Incompetence and corruption in procurement auctions

Abstract

This paper analyzes the impact of incompetence and corruption in procurements. The government conducts a procurement auction where a firm that quotes the lowest bid wins the contract to construct a public good (subject to the fulfillment of the minimum quality requirement). Incompetence leads to measurement errors. There is also corruption in the system: if the measured quality falls short of the minimum stipulated level, the firm can pay a bribe to inflate the reported quality. We show that higher levels of corruption parameters unambiguously reduce the actual quality produced and the expected welfare. The effects of greater incompetence are more complicated. We show greater incompetence may lead to an increase in both equilibrium quality and welfare. This is counter-intuitive and goes against conventional wisdom. We also demonstrate that the winning firm chooses a quality that is strictly lower than the welfare-maximizing quality.

Appendix

Proof of Proposition 1

Note that from the first order condition (5) we get

$$\begin{aligned}{} & {} -c_{q}(q,\, \theta )+\frac{1}{2\varepsilon \left[ \alpha +\beta \varepsilon \right] }\left( \frac{k}{q-\varepsilon }-1\right) =0\\{} & {} \quad \Rightarrow k-\left( q-\varepsilon \right) \left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] =0 \end{aligned}$$

Write the above equation as $J\left( q;\theta ,k,\varepsilon \right) =0$. That is,

$$\begin{aligned} J\left( q;\; \theta ,k,\varepsilon \right) =k-\left( q-\varepsilon \right) \left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] =0 \end{aligned}$$

(13)

Note that

$$\begin{aligned} J_{q}\left( .\right) =\frac{\partial J\left( .\right) }{\partial q}=-\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta ) \right] -\left( q-\varepsilon \right) c_{qq}\left( .\right) <0 \end{aligned}$$

(i) Note that we get the optimal quality, $q^{*}$, by solving $J\left( q;\theta ,k,\varepsilon \right) =0$ (see 13). Using implicit function theorem, we get

$$\begin{aligned} \frac{\partial q^{*}}{\partial \theta }=-\frac{J_{\theta }}{J_{q}} \end{aligned}$$

Since $J_{q}<0$, $\frac{\partial q^{*}}{\partial \theta }$ has the same sign as $J_{\theta }$. Now, since $c_{q\theta }\left( .\right) \ge 0$ (assumption 1), we have the following.

$$\begin{aligned} J_{\theta }\left( .\right) =\frac{\partial J\left( .\right) }{\partial \theta }=-2\left( q-\varepsilon \right) \varepsilon \left( \alpha +\beta \varepsilon \right) c_{q\theta }(q,\theta )\le 0 \end{aligned}$$

Hence, $\frac{\partial q^{*}}{\partial \theta }\le 0$. Note the following.

$$\begin{aligned} J_{\alpha }= & {} -2\varepsilon \left( q-\varepsilon \right) c_{q}(q,\theta )<0 \\ J_{\beta }= & {} -2\varepsilon ^{2}\left( q-\varepsilon \right) c_{q}(q,\theta )<0 \end{aligned}$$

Consequently, using a similar logic as above we get that $\frac{\partial q^{*}}{\partial \alpha },\ \frac{\partial q^{*}}{\partial \beta }<0$.

(ii) Note that

$$\begin{aligned} \frac{\partial q^{*}}{\partial k}=-\frac{J_{k}}{J_{q}} \end{aligned}$$

Now,

$$\begin{aligned} J_{k}\left( .\right) =\frac{\partial J\left( .\right) }{\partial k}=1>0\text { .} \end{aligned}$$

Hence, $\frac{\partial q^{*}}{\partial k}>0$.

