Sequential procurement auctions with risk-averse suppliers

Abstract

We compare two procurement mechanisms, bundling and unbundling, in a two-stage auction model with risk-averse suppliers. The mechanisms differ in that the two tasks of investment and production are procured through a single auction or two sequential auctions. Suppliers’ production costs are affected by two risk factors, aggregate risk and idiosyncratic risk, as well as by the cost-reducing investment. We show that the bundling mechanism is optimal for a buyer and socially desirable if the aggregate risk is below certain thresholds. The result may not hold true for idiosyncratic risk.

Appendix

Proof of Proposition 1

First, the efficient investment level $\tilde{a}$ minimizes $\psi (a) + E[\theta _{(n)} + \omega ] - \delta a$; thus, the level is uniquely determined by the first-order condition

$$\begin{aligned} \frac{d \psi }{da}(\tilde{a}) = \delta . \end{aligned}$$

(10)

In the bundling mechanism, a winner chooses an investment level to maximize the certainty equivalent $p^1_i - \psi (a) + \delta a - \rho ^*$. Since the risk premium $\rho ^*$ does not depend on $a$, the equilibrium investment level is the same as the efficient level $\tilde{a}$. Next, the auction is equivalent to the Bertrand competition among symmetric suppliers. Thus, in any symmetric equilibrium, all suppliers submit the same bid $\psi (\tilde{a}) -\delta \tilde{a} + \rho ^{*}$. $\square$

Proof of Lemma 1

We must show that for any $a$ and $\omega$, a supplier with $\theta$ cannot gain by deviating from a bid $p^2(a, \theta , \omega )$ when the other suppliers follow the strategy $p^2(\cdot )$. Note that $p^2(\cdot )$ is increasing and continuous in $\theta$.

First, we can assume without loss of generality that no supplier submits a bid $p \not \in [p^2(a, \underline{\theta }, \omega ), p^2(a, \bar{\theta }, \omega )]$. If a supplier bids $p > p^2(a, \bar{\theta }, \omega )$, then he loses with probability one. By bidding $p^2(a, \bar{\theta }, \omega )$, he can obtain the same utility. If a supplier bids $p < p^2(a, \underline{\theta }, \omega )$, then he wins with probability one. By bidding $p^2(a, \underline{\theta }, \omega )$, he can win with probability one and obtain higher utility.

Second, we show that it is optimal for a supplier with $\theta$ to bid $p = p^2(a, \theta , \omega )$. Consider the following two cases: The supplier wins or loses in the first stage. If the supplier loses in the first-stage auction, then his expected utility from bidding $p^2(a, \hat{\theta }, \omega )$ is

$$\begin{aligned}&\ (1-F_{(n-1)}(\hat{\theta })) u(p^2(a, \hat{\theta }, \omega ) - c(a, \theta ,\omega ) ) + F_{(n-1)}(\hat{\theta }) u(0) \\&\quad = (1-F_{(n-1)}(\hat{\theta })) \int \limits _{\hat{\theta }}^{\bar{\theta }} \left[ 1 - \exp (-r (s - \theta )) \right] \frac{f_{(n-1)}(s)}{1-F_{(n-1)}(\hat{\theta })} ds \\&\quad = \int \limits _{\hat{\theta }}^{\bar{\theta }} u(s - \theta ) f_{(n-1)}(s) ds. \end{aligned}$$

The equalities follow from the definitions of $u(\cdot )$ and $p^2(\cdot )$. The difference between the expected utility from bidding $p^2(a, \theta , \omega )$ and that from bidding $p^2(a, \hat{\theta }, \omega ) \not = p^2(a, \theta , \omega )$ is

$$\begin{aligned} \int \limits _{\theta }^{\hat{\theta }} u(s - \theta ) f_{(n-1)}(s) ds > 0. \end{aligned}$$

