Reverse auctions with regret-anticipated bidders

Abstract

Suppliers may experience emotional/behavioral consequences of anticipated regrets that consist of winner and loser regrets in first- and second-price sealed-bid reverse auctions. Constructing mathematical models that incorporate regret theory to derive closed-form solutions of regret-anticipated suppliers’ bid decisions, this paper theoretically examines the effects of anticipated regrets on suppliers’ bid prices, buyer’s expected procurement cost and auction format decision. Comparing with the no regret scenario, we find that winner regret has adverse effects on the buyer’s expected procurement cost in first-price sealed-bid reverse auctions with regret-anticipated suppliers. To mitigate the adverse effects, we propose using the reserve price strategy for the buyer with theoretical analysis and numerical supports. An interesting analysis reveals that as the number of suppliers increases, the optimal reserve price increases or decreases depending on the degree of winner regret is lower or higher than that of loser regret. Also, the classical revenue equivalence theorem no longer holds when the degree of winner regret differs from that of loser regret.

Appendix

Proof of Lemma 1

For an IC symmetric bidding strategy $b(\cdot \,;\alpha ,\beta )$, let $c_1,c_2\in [\underline{c},\overline{c}]$ with $c_1<c_2$, $b_1=b(c_1\,;\alpha ,\beta )$ and $b_2=b(c_2\,;\alpha ,\beta )$ be the corresponding bids. With the incentive compatibility $b(c;\alpha ,\beta ) = \mathop {\arg \max }\nolimits _{b}\{U^f(b,c\,;c,\alpha ,\beta )\}$, we have $U^f(b,c_1\,;c_1,\alpha ,\beta ) \ge U^f(b,c_2\,;c_1,\alpha ,\beta )$ and $U^f(b,c_2\,;c_2,\alpha ,\beta )\ge U^f(b,c_1\,;c_2,\alpha ,\beta )$. Adding up the two inequalities, we obtain ${\text {E}}[\{b_1-c_1-\alpha (B^w-b_1)\}\mathbf {1}_{\{ B^w> b_1\}}] +{\text {E}}[\{0-\beta (B^w-c_1)\}\mathbf {1}_{\{ c_1 \le B^w<b_1\}}] +{\text {E}}[\{b_2-c_2-\alpha (B^w-b_2)\}\mathbf {1}_{\{ B^w> b_2\}}] +{\text {E}}[\{0-\beta (B^w-c_2)\}\mathbf {1}_{\{ c_2 \le B^w<b_2\}}] \ge {\text {E}}[\{b_2-c_1-\alpha (B^w-b_2)\}\mathbf {1}_{\{ B^w> b_2\}}] +{\text {E}}[\{0-\beta (B^w-c_1)\}\mathbf {1}_{\{ c_1 \le B^w<b_2\}}]+ {\text {E}}[\{b_1-c_2-\alpha (B^w-b_1)\}\mathbf {1}_{\{ B^w > b_1\}}]+ {\text {E}}[\{0-\beta (B^w-c_2)\}\mathbf {1}_{\{ c_2 \le B^w<b_1\}}]$. Rearranging the terms, we derive $(1+\beta )(c_1-c_2)[\Pr (B^w \le b_1)-\Pr (B^w \le b_2)] \ge 0$. Since $\beta \ge 0$ and $c_1<c_2$, we have $b_1 \le b_2$. Suppose that there exists an interval $[c_1,c_2]$ such that $b_1=b_2 = b(c\,;\alpha ,\beta )$ for any $c \in [c_1,c_2]$. In this case, the buyer will randomly choose one of these two suppliers with cost $c_1$ and $c_2$, respectively. However, there exists a sufficiently small $0< \varepsilon < \frac{b_2-c}{2}$ such that $\widetilde{b} \triangleq b_2-\varepsilon$ and $\widetilde{b}-c>\frac{b_2-c}{2}$. This implies that bid price $\widetilde{b}$ is a profitable deviation for the supplier with cost $c_1$, which contradicts the incentive compatibility. Thus, we have $b_1>b_2$. $\square$