(iii) Note that since $J_{q}<0$, $\frac{\partial q^{*}}{\partial \varepsilon }$ has the same sign as $J_{\varepsilon }$. Now,

$$\begin{aligned} J_{\varepsilon }\left( .\right) =\frac{\partial J\left( .\right) }{\partial \varepsilon }=\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] -2\left( q-\varepsilon \right) c_{q}(q,\theta )\left[ \alpha +2\varepsilon \beta \right] \end{aligned}$$

Using $J\left( q;\theta ,k,\varepsilon \right) =0$ (from (13)) we get $\left( q-\varepsilon \right) =\frac{k}{\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] }$. Hence,

$$\begin{aligned} J_{\varepsilon }\left( .\right) =\frac{\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] ^{2}-2kc_{q}(q,\theta ) \left[ \alpha +2\varepsilon \beta \right] }{\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] } \end{aligned}$$

(14)

Note that since $\alpha >0$, when k is small enough, the term, $2kc_{q}(q,\theta )\left[ \alpha +2\varepsilon \beta \right]$, will be small enough. This implies $\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] ^{2}-2kc_{q}(q,\theta )\left[ \alpha +2\varepsilon \beta \right] >0$. Consequently, using (14) we get that for small enough k, $J_{\varepsilon }\left( .\right) >0$. Hence, $\frac{ \partial q^{*}}{\partial \varepsilon }>0$.

(iv) Since $\beta \ge 0$, from (14) we get

$$\begin{aligned} J_{\varepsilon }\left( .\right) \le \frac{\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] ^{2}-2kc_{q}(q, \theta )\alpha }{\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta )\right] } \end{aligned}$$

(15)

Note that $\alpha >0$, $q_{k}^{*}>0$ and $c_{qq}\left( .\right) \ge 0$. Hence, if $\varepsilon$ is small enough and k is large enough, $\left[ 1+2\varepsilon \left( \alpha +\beta \varepsilon \right) c_{q}(q,\theta ) \right] ^{2}-2kc_{q}(q,\theta )\alpha <0$. Consequently, using (15) we get that if $\varepsilon$ small enough and k is large enough then $J_{\varepsilon }\left( .\right) <0$. Hence, $\frac{\partial q^{*}}{ \partial \varepsilon }<0$.$\square$

Proof of Proposition 2

Note that

$$\begin{aligned} p(\theta )= & {} \int _{q^{*}\left( \theta \right) -\varepsilon }^{k}\frac{1 }{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{T}-1\right) \frac{1}{2\varepsilon }dT+c(q^{*}\left( \theta \right) ,\theta )+I(\theta ) \\ \text {where }I(\theta )= & {} \int _{\theta }^{{\overline{\theta }}}c_{\theta }\left( q^{*}\left( u\right) ,u\right) \frac{(1-F(u))^{n-1}}{(1-F(\theta ))^{n-1}}du. \end{aligned}$$

Note that

$$\begin{aligned} I^{\prime }\left( \theta \right)= & {} \left[ \begin{array}{c} -c_{\theta }(q^{*}\left( \theta \right) ,\theta ) \\ +\int _{\theta }^{{\overline{\theta }}}\frac{\left( n-1\right) f\left( \theta \right) \left( 1-F\left( \theta \right) \right) ^{n-2}c_{\theta }\left( q^{*}\left( u\right) ,u\right) (1-F(u))^{n-1}}{(1-F(\theta ))^{2n-2}}du \end{array} \right] \nonumber \\= & {} \left[ \begin{array}{c} -c_{\theta }(q^{*}\left( \theta \right) ,\theta ) \\ +\int _{\theta }^{{\overline{\theta }}}\frac{\left( n-1\right) f\left( \theta \right) c_{\theta }\left( q^{*}\left( u\right) ,u\right) (1-F(u))^{n-1}}{ (1-F(\theta ))^{n}}du \end{array} \right] \nonumber \\= & {} \left[ \begin{array}{c} -c_{\theta }(q^{*}\left( \theta \right) ,\theta ) \\ +\left( n-1\right) \frac{f\left( \theta \right) }{1-F\left( \theta \right) } \int _{\theta }^{{\overline{\theta }}}c_{\theta }\left( q^{*}\left( u\right) ,u\right) \frac{(1-F(u))^{n-1}}{(1-F(\theta ))^{n-1}}du \end{array} \right] \nonumber \\= & {} -c_{\theta }(q^{*}\left( \theta \right) ,\theta )+\left( n-1\right) \frac{f\left( \theta \right) }{1-F\left( \theta \right) }I\left( \theta \right) \end{aligned}$$