In contrast, if the supplier wins with a bid $p^1$ in the first-stage auction, then his expected utility from bidding $p^2(a, \hat{\theta }, \omega )$ is given by

$$\begin{aligned}&\ (1-F_{(n-1)}(\hat{\theta })) u(p^1 - \psi (a) + p^2(a, \hat{\theta }, \omega ) - c(a, \theta , \omega ) ) {+} F_{(n-1)}(\hat{\theta }) u(p^1 - \psi (a)) \\&\quad = u(p^1 - \psi (a)) + \exp (-r(p^1 - \psi (a))) \left\{ \int \limits _{\hat{\theta }}^{\bar{\theta }} u(s - \theta ) f_{(n-1)}(s) ds \right\} . \end{aligned}$$

The equality follows from some additional calculations. The difference between the expected utility from bidding $p^2(a, \theta , \omega )$ and that from bidding $p^2(a, \hat{\theta }, \omega ) \not = p^2(a, \theta , \omega )$ is

$$\begin{aligned} \exp (-r(p^1 - \psi (a))) \left\{ \int \limits _{\theta }^{\hat{\theta }} u(s - \theta ) f_{(n-1)}(s) ds \right\} > 0. \end{aligned}$$

Therefore, it is optimal for a supplier with $\theta$ to bid $p = p^2(a, \theta , \omega )$ regardless of whether he wins or loses in the first stage. $\square$

Proof of Proposition 2

(i) First, Lemma 1 implies that $p^2(\cdot )$ constitutes a symmetric equilibrium in the second-stage auction. Second, a winner in the first-stage auction chooses an investment level to maximize the certainty equivalent (6). Since the certainty equivalent decreases in $a$, the winner optimally chooses $a = 0$. Third, in the first-stage auction, every supplier submits a bid which makes him indifferent between winning and losing, in any symmetric equilibrium. Even if a supplier loses the first-stage auction, he can obtain the positive expected utility in the second-stage auction. The certainty equivalent is given by $E[ (1-F_{(n-1)}(\theta _i)) (b(\theta _i) - \theta _i) ] - \rho ^{**}$. Hence, all suppliers submit the same bid $\psi (0)$ in the first-stage auction.

(ii) Since a supplier with $\theta _{(n)}$ wins the second-stage auction in equilibrium, the buyer’s expected utility is $EU_B^{**} = v - \psi (0) -E[p^2(0, \theta _{(n)}, \omega )]$. The expected payment in the second-stage auction $E[p^2(0, \theta _{(n)}, \omega )]$ can be rewritten as

$$\begin{aligned} E [ b(\theta _{(n)}) ] + E[\omega ]&= \int \limits _{\underline{\theta }}^{\bar{\theta }} b(s) f_{(n)}(s) ds + E [\theta _{(n)}] - \int \limits _{\underline{\theta }}^{\bar{\theta }} s f_{(n)}(s) ds \\&= E [\theta _{(n)}] + \int \limits _{\underline{\theta }}^{\bar{\theta }} [b(s) - s] \cdot n (1-F_{(n-1)}(s)) f(s) ds \\&= E [\theta _{(n)}] + n E [(1-F_{(n-1)}(\theta _i)) (b(\theta _i) - \theta _i)]. \end{aligned}$$

The second equality follows from the definition of $f_{(n)}(\cdot )$. Thus, $EU_B^{**} = v - \bigl \{ \psi (0) + E[\theta _{(n)}] + n E[(1-F_{(n-1)}(\theta _i)) (b(\theta _i) - \theta _i)] \bigr \}$. Since every supplier wins the first-stage auction with equal probability, his equilibrium expected utility is given by $EU_i^{**} = u(E[ (1-F_{(n-1)}(\theta _i)) (b(\theta _i) - \theta _i) ] - \rho ^{**})$. Finally, the social welfare is given by $W^{**} = EU_B^{**} + n \cdot u^{-1}(EU_i^{**}) = v - \left\{ \psi (0) + E[\theta _{(n)}] + n \rho ^{**} \right\}$. $\square$