Proof of Theorem 1

Since $U^{f}(b,z\,;c,\alpha ,\beta ) =$ $(b(z\,;\alpha ,\beta )-c)(1-F(z))-\int _{y:z<y\le \overline{c}}\alpha (b(y\,;\alpha ,\beta )$ $-b(z\,;\alpha ,\beta ))dF(y)$ $-\int _{y:b^{-1}(c\,;\alpha ,\beta )\le y< z} \beta (b(y\,;\alpha ,\beta )-c)dF(y)$, we have $U^f_{z}(b,z;c,\alpha ,\beta ) = b_z(z\,;\alpha ,\beta )(1-F(z))- (b(z\,;\alpha , \beta )-c)f(z) + \alpha b_z(z\,;\alpha ,\beta ) \int _{z}^{\overline{c}} f(y)dy - \beta (b(z\,;\alpha ,\beta )-c)f(z)$. According to Definition 3, $U^f_{z}(b^{f},z\,;c,\alpha ,\beta )|_{z=c}=0$ is necessary to characterize $b^f$. That is, $(b_z^{f}(z\,;\alpha ,\beta ) (1-F(z))- (b^f(z\,;\alpha ,\beta )-z)f(z) + \alpha b_z^{f}(z\,;\alpha ,\beta ) (1-F(z)) - \beta (b^{f}(z\,;\alpha ,\beta )-z)f(z)]|_{z=c} = 0$. Next, we take $c\in [\underline{c},\overline{c}]$ as a variable.

On one hand, we obtain an equivalent form $-([b^{f}(c\,;\alpha ,\beta )(1-F(c))])^{'} = cf(c) + \alpha b_c^{f}(c\,;\alpha ,\beta ) (1-F(c)) - \beta (b^{f}(c\,;\alpha ,\beta )-c)f(c)$. After solving this differential equation, we derive $b^f(c\,;\alpha ,\beta ) = E_{C}[C\mathbf {1}_{\{ C>c\}}] + \alpha E_{C}[\{b^{f}(C\,;\alpha ,\beta )-b^{f}(c\,;\alpha ,\beta )\}\mathbf {1}_{\{ C>c\}}] - \beta E_{C}[\{b^{f}(C\,;\alpha ,\beta )-C\}\mathbf {1}_{\{ C>c\}}]$.

On the other hand, rearranging the terms, we have $b^f(c\,;\alpha ,\beta ){=}c+\frac{1+\alpha }{1+\beta }\frac{1-F(c)}{f(c)}b_c^{f}(c\,;\alpha ,\beta )$, which can be equivalently written as $(b_c^{f}(c\,;\alpha ,\beta ))^{'}-b^{f}(c\,;\alpha ,\beta ) \frac{1+\beta }{1+\alpha }\frac{f(c)}{1-F(c)} {=} -\frac{1{+}\beta }{1+\alpha }\frac{f(c)}{1-F(c)}c$. Multiplying $e^{\frac{1+\beta }{1+\alpha }\ln [1-F(c)]}$ on both sides, we have $\left( b^{f}(c\,;\alpha ,\beta )e^{\frac{1{+}\beta }{1{+}\alpha } \ln [1-F(c)]}\right) '{=}$ $-\frac{1{+}\beta }{1{+}\alpha }\frac{f(c)}{1-F(c)}ce^{\frac{1+\beta }{1+\alpha } \ln [1-F(c)]}$. Consequently, $b^{f}(c\,;\alpha ,\beta )=c + e^{-\frac{1+\beta }{1+\alpha } \ln [1-F(c)]} \int _c^{\overline{c}} {e^{\frac{1+\beta }{1+\alpha } \ln [1-F(t)]} dt} =c+\int _c^{\overline{c}}{[\frac{1-F(t)}{1-F(c)}]^{\frac{1+\beta }{1+\alpha }}dt}$, $\forall c\in [\underline{c},\overline{c}]$.

