where

$$\begin{aligned} T(a) \equiv 2 (1-\gamma ) (1+ 3 \gamma (1- a) )+ \gamma (4-3 a) \ln \left(\frac{\gamma }{2-\gamma }\right). \end{aligned}$$

For all $\gamma \in (0,1)$, the first two terms in $N^{\prime \prime }$ are clearly positive. The third appeared in $\varDelta ^{\prime }(a)$ and was shown to be negative. To show that the last term $T$ is positive, we take its derivative w.r.t. $a$

$$\begin{aligned} T^{\prime }(a) = - 3 \gamma \left(2(1 - \gamma ) + \ln \left(\frac{\gamma }{2-\gamma }\right) \right). \end{aligned}$$

Again, the last term is negative, so $T^{\prime }(a) > 0$ for any $\gamma \in (0,1)$. Thus, $T$ is increasing in $a$ and has its minimum at $a = 0$. There, it takes a value of $T(0) = 2+4 \gamma -6 \gamma ^2+4 \gamma \ln \left(\frac{\gamma }{2-\gamma }\right)$. If we take the second derivative of $T(0)$ w.r.t. $\gamma$, we obtain

$$\begin{aligned} T_{\gamma \gamma } (0) = \frac{16}{\gamma (2 - \gamma )^2} - 12, \end{aligned}$$

which is strictly positive since $\gamma (2 - \gamma )^2$ is largest at $\gamma = 2/3$; there, $T_{\gamma \gamma } (0) = 3/2$, with all other values being strictly greater. Since $T_{\gamma \gamma } (0) > 0$, then $T_{\gamma } (0)$ must take its largest value at $\gamma = 1$, where it is equal to 0. In other words, $T_{\gamma } (0) < 0$, and hence, $T(0)$ takes its smallest value at $\gamma = 1$, where $T(0) = 0$. Thus, we conclude that $T(a) > 0$ for all $a \in (0,1)$ and all $\gamma \in (0,1)$, and hence $N^{\prime \prime } < 0$ on the same domain.

Since $N^{\prime \prime }$ is negative, $N^{\prime }$ must be decreasing in $a$ and, hence, reaches its smallest value at $a=1$. This minimum value computes to $N^{\prime }(1) = 4 (1-\gamma ) + 2 \gamma (2- \gamma ) \ln (\frac{\gamma }{2-\gamma })$. Note that $N^{\prime }_\gamma (1) = 4 (1 - \gamma ) \ln (\frac{\gamma }{2-\gamma }) < 0$. Thus, $N^{\prime }(1)$ takes its smallest value at $\gamma = 1$, where $N^{\prime }(1) = 0$. Therefore, $N^{\prime }(1) > 0$ for all $\gamma \in (0,1)$, and by extension, $N^{\prime } > 0$ for all $\gamma \in (0,1)$ and $a \in (0,1)$.

This means that $N$ is increasing in $a$ and reaches its minimum at $a=0$, which computes to

$$\begin{aligned} N(0)&= 2 \gamma \left(-2 \left(1+\gamma -3 \gamma ^2+\gamma ^3\right)\right.\\&\left.+(2-\gamma ) \left(1-3 \gamma +\ln \left(\frac{\gamma }{2-\gamma }\right)\right) \ln \left(\frac{\gamma }{2-\gamma }\right) \right). \end{aligned}$$

Note that $-2 \left(1+\gamma -3 \gamma ^2+\gamma ^3\right) > - 4(1 - \gamma )$, because this is equivalent to $2(1 - \gamma )^3 > 0$, which holds for all $\gamma < 1$. Also, we have previously noted that $2( \gamma - 1) > \ln (\frac{\gamma }{2-\gamma })$, which rearranges to $1-3 \gamma +\ln (\frac{\gamma }{2-\gamma }) < -1 - \gamma$. Therefore, we can use all three substitutions to obtain

$$\begin{aligned} N(0) > 2 \gamma \left(- 4(1 - \gamma ) + (2-\gamma ) (-1 - \gamma ) 2( \gamma - 1) \right). \end{aligned}$$

Note that $N_\gamma (0) = 0$ at $\gamma = 0,\,\frac{1}{2},\, \text{ and} \ 1$, taking values of $0,\, \frac{1}{4},\, \text{ and} \ 0$, respectively. Therefore, $N_\gamma (0) > 0$ for all $\gamma \in (0,1)$, and by extension, $N > 0$ and $\frac{\partial R_1}{\partial \gamma } < 0$ for all $\gamma \in (0,1)$ and $a \in (0,1)$.