Proof of Theorem 1

(i) It follows from the first-order condition (10) that $\tilde{a}$ does not depend on $\lambda$. Since $G(\cdot ; \lambda )$ second-order stochastically dominates $G(\cdot ; \lambda ')$ if $\lambda ' > \lambda$, the risk premium $\rho ^*$ under bundling increases in $\lambda$. Thus, both $EU_B^*$ and $W^*$ decrease in $\lambda$. On the other hand, under unbundling, the second-stage winner’s profit $b(\theta _i) - \theta _i$ does not depend on $\lambda$; therefore, the risk premium $\rho ^{**}$ is invariant to $\lambda$. Thus, both $EU_B^{**}$ and $W^{**}$ are invariant to $\lambda$. Proposition 2 (ii) also implies that $EU_B^{**} < W^{**}$.

Then, we have thresholds $\underline{\lambda }$ and $\bar{\lambda }$ with the properties in part (i) of the theorem. If $EU_B^* > EU_B^{**}$ (resp., $W^* > W^{**}$) for each $\lambda \in (0,1)$, then we set $\bar{\lambda } = 1$ (resp., $\underline{\lambda } = 1$). If $EU_B^* < EU_B^{**}$ (resp., $W^* < W^{**}$) for each $\lambda \in (0,1)$, then we set $\bar{\lambda } = 0$ (resp., $\underline{\lambda } = 0$).

(ii) First, we claim that $\underline{\lambda }$ is greater than zero in the limit as $\kappa \rightarrow 0$. We show that as $\kappa \rightarrow 0$ and $\lambda \rightarrow 0$, the difference between $W^* (= EU_B^*)$ and the first-best utility $v - \{ \psi (\tilde{a}) + E[\theta _{(n)}] - \delta \tilde{a} \}$ converges to zero. According to the definition of the convergence in law, for any $\theta > 0,F(\theta ; \kappa ) \rightarrow 1$ as $\kappa \rightarrow 0$. Also, for any $\kappa ,\lim _{m \rightarrow \infty } \int \limits _m^{\infty } \exp (r \theta ) d F(\theta ; \kappa ) = 0$. This implies that, for any $\varepsilon > 0$, there exist positive numbers $\delta , m$ with $0 < \delta < m$ and $\bar{\kappa } > 0$ such that for any $\kappa < \bar{\kappa }$, the following inequalities remain true:¹²

$$\begin{aligned} \int \limits _{\underline{\theta }}^{\bar{\theta }} \exp (r \theta ) d F(\theta ; \kappa )&= \int \limits _{-\infty }^\delta \exp (r \theta ) d F(\theta ; \kappa ) + \int \limits _\delta ^m \exp (r \theta ) d F(\theta ; \kappa ) \\&\quad {+}\!\! \int \limits _m^{\infty } \exp (r \theta ) d F(\theta ; \kappa ){<} \exp (r \delta ) {+} \exp (r m) \left[ F(m; \kappa ){-} F(\delta ; \kappa )\!\right] \\&\quad + \int \limits _m^{\infty } \exp (r \theta ) f(\theta ; \kappa ) d\theta < \left( 1 + \frac{\varepsilon }{3} \right) + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = 1 + \varepsilon . \end{aligned}$$

Since $E [\exp (r \theta _i)] > \exp (r E[\theta _i]) = 1$ always remains true, $\ln E[\exp (r \theta _i)] \rightarrow 0$ as $\kappa \rightarrow 0$. Similarly, $\ln E[\exp (r \omega )] \rightarrow 0$ as $\lambda \rightarrow 0$. Hence, the risk premium $\rho ^*$ converges to zero as $\kappa , \lambda \rightarrow 0$. Also, according to the definition of the convergence in law, for any $\theta < 0,F(\theta ; \kappa ) \rightarrow 0$ as $\kappa \rightarrow 0$. Hence, a similar argument shows that $E[\theta _{(n)}] \rightarrow 0$ as $\kappa \rightarrow 0$. On the other hand, the difference between $W^{**}$ and the first-best utility is strictly smaller than zero even in the limit as $\kappa , \lambda \rightarrow 0$. This is because $0 < \tilde{a}$. Therefore, $\underline{\lambda } > 0$ in the limit as $\kappa \rightarrow 0$.