For sufficiency, we need to show that for a supplier with any fixed cost $c \in [\underline{c},\overline{c}]$, $U^{f}(b^f,c\,;c,\alpha ,\beta )$ $\ge U^{f}(b^f,z\,;c,\alpha ,\beta )$ for $z\in [\underline{c},\overline{c}]$. Taking cross partial derivative of $U^{f}(b,z\,;c,\alpha ,\beta )$ with respect to z and c, we have $U_{zc}^{f}(b,z\,;c,\alpha ,\beta )=(1+\beta )f(z)\ge 0$. Recall the first-order condition $U_{z}^{f}(b^f,z\,;c,\alpha ,\beta )|_{z=c}=0$. For all $z\in [\underline{c},c]$, we see $U^{f}(b^f,c\,;c,\alpha ,\beta )- U^{f}(b^f,z\,;c,\alpha ,\beta ) = \int _{z}^{c} {U_{s}^{f}(b^f,s\,;c,\alpha ,\beta )ds} = \int _{z}^{c} [U_{s}^{f}(b^f,s\,;c,\alpha ,\beta )- U_{s}^{f}(b^f,s\,;s,\alpha ,\beta )]ds = \int _{z}^{c} \int _{s}^{c} U_{sk}^{f}(b^f,s\,;k,\alpha ,\beta )dkds \ge 0$. For all $z\in [c,\overline{c}]$, we see $U^{f}(b^f,c\,;c,\alpha ,\beta ) - U^{f}(b^f,z\,;c,\alpha ,\beta ) = -\int _{c}^{z} {U_{s}^{f}(b^f,s\,;c,\alpha ,\beta )ds} = -\int _{c}^{z} [U_{s}^{f}(b^f,s\,;c,\alpha ,\beta )- U_{s}^{f}(b^f,s\,;s,\alpha ,\beta )]ds = \int _{c}^{z} {\int _{c}^{s} U_{sk}^{f}(b^f,s\,;k,\alpha ,\beta )dkds\ge 0}$. Hence, we have $U^{f}(b^f,c\,;c,\alpha ,\beta )\ge U^{f}(b^f,z\,;c,\alpha ,\beta )$ for all $z\in [\underline{c},\overline{c}]$.

We check that $b^f$ satisfies first-order necessary condition with respect to b. Since $U^f_{z}(b,z\,;c,\alpha ,\beta )$ $=U^f_{b(z\,;\alpha ,\beta )}(b(z\,;\alpha ,\beta ),z\,;c,\alpha ,\beta )b'(z\,;\alpha ,\beta )$, and $b'(z\,;\alpha ,\beta )>0$ from Lemma 1, $U^f_{z}(b^{f},z\,;c,\alpha ,\beta )|_{z=c} =0$ is equivalent to $U^f_{b(z\,;\alpha ,\beta )}(b^f(z\,;\alpha ,\beta ),z\,;c,\alpha ,\beta )|_{z=c}=0$. In addition, $U^{f}(b,c\,;c,\alpha ,\beta )\ge U^{f}(b,z\,;c,\alpha ,\beta )$ for all $z\in [\underline{c},\overline{c}]$ implies that for any $\widehat{b} \in [b(\underline{c}\,;\alpha ,\beta ),\bar{c}]$, bidding $b(c;\alpha ,\beta )$ would not make the supplier achieve less expected utility than bidding $\widehat{b}$. Given that all other suppliers follow the equilibrium bidding strategy $b^f(\cdot \,;\alpha ,\beta )$, any supplier with fixed cost $c\in [\underline{c},\bar{c}]$ would not achieve higher utility by bidding $\widehat{b} \in [\underline{c}, b(\underline{c}\,;\alpha ,\beta )]$ than by bidding $b(\underline{c}\,;\alpha ,\beta )$ due to increased winner regret and decreased profit. Therefore, we have proved that $b^f(c;\alpha ,\beta ) = \mathop {\arg \max }\nolimits _{b}\{U^f(b,c\,;c,\alpha ,\beta )\}$ for all $c \in [\underline{c},\overline{c}]$; that is, $b^f(\cdot \,;\alpha ,\beta )$ is the (unique) incentive compatible symmetric bidding strategy. It would be interesting to explore the existence and uniqueness of equilibrium solution under more general assumptions on bidding strategies. $\square$

Proof of Corollary 1

• (1) Recall that $b^f(c\,;\alpha ,\beta )= c + \int _c^{\overline{c}} {[\frac{1-F(t)}{1-F(c)}]^{\frac{1+\beta }{1+\alpha }} dt}$ and $b^n(c)=c+\frac{\int _c^{\overline{c}} {[1-F(t)]dt}}{1-F(c)}$. For any two integers $1<n_1<n_2$, let $b_1=b^f(c\,;\alpha ,\beta )$ when $n=n_1$, and $b_2=b^f(c\,;\alpha ,\beta )$ when $n=n_2$. Since $0<\frac{1-F(t)}{1-F(c)} <1$ for all $c<t<\overline{c}$ and $F(x)=1-(1-G(x))^n$, we see $b_1>b_2$.

• (2) Since $0<\frac{1-F(t)}{1-F(c)} <1$ for all $c<t<\overline{c}$, we have $b^f(c\,;\alpha ,\beta )>b^n(c)$ if $\alpha >\beta$, and $b^f(c\,;\alpha ,\beta )<b^n(c)$ if $\alpha <\beta$.

• (3) Now we take $\alpha$ and $\beta$ as variables. Taking the first-order partial derivative of $b^f(c\,;\alpha ,\beta )$ with respect to $\alpha$, we have $b^f_{\alpha }(c\,;\alpha ,\beta ) =$ $[1-F(c)]^{-\frac{1+\beta }{1+\alpha }}[\frac{(1+\beta )\ln [1-F(c)]}{(1+\alpha )^2} \!\int _c^{\overline{c}}{e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt} - \int _c^{\overline{c}}\frac{(1+\beta )\ln [1-F(t)]}{(1+\alpha )^2} {e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt}] \ge [ \int _c^{\overline{c}}{e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt} - \int _c^{\overline{c}}{e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt}] [1 - F(c)]^{-\frac{1+\beta }{1+\alpha }} \frac{(1+\beta )\ln [1-F(c)]}{(1+\alpha )^2}=0$ , where the inequality is obtained by the fact that function $\ln [1-F(c)]$ is decreasing in c. Thus, with c and $\beta$ being fixed, $b^f(c\,;\alpha ,\beta )$ increases as $\alpha$ increases in $[0,\infty )$. Similarly, taking the first-order partial derivative of $b^f(c\,;\alpha ,\beta )$ with respect to $\beta$, we have $b^f_{\beta }(c\,;\alpha ,\beta ) = [1-F(c)]^{-\frac{1+\beta }{1+\alpha }}[-\frac{\ln [1-F(c)]}{1+\alpha } \int _c^{\overline{c}}{e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt} + \int _c^{\overline{c}}\frac{\ln [1-F(t)]}{1+\alpha } {e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt}] \le [1-F(c)]^{-\frac{1+\beta }{1+\alpha }}\frac{\ln [1-F(c)]}{1+\alpha }[ -\int _c^{\overline{c}}{e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt} + \int _c^{\overline{c}}{e^{\frac{1+\beta }{1+\alpha }\ln [1-F(t)]}dt}] = 0$. Therefore, with c and $\alpha$ being fixed, $b^f(c\,;\alpha ,\beta )$ decreases as $\beta$ increases in $[0,\infty )$.

$\square$

Proof of Theorem 2

Denote $\widehat{g}$ and $\widetilde{g}$ as the probability density of $\widehat{G}$ and $\widetilde{G}$, respectively. Using the property of failure rate [see Müller and Stoyan (2002)], we have $\frac{1-\widehat{G}(c+s)}{1-\widehat{G}(c)} \ge \frac{1-\widetilde{G}(c+s)}{1-\widetilde{G}(c)}$ for all $s\in [0,\overline{c}-c]$, and, consequently, $\frac{1-\widehat{F}(c+s)}{1-\widehat{F}(c)} \ge \frac{1-\widetilde{F}(c+s)}{1-\widetilde{F}(c)}$. From $b^f(c\,;\alpha ,\beta )= c + \int _c^{\overline{c}}{[\frac{1-F(t)}{1-F(c)}]^{\frac{1+\beta }{1+\alpha }} dt}$, we see $\widehat{b}^f(c\,;\alpha ,\beta ) \ge \widetilde{b}^f(c\,;\alpha ,\beta )$. $\square$

Proof of Theorem 3

We show that $b(c\,;\alpha ,\beta )=c$ is an IC symmetric bidding strategy for all $c\in [\underline{c},\overline{c}]$. It suffices to show that given any cost $c\in [\underline{c},\overline{c}]$, a supplier cannot do better than submitting a bid price of c. We discuss on $b^w$ that represents the minimum bid from the supplier’s $n-1$ competitors. Consider $b^w<c$. If bidding lower than $b^w$, the supplier wins the auction though, but ends up with a negative profit $b^w-c$. Otherwise, the supplier loses the auction with no loser regret and the utility is zero. Consider $b^w\ge c$. The supplier can win the auction by bidding c and get paid at $b^w$, resulting in a payoff of $b^w-c$. Any bid below $b^w$ does not change this result. Bidding above $b^w$, however, would lead the supplier to lose the game with a non-positive utility $-\beta (b^w-c)$ due to loser regret. $\square$