Next, we claim that $\bar{\lambda }$ is smaller than one in the limit as $n \rightarrow \infty$ and $\delta \rightarrow 0$. Using integration by parts, we transform the bidding function $b(\cdot )$ in Lemma 1 into

$$\begin{aligned} b(\theta _i)&=\! - \frac{1}{r} \ln \left\{ - \left[ \exp (- r s) \frac{1 - F_{(n-1)} (s)}{1 - F_{(n-1)}(\theta _i)} \right] _{\theta _i}^{\bar{\theta }} {-} r \int \limits _{\theta _i}^{\bar{\theta }} \exp (- r s) \frac{1 - F_{(n-1)} (s)}{1 - F_{(n-1)}(\theta _i)} d s \right\} \nonumber \\&= - \frac{1}{r} \ln \left\{ \exp (- r \theta _i) - r \int \limits _{\theta _i}^{\bar{\theta }} \exp (- r s) \left( \frac{1 - F (s)}{1 - F(\theta _i)} \right) ^{n-1} d s \right\} . \end{aligned}$$

(11)

Lebesgue’s dominated convergence theorem implies that as the number $n$ of suppliers goes to infinity, the term $b(\theta _i)$ converges to $\theta _i$ for each $\theta _i$. Based on this fact, we can easily verify that $E[n(1-F_{(n-1)}(\theta _i)) (b(\theta _i)-\theta _i)]$ converges to zero as $n \rightarrow \infty$. Moreover, $E[\theta _{(n)}]$ converges to $\underline{\theta }$ as $n \rightarrow \infty$. Thus, $EU_B^{**}$ converges to $v - \{ \psi (0) + \underline{\theta } \}$ as $n \rightarrow \infty$. In contrast, $EU_B^*$ is invariant to $n$. It follows from the first-order condition (10) that $\tilde{a} \rightarrow 0$ as $\delta \rightarrow 0$. Therefore, $\bar{\lambda } = 0$ for sufficiently large $n$ and small $\delta$. This establishes the claim. $\square$

Proof of Example 1

Assume that $F(\theta _i; \kappa ) = 1 - \exp (-\theta _i/\kappa -1)$ for each $\theta _i \ge - \kappa$. From the distribution, the bidding function $b(\cdot )$ in the second-stage auction is given as follows:

$$\begin{aligned} b(\theta _i)&= -\frac{1}{r} \ln \int \limits _{\theta _i}^\infty \exp (-rs) \frac{n-1}{\kappa } \exp \left( {-}(n{-}1) \left( \frac{s}{\kappa }{+}1 \right) \right) \exp \left( (n{-}1) \left( \frac{\theta _i}{\kappa }{+}1 \right) \right) ds \\&= \theta _i + \frac{1}{r} \left( \ln (\kappa r + n-1) - \ln (n-1) \right) . \end{aligned}$$

Therefore, a simple calculation shows that

$$\begin{aligned} E[(1-F_{(n-1)}(\theta _i)) (b(\theta _i) - \theta _i)] = \frac{1}{nr} \left( \ln (\kappa r + n-1) - \ln (n-1) \right) , \end{aligned}$$

which increases in $\kappa$. The equality follows from $E[(1-F_{(n-1)}(\theta _i))] = 1/n$.

We next show that $\rho ^* > n \rho ^{**}$. From the distribution of $\theta _i,\rho ^*$ is given as follows:

$$\begin{aligned} \rho ^*&= \frac{1}{r} \ln \int \limits _{-\kappa }^\infty \exp (r s) \frac{1}{\kappa } \exp \left( -\frac{s}{\kappa } - 1 \right) ds + \frac{1}{r} \ln E[\exp (r \omega )] \\&= - \frac{1}{r} \ln (1- \kappa r) - \kappa + \frac{1}{r} \ln E[\exp (r \omega )]. \end{aligned}$$