Proof of Corollary 2

Since $\frac{1-F(t)}{1-F(c)} \ge 0$ for all $t \in [c,\overline{c}]$, according to Theorems 1 and  3, we have $b^{f}(c\,;\alpha ,\beta )= c + \int _c^{\overline{c}} {[\frac{1-F(t)}{1-F(c)}]^{\frac{1+\beta }{1+\alpha }} dt} \ge c = b^{s}(c\,;\alpha ,\beta )$. $\square$

Proof of Theorem 4

In SPRA-AR, given $\alpha$, $\beta \ge 0$, the expected revenue of a supplier with cost $c\in [\underline{c},\overline{c}]$ is represented as $R^{s}(c\,;\alpha ,\beta )=\Pr (B^w{>}\,b^s(c\,;\alpha ,\beta )){\text {E}}[B^w\mathbf {1}_{\{ B^w> b^s(c\,;\alpha ,\beta )\}}]=(1-F(c)){\text {E}}[Y\mathbf {1}_{\{Y>c\}}]=\int _c^{\overline{c}}{tf(t)dt}$. Consequently, the expected procurement cost of the buyer is the sum of n suppliers’ individual expected revenues, i.e., $PC^{s}(\alpha ,\beta ) = n{\text {E}}[R^{s}(C\,;\alpha ,\beta )]= n \int _{\underline{c}}^{\overline{c}} {\left( \int _x^{\overline{c}} {tf(t)dt}\right) g(x)dx} = \int _{\underline{c}}^{\overline{c}} {t {n(n-1)G(t)[1-G(t)]^{n-2}g(t)}dt}$, where the last equality holds for integration by parts. Notice that $n(n-1)G(t)[1-G(t)]^{n-2}g(t)$ is the probability density of the second order statistics of n random variables with identical distribution G. Therefore, the buyer’s expected procurement cost in SPRA-AR is exactly the same as the expectation of the second lowest cost of n suppliers.

In FPRA-AR, given $\alpha$, $\beta \ge 0$, the expected revenue of a supplier with cost $c\in [\underline{c},\overline{c}]$ is represented as $R^{f}(c\,;\alpha ,\beta ) = \Pr (B^w>b^f(c\,;\alpha ,\beta ))b^f(c\,;\alpha ,\beta )= (1-F(c))(c + \int _c^{\overline{c}} {[\frac{1-F(t)}{1-F(c)}]^{\frac{1+\beta }{1+\alpha }} dt})$. Notice that when $\alpha =\beta$, $R^{f}(c\,;\alpha ,\beta ) = (1-F(c))(c+\frac{\int _c^{\overline{c}} {[1-F(t)]dt}}{1-F(c)}) = \int _c^{\overline{c}} {tf(t)dt} = R^{s}(c\,;\alpha ,\beta )$, where the second equality holds from integration by parts. It follows that when $\alpha > \beta$, $R^{f}(c\,;\alpha ,\beta )>R^{s}(c\,;\alpha ,\beta )$, and when $\alpha < \beta$, $R^{f}(c\,;\alpha ,\beta ) < R^{s}(c\,;\alpha ,\beta )$. Correspondingly, the expected procurement cost of the buyer is the sum of n suppliers’ individual expected profits, i.e., $PC^{f}(\alpha ,\beta )=n{\text {E}}[R^{f}(C\,;\alpha ,\beta )]= n \int _{\underline{c}}^{\overline{c}}R^{f}(x\,;\alpha ,\beta )g(x)dx$. Therefore, we conclude that $PC^{f}(\alpha ,\beta )=PC^{s}(\alpha ,\beta )$ if $\alpha = \beta$, $PC^{f}(\alpha ,\beta )>PC^{s}(\alpha ,\beta )$ if $\alpha > \beta$, and $PC^{f}(\alpha ,\beta )<PC^{s}(\alpha ,\beta )$ if $\alpha < \beta$. $\square$

Proof of Corollary 3

According to Corollary 1, we know that $b^f_{\alpha }(x\,;\alpha ,\beta )\ge 0$ and $b^f_{\beta }(x\,;\alpha ,\beta ) \le 0$ for all $c\in [\underline{c},\overline{c}]$ and $\alpha ,\beta \ge 0$. Taking the first-order partial derivative of $PC^f(\alpha ,\beta )$ with respect to $\alpha$, we get $n \int _{\underline{c}}^{\overline{c}}[(1-F(x))b^f_{\alpha }(x\,;\alpha ,\beta )]g(x)dx\ge 0$. Therefore, with $\beta$ being fixed, $PC^f(\alpha ,\beta )$ increases as the winner regret parameter $\alpha$ increases in $[\,0,\infty )$. Taking the first-order partial derivative of $PC^f(\alpha ,\beta )$ with respect to $\beta$, we get $n \int _{\underline{c}}^{\overline{c}}[(1-F(x))b^f_{\beta }(x\,;\alpha ,\beta )]g(x)dx\le 0$. With $\alpha$ being fixed, $PC^f(\alpha ,\beta )$ decreases as the loser regret parameter $\beta$ increases in $[\,0,\infty )$. $\square$

Proof of Lemma 2

Proof follows the same logic as in Lemma 1. $\square$

Proof of Theorem 5

Proof follows the same logic as in Theorem 1. $\square$

Proof of Corollary 4

Proof follows the same logic as in Corollary 1. $\square$

Proof of Corollary 5

Taking the first-order partial derivative of $b^{fp}(c\,;r,\alpha ,\beta )$ with respect to r, we have $b_r^{fp}(c\,;r,\alpha ,\beta ) = \left[ \frac{1-F(r)}{1-F(c)}\right] ^{\frac{1+\beta }{1+\alpha }} > 0$ for all $r\in [\underline{c},\overline{c}]$. Therefore, $b^{fp}(c\,;r,\alpha ,\beta )$ increases as r increases in $[\underline{c},\overline{c}]$. $\square$

Derivation of the expected payment $PC^{fp}$. The expected payment to a specific supplier in FPRA-ARP is $\Pr (\underline{c}\le C\le r){\text {E}}[(1-F(C))b^{fp}(C\,;r,\alpha ,\beta ) \mathbf {1}_{\{\underline{c}\le C\le r\}}] = \int _{\underline{c}}^r (1-F(x))b^{fp}(x)g(x)dx$ $= \int _{\underline{c}}^r {[1-F(x)]^{\frac{\alpha -\beta }{1+\alpha }} \left( \int _x^r {[1-F(t)]^{\frac{1+\beta }{1+\alpha }}}dt\right) g(x)dx}+\int _{\underline{c}}^r {(1-F(x)) x g(x) dx}= \int _{\underline{c}}^r [1-F(t)]^{\frac{1+\beta }{1+\alpha }} \left( \int _{\underline{c}}^t {[1-F(x)]^{\frac{\alpha -\beta }{1+\alpha }} g(x) dx}\right) dt + \int _{\underline{c}}^r {(1-F(x)) x g(x) dx}$, where the last equality is obtained by interchanging the order of integration. Correspondingly, the expected procurement cost of the buyer is $PC^{fp}(r\,;\alpha ,\beta )=n\Pr (\underline{c}\le C\le r) {\text {E}}[(1-F(C))b^{fp}(C\,;r,\alpha ,\beta )\mathbf {1}_{\{\underline{c}\le C\le r\}}]+(1-G(r))^nc_0$, where the first term equals the sum of n suppliers’ individual expected payments, and the second term is the expected procurement cost for using the outside option. Hence, the expected procurement cost of the buyer is shown as in Eq. (8). $\square$

Proof of Theorem 6

Taking the first-order derivative of Eq. (8) with respect to r, we have $PC_r^{fp}(r\,;\alpha ,\beta ) = -n (1-F(r)) g(r)c_0 + n(1-F(r))rg(r) + n [1-F(r)]^{\frac{1+\beta }{1+\alpha }} \int _{\underline{c}}^r [1-F(x)]^{\frac{\alpha -\beta }{1+\alpha }} g(x) dx$. Solving $PC_r^{fp}(r^{fp}\,;\alpha ,\beta ) = 0$ leads to $c_0=r^{fp}+\int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1+\alpha }} \frac{g(x)}{g(r^{fp})} dx}$.