This is well-defined because $\kappa r \in (0,1)$. We can compute $n \rho ^{**}$ as follows:

$$\begin{aligned} n \rho ^{**}&= \frac{n}{r} \ln E \left[ (1-F_{(n-1)}(\theta _i)) \frac{n-1}{\kappa r + n-1} + F_{(n-1)}(\theta _i) \right] \\&\ \ \ + n E[(1-F_{(n-1)}(\theta _i))] \frac{1}{r}(\ln (\kappa r + n-1) - \ln (n-1) ) \\&= \frac{n-1}{r} \left[ \ln (n-1) - \ln (\kappa r + n-1) \right] + \frac{n}{r} \left[ \ln (\kappa r + n) - \ln (n) \right] . \end{aligned}$$

It is easy to show that $\rho ^* \rightarrow \frac{1}{r} \ln E[\exp (r \omega )] > 0$ and $n \rho ^{**} \rightarrow 0$ as $\kappa \rightarrow 0$. Also, a simple calculation shows that

$$\begin{aligned} \frac{\partial \rho ^*}{\partial \kappa } = \frac{\kappa r}{1-\kappa r} > \frac{\kappa r}{(\kappa r + n)(\kappa r + n-1)} = \frac{\partial (n \rho ^{**})}{\partial \kappa } \end{aligned}$$

for each $\kappa , r$ with $\kappa r \in (0,1)$ and $n \ge 2$. Therefore, it holds that $\rho ^* > n \rho ^{**}$. $\square$

Proof of Remark 1

We claim that if the idiosyncratic risks $\theta _i$’s are exponentially distributed as in Example 1, then for sufficiently small $\lambda ,\kappa$ and $n$, the buyer’s expected utility $EU_B^*$ in the bundling mechanism is greater than the following expected utility in the unbundling mechanism with the sophisticated contract; i.e.,

$$\begin{aligned}&\ v - \left\{ \psi (\tilde{a}) - \delta \tilde{a} + E[\theta _{(n)}] + n E[(1-F_{(n-1)}(\theta _i)) (b(\theta _i)-\theta _i)] \right\} \nonumber \\&\quad = v - \left\{ \psi (\tilde{a}) - \delta \tilde{a} - \left( \frac{n-1}{n} \right) \kappa + \frac{1}{r} \left( \ln (\kappa r + n-1) - \ln (n-1) \right) \right\} . \end{aligned}$$

(12)

The difference between the expected utility (12) and the first-best utility converges to zero as $\kappa \rightarrow 0$. For the expected utility (12), the derivative with respect to $\kappa$ is given by

$$\begin{aligned} \frac{n-1}{n} - \frac{1}{\kappa r + n -1}. \end{aligned}$$

(13)

As shown in the proof of Theorem 1, the difference between the buyer’s expected utility $EU_B^*$ in the bundling mechanism and the first-best utility converges to zero as $\lambda \rightarrow 0$ and $\kappa \rightarrow 0$. The derivative with respect to $\kappa$ is

$$\begin{aligned} \frac{\partial EU_B^*}{\partial \kappa } = - \frac{\partial \rho ^*}{\partial \kappa } = - \frac{\kappa r}{1 - \kappa r}. \end{aligned}$$

(14)

The derivative (14) is greater than (13) when $\kappa$ is close to zero and $n=2$. Therefore, $EU_B^*$ is greater than the expected utility (12) for sufficiently small $\lambda ,\kappa$ and $n = 2$. $\square$

Proof of Proposition 3

First, using the envelope theorem, we can show that $\frac{\partial EU_B^*}{\partial \delta } = \frac{\partial W^*}{\partial \delta } = \tilde{a}$. Next, the assumption of additively separable production cost implies that neither the second-stage winner’s profit $b(\theta _i) - \theta _i$ nor the risk premium $\rho ^{**}$ under unbundling depends on $\delta$. Hence, $\frac{\partial EU_B^{**}}{\partial \delta } = \frac{\partial W^{**}}{\partial \delta } = 0$. Since $\tilde{a} > 0,\frac{\partial EU_B^*}{\partial \delta } = \frac{\partial W^*}{\partial \delta } = \tilde{a} > 0 = \frac{\partial EU_B^{**}}{\partial \delta } = \frac{\partial W^{**}}{\partial \delta }$. Proposition 2 (ii) also implies that $EU_B^{**} < W^{**}$.

Then, we have thresholds $\underline{\delta }$ and $\bar{\delta }$ with the properties in the proposition. If $EU_B^* > EU_B^{**}$ (resp., $W^* > W^{**}$) for any $\delta > 0$, then we set $\underline{\delta } = 0$ (resp., $\bar{\delta } = 0$). Also, $\underline{\delta }, \bar{\delta } < \infty$ because $W_B^* > W_B^{**}$ and $EU_B^* > EU_B^{**}$ for some sufficiently large $\delta$. This completes the proof. $\square$

Proof of Lemma 3

(i) First, Lemma 2 implies that $p(\cdot )$ is an equilibrium bidding strategy in the subcontracting auction. Second, a first-stage winner chooses an investment level to maximize the certainty equivalent (9). Differentiating the certainty equivalent yields the first-order condition (10). Hence, the winner chooses the efficient investment level. Third, in the first-stage auction, every supplier submits a bid which makes him indifferent between winning and losing, in any symmetric equilibrium. Even if supplier $i$ loses the first-stage auction, he can obtain the positive expected utility in the subcontracting auction. The certainty equivalent is given by

$$\begin{aligned}&\ E \left[ F(\theta _j) E_{\theta _i} [(1-F_{(n-2)}(\theta _i)) (b(\theta _i, \theta _j) - \theta _i) \mid \theta _i < \theta _j] \right] - \rho _l. \end{aligned}$$

Hence, all suppliers submit the following bid:

$$\begin{aligned}&\ \psi (\tilde{a}) + E \left[ (1-F_{(n-1)}(\theta _i))\theta _i + \int \limits _{\underline{\theta }}^{\theta _i} b(s,\theta _i) f_{(n-1)}(s)ds + \omega - \delta \tilde{a} \right] + \rho _w \\&\qquad +E \left[ \int \limits _{\underline{\theta }}^{\theta _i} (1-F_{(n-2)}(s)) (b(s, \theta _i)-s) f(s) ds \right] - \rho _l \\&\quad = \psi (\tilde{a}) -\delta \tilde{a} + E \left[ \frac{n}{n-1} \int \limits _{\underline{\theta }}^{\theta _i} b(s, \theta _i) f_{(n-1)}(s) ds \right] + \rho _w - \rho _l. \end{aligned}$$

The equality follows from the interchange of the integrals.

(ii) It is obvious that $EU_B^s$ and $EU_i^s$ are given as in the proposition. The social welfare $W^s = EU_B^s + n \cdot u^{-1}(EU_i^s)$ can be rewritten as

$$\begin{aligned}&\ v - \left\{ \psi (\tilde{a}) -\delta \tilde{a} + E \left[ \frac{n}{n-1} \int \limits _{\underline{\theta }}^{\theta _i} b(s, \theta _i) f_{(n-1)}(s) ds \right] + \rho _w - \rho _l \right\} \\&\qquad + n E \left[ \int \limits _{\underline{\theta }}^{\theta _i} (1-F_{(n-2)}(s)) (b(s, \theta _i)-s) f(s) ds \right] - n \rho _l \\&\quad = v - \left\{ \psi (\tilde{a}) -\delta \tilde{a} + E[\theta _{(n)}] + \rho _w + (n-1) \rho _l \right\} . \end{aligned}$$