For sufficiency, define $\varphi (r\,;\alpha ,\beta ) = r+\int _{\underline{c}}^r \left[ \frac{1-F(x)}{1-F(r)}\right] ^{\frac{\alpha -\beta }{1+\alpha }} \frac{g(x)}{g(r)} dx$. Taking the first-order derivative of $\varphi$ with respect to r, we have $\varphi _r(r\,;\alpha ,\beta ) = 2 + \int _{\underline{c}}^r \left[ \frac{1-F(x)}{1-F(r)}\right] ^{\frac{\alpha -\beta }{1+\alpha }} \left[ \frac{\alpha -\beta }{1+\alpha }\frac{f(r)}{1-F(r)} - \frac{g'(r)}{g(r)}\right] \frac{g(x)}{g(r)} dx = \int _{\underline{c}}^r \left[ \frac{1-F(x)}{1-F(r)}\right] ^{\frac{\alpha -\beta }{1+\alpha }} \left[ \frac{2}{c_0-r} + \frac{\alpha -\beta }{1+\alpha }\frac{f(r)}{1-F(r)}-\frac{g'(r)}{g(r)}\right] \frac{g(x)}{g(r)} dx$. Suppose that $\frac{2}{c_0-r}+ \frac{\alpha -\beta }{1+\alpha }\frac{(n-1)g(r)}{1-G(r)}-\frac{g'(r)}{g(r)}>0$ holds for all $r\in [\underline{c},\overline{c}]$. Consequently, $\varphi _r(r\,;\alpha ,\beta )>0$ for all $r\in [\underline{c},\overline{c}]$. Therefore, $\varphi (r\,;\alpha ,\beta )$ is increasing in r. Then, when $\underline{c}\le r<r^{fp}$, we have $c_0<r^{fp}+\int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1+\alpha }} \frac{g(x)}{g(r^{fp})} dx}$, which implies $PC_r^{fp}(r\,;\alpha ,\beta )<0$. Similarly, when $r^{fp}<r\le \overline{c}$, $PC_r^{fp}(r\,;\alpha ,\beta )>0$. Hence, with the technical assumption, we conclude that Eq. (9) is sufficient for optimality. $\square$

Proof of Corollary 6

Define $\phi (\alpha ,\beta )=\int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1+\alpha }} \frac{g(x)}{g(r^{fp})} dx} + r^{fp}(\alpha ,\beta ) \equiv c_0$. Taking the first-order partial derivative of $\phi (\alpha ,\beta )$ with respect to $\alpha$, we have $\phi _{\alpha } (\alpha ,\beta ) = r^{fp}_{\alpha }(\alpha ,\beta ) \{2 + \int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \left[ \frac{\alpha -\beta }{1+\alpha } \frac{f(r^{fp})}{1-F(r^{fp})} - \frac{g'(r^{fp})}{g(r^{fp})}\right] \frac{g(x)}{g(r^{fp})} }dx\} + \int _{\underline{c}}^{r^{fp}} \left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \frac{1+\beta }{(1+\alpha )^2} \ln [\frac{1-F(x)}{1-F(r^{fp})}] \frac{g(x)}{g(r^{fp})} dx$ $=r^{fp}_{\alpha }(\alpha ,\beta ) \int _{\underline{c}}^{r^{fp}} \left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \left[ \frac{2}{c_0-r^{fp}}+ \frac{\alpha -\beta }{1+\alpha } \frac{f(r^{fp})}{1-F(r^{fp})} \right. \left. - \frac{g'(r^{fp})}{g(r^{fp})}\right] \frac{g(x)}{g(r^{fp})} dx + \int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \frac{1+\beta }{(1+\alpha )^2} \ln [\frac{1-F(x)}{1-F(r^{fp})}] \frac{g(x)}{g(r^{fp})} }dx=0$, where the last equality is a result of zero-order information. Since $\int _{\underline{c}}^{r^{fp}} \left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \frac{1+\beta }{(1+\alpha )^2} \mathrm{ln} [\frac{1-F(x)}{1-F(r^{fp})}] \frac{g(x)}{g(r^{fp})} dx > 0$ and $\frac{2}{c_0-r^{fp}} + \frac{\alpha -\beta }{1+\alpha }\frac{(n-1)g(r^{fp})}{1-G(r^{fp})}-\frac{g'(r^{fp})}{g(r^{fp})} > 0$. Hence, $r^{fp}_{\alpha }(\alpha ,\beta )<0$ for all $\alpha \ge 0$. We conclude that $r^{fp}(\alpha ,\beta )$ decreases as $\alpha$ increases in $[\,0,\infty )$.