$\square$

Proof of Proposition 4

As a preliminary, we prove that $EU_i^s = EU_i^{**}$; thus, $W^s - EU_B^s = W^{**} - EU_B^{**}$. From Proposition 2 (ii) and Eq. (11), the supplier’s expected utility $EU_i^{**}$ in the unbundling mechanism can be rewritten as

$$\begin{aligned}&\ 1 - E[ (1-F_{(n-1)}(\theta _i)) \exp (-r (b(\theta _i) - \theta _i)) + F_{(n-1)}(\theta _i)] \nonumber \\&\quad = 1 - \int \limits _{\underline{\theta }}^{\bar{\theta }} \left\{ (1-F_{(n-1)}(\theta _i)) \left[ 1 - r \int \limits _{\theta _i}^{\bar{\theta }} \frac{\exp (-rs)}{\exp (-r \theta _i)} \left( \frac{1-F(s)}{1-F(\theta _i)} \right) ^{n-1} ds \right] \right. \nonumber \\&\qquad \left. + F_{(n-1)}(\theta _i) \right\} f(\theta _i) d \theta _i = r \int \limits _{\underline{\theta }}^{\bar{\theta }} \int \limits _{\theta _i}^{\bar{\theta }} \frac{\exp (-rs)}{\exp (-r \theta _i)} \left( 1-F(s) \right) ^{n-1} f(\theta _i) ds d \theta _i.\quad \end{aligned}$$

(15)

Using integration by parts, we transform the bidding function $b(\cdot , \cdot )$ in Lemma 2 into

$$\begin{aligned} b(\theta _i, \theta )&= - \frac{1}{r} \ln \left\{ \int \limits _{\theta _i}^{\theta } \exp (- r s) \frac{f_{(n-2)} (s)}{1 - F_{(n-2)}(\theta _i)} d s {+} \int \limits _{\theta }^{\bar{\theta }} \exp (- r \theta ) \frac{f_{(n-2)} (s)}{1 - F_{(n-2)}(\theta _i)} d s \right\} \nonumber \\&= - \frac{1}{r} \ln \left\{ \int \limits _{\theta _i}^{\theta } \exp (- r s) \frac{f_{(n-2)} (s)}{1 - F_{(n-2)}(\theta _i)} d s + \exp (- r \theta ) \left( \frac{1 - F (\theta )}{1 - F(\theta _i)} \right) ^{n-2} \right\} \nonumber \\&= - \frac{1}{r} \ln \left\{ \exp (- r \theta _i) - r \int \limits _{\theta _i}^{\theta } \exp (- r s) \left( \frac{1 - F (s)}{1 - F(\theta _i)} \right) ^{n-2} d s \right\} . \end{aligned}$$

(16)

From Lemma 3 (ii) and Eq. (16), the supplier’s expected utility $EU_i^s$ in the bundling mechanism with subcontracting can be rewritten as

$$\begin{aligned}&\ 1 - E \bigl [ F(\theta _j) E_{\theta _i} [(1-F_{(n-2)}(\theta _i)) \exp ( -r (b(\theta _i, \theta _j)-\theta _i)) + F_{(n-2)}(\theta _i) \mid \theta _i \nonumber \\&\quad < \theta _j] + (1-F(\theta _j)) \bigr ] = \frac{1}{2} - \int \limits _{\underline{\theta }}^{\bar{\theta }} \int \limits _{\underline{\theta }}^{\theta _j} \left\{ (1-F_{(n-2)}(\theta _i)) \right. \nonumber \\&\qquad \left. \times \left[ 1 {-} r \int \limits _{\theta _i}^{\theta _j} \frac{\exp (-rs)}{\exp (-r \theta _i)} \left( \frac{1-F(s)}{1-F(\theta _i)} \right) ^{n-2} ds \right] {+} F_{(n-2)}(\theta _i) \right\} \ f(\theta _i) d \theta _i f(\theta _j) d \theta _j \nonumber \\&\quad = r \int \limits _{\underline{\theta }}^{\bar{\theta }} \int \limits _{\underline{\theta }}^{\theta _j} \int \limits _{\theta _i}^{\theta _j} \frac{\exp (-rs)}{\exp (-r \theta _i)} \left( 1-F(s) \right) ^{n-2} f(\theta _i) f(\theta _j) ds d \theta _i d \theta _j\nonumber \\&\quad = r \int \limits _{\underline{\theta }}^{\bar{\theta }} \int \limits _{\theta _i}^{\bar{\theta }} \frac{\exp (-rs)}{\exp (-r \theta _i)} \left( 1-F(s) \right) ^{n-1} f(\theta _i) ds d \theta _i. \end{aligned}$$

(17)

The first two equalities follows from $E[1-F(\theta _j)] = 1/2$, and the third equality follows from Fubini’s theorem. Hence, it follows from (15) and (17) that $EU_i^{**} = EU_i^s$.