Similarly, taking the first-order partial derivative of $\phi (\alpha ,\beta )$ with respect to $\beta$, we have $\phi _{\beta } (\alpha ,\beta ) = r^{fp}_{\beta }(\alpha ,\beta ) \int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \left[ \frac{2}{c_0-r^{fp}} + \frac{\alpha -\beta }{1+\alpha } \frac{f(r^{fp})}{1-F(r^{fp})} - \frac{g'(r^{fp})}{g(r^{fp})}\right] \frac{g(x)}{g(r^{fp})} }dx - \int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \frac{1}{1+\alpha } \ln [\frac{1-F(x)}{1-F(r^{fp})}] \frac{g(x)}{g(r^{fp})} }dx$. Since $\phi (\alpha ,\beta ) \equiv c_0$, we have $\phi _{\beta } (\alpha ,\beta )=0$. Moreover, $\int _{\underline{c}}^{r^{fp}} {\left[ \frac{1-F(x)}{1-F(r^{fp})}\right] ^{\frac{\alpha -\beta }{1 + \alpha }} \frac{1}{1+\alpha } \ln [\frac{1-F(x)}{1-F(r^{fp})}] \frac{g(x)}{g(r^{fp})} }dx>0$ and $\frac{2}{c_0-r^{fp}} + \frac{\alpha -\beta }{1+\alpha }\frac{(n-1)g(r^{fp})}{1-G(r^{fp})}-\frac{g'(r^{fp})}{g(r^{fp})} > 0$. Hence, $r_{\beta }^{fp}(\alpha ,\beta )>0$ for all $\beta \ge 0$. We conclude that $r^{fp}(\alpha ,\beta )$ increases as $\beta$ increases in $[\,0,\infty )$.

For any two integers $1<n_1<n_2$, let $r^{fp}_1=r^{fp}(\alpha ,\beta )$, $\phi _1=\phi (\alpha ,\beta )$ when $n=n_1$, and $r^{fp}_2=r^{fp}(\alpha ,\beta )$, $\phi _2=\phi (\alpha ,\beta )$ when $n=n_2$. Since $\phi _1=r^{fp}_1+\int _{\underline{c}}^{r^{fp}_1} {\left[ \frac{1-G(x)}{1-G(r^{fp}_1)}\right] ^{\frac{\alpha -\beta }{1+\alpha }(n_1-1)} \frac{g(x)}{g(r^{fp}_1)} dx}=c_0=r^{fp}_2+\int _{\underline{c}}^{r^{fp}_2} {\left[ \frac{1-G(x)}{1-G(r^{fp}_2)}\right] ^{\frac{\alpha -\beta }{1+\alpha }(n_2-1)} \frac{g(x)}{g(r^{fp}_2)} dx} =\phi _2$, we see that $r^{fp}_1>r^{fp}_2$ if $\alpha >\beta$, and $r^{fp}_1<r^{fp}_2$ if $\alpha <\beta$. $\square$

Proof of Theorem 7

Proof follows the same logic as in Theorem 2. $\square$

Derivation of the expected payment $PC^{sp}$. Proof follows the same logic as in derivation of the expected payment $PC^{fp}$. $\square$

Proof of Theorem 8

Proof follows the same logic as in Theorem 6. $\square$

Proof of Theorem 9

According to Eq. (9) in Theorem 6 and Eq. (12) in Theorem 8, we know that the optimal reserve price in FPRA-ARP is the same as that in SPRA-ARP when the winner and loser regret parameters are equal. Hence, Theorem 9 is a direct consequence of Corollary 6. $\square$

Proof of Corollary 7

Since the expression of the expected procurement cost in FPRA-AR is the same as that in SPRA-AR when winner and loser regret parameters are equal. Corollary 7 is a direct consequence of expressions of Eqs. (8) and (10). $\square$