(i) It follows from the first-order condition (10) and the definition of the first-stage loser’s risk premium $\rho _l$ that neither $\tilde{a}$ nor $\rho _l$ depends on the parameter $\lambda$ of the aggregate risk $\omega$. According to the definition of the second-order stochastic dominance, the first-stage winner’s risk premium $\rho _w$ increases in $\lambda$. Hence, both $EU_B^s$ and $W^s$ decrease in $\lambda$. Now, the above preliminary result implies that $EU_B^s > EU_B^{**}$ iff $W^s > W^{**}$. Therefore, we have a threshold $\hat{\lambda }$ such that $EU_B^s > EU_B^{**}$ iff $\lambda < \hat{\lambda }$, and $W^s > W^{**}$ iff $\lambda < \hat{\lambda }$.

Now, we claim that $\hat{\lambda } > 0$ in the limit as $\kappa \rightarrow 0$. From Eq. (16), it is easy to see that $b(\theta _i, \theta )$ increases in $\theta _i$ and $b(\theta , \theta ) = \theta$ for each $\theta$. Using this fact, a similar argument to that for the proof of Theorem 1 shows that as $\kappa \rightarrow 0$ and $\lambda \rightarrow 0$, the difference between $W^s$ and the first-best utility converges to zero. The difference between $W^{**}$ and the first-best utility is strictly smaller than zero, even in the limit, because $0 < \tilde{a}$. Hence, it holds that $\hat{\lambda } > 0$ in the limit as $\kappa \rightarrow 0$.

Next, we claim that $\hat{\lambda } < 1$ in the limit as $\delta \rightarrow 0$ and $n \rightarrow \infty$. From Eq. (16), Lebesgue’s dominated convergence theorem implies that as the number of suppliers, $n$, goes to infinity, the term $b(\theta _i, \theta )$ converges to $\theta _i$ for any $\theta _i$ and $\theta$ with $\theta _i \le \theta$. In the limit as $n \rightarrow \infty$, subcontracting occurs with probability one, and the subcontracting cost is $\underline{\theta } + \omega - \delta \tilde{a}$. Hence, as $n \rightarrow \infty ,\rho _l \rightarrow 0$ and $\rho _w \rightarrow \frac{1}{r} \ln E[\exp (r \omega )]$, so that $EU_B^s$ converges to $v - \{ \psi (\tilde{a}) - \delta \tilde{a} + \underline{\theta } + \frac{1}{r} \ln E[\exp (r \omega )]\}$. The proof of Theorem 1 shows that $EU_B^{**}$ converges to $v - \{ \psi (0) + \underline{\theta } \}$ as $n \rightarrow \infty$. Also, it follows from the first-order condition (10) that $\tilde{a} \rightarrow 0$ as $\delta \rightarrow 0$. Since $\frac{1}{r} \ln E[\exp (r \omega )] > 0$, it holds that $\hat{\lambda } = 0$ for sufficiently large $n$ and small $\delta$. Thus, this establishes the claim.

(ii) The envelope theorem implies that $\frac{\partial EU_B^s}{\partial \delta } = \frac{\partial W^s}{\partial \delta } = \tilde{a}$. It follows from the proof of Proposition 3 that $\frac{\partial EU_B^{**}}{\partial \delta } = \frac{\partial W^{**}}{\partial \delta } = 0$. Therefore, we can obtain the threshold $\hat{\delta }$ in the same way as part (i) of this proposition. $\square$