(16)

Differentiating $p(\theta )$ w.r.t. $\theta$ we get

$$\begin{aligned} p^{\prime }\left( \theta \right)= & {} \left[ \begin{array}{c} -\frac{\partial q^{*}\left( .\right) }{\partial \theta }\frac{1}{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{q^{*}\left( \theta \right) -\varepsilon }-1\right) \frac{1}{2\varepsilon } \\ +c_{q}(q^{*}\left( \theta \right) ,\theta )\frac{\partial q^{*}\left( .\right) }{\partial \theta }+c_{\theta }(q^{*}\left( \theta \right) ,\theta )+I^{\prime }(\theta ) \end{array} \right] \nonumber \\= & {} \frac{\partial q^{*}\left( .\right) }{\partial \theta }\left[ \begin{array}{c} c_{q}(q^{*}\left( \theta \right) ,\theta ) \\ -\frac{1}{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{q^{*}\left( \theta \right) -\varepsilon }-1\right) \frac{1}{2\varepsilon } \end{array} \right] +\left[ \begin{array}{c} c_{\theta }(q^{*}\left( \theta \right) ,\theta ) \\ +I^{\prime }(\theta ) \end{array} \right] \end{aligned}$$

(17)

From (7) we know that at $q^{*}\left( \theta \right)$ we have

$$\begin{aligned} c_{q}(q^{*}\left( \theta \right) ,\theta )-\frac{1}{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{q^{*}\left( \theta \right) -\varepsilon }-1\right) \frac{1}{2\varepsilon }=0. \end{aligned}$$

And, from (16) we have

$$\begin{aligned} c_{\theta }(q^{*}\left( \theta \right) ,\theta )+I^{\prime }(\theta )=\left( n-1\right) \frac{f\left( \theta \right) }{1-F\left( \theta \right) } I\left( \theta \right) . \end{aligned}$$

Using the above two in (17) we get

$$\begin{aligned} p^{\prime }(\theta )=\left( n-1\right) \frac{f\left( \theta \right) }{ 1-F\left( \theta \right) }I\left( \theta \right) \end{aligned}$$

(18)

Thus, $p^{\prime }(\theta )>0$. Let firms 2, 3, ...n adopt the strategy p(.) given as above. Note that $p(\theta _{j})\in [p(\underline{ \theta }),p({\overline{\theta }})]$, $\forall j\ne 1$. Firm 1’s type is $\theta _{1}=\theta$ and let it choose $p_{1}=p\left( z\right)$. Firm 1 will win if and only if $p(z)<min\left\{ p(\theta _{2}),p(\theta _{3}).....,p(\theta _{n})\right\}$. Since $p^{\prime }(.)>0$, firm 1 wins iff $z<min\left\{ \theta _{2},\theta _{3.}....,\theta _{n}\right\}$. Probability that 1 wins is therefore $(1-F(z))^{n-1}$. Firm 1’s expected payoff when other firms are following the above mentioned strategy is as follows.

$$\begin{aligned} \Pi _{1}(\theta ,z)=(1-F(z))^{n-1}\left[ p(z)-\left( \begin{array}{c} c(q^{*}\left( \theta \right) ,\theta ) \\ +\int _{q^{*}\left( \theta \right) -\varepsilon }^{k}\frac{1}{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{T}-1\right) \frac{1}{ 2\varepsilon }dT \end{array} \right) \right] \end{aligned}$$

(19)

Note that $\left( c(q^{*}\left( \theta \right) ,\theta )+\int _{q^{*}\left( \theta \right) -\varepsilon }^{k}\frac{1}{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{T}-1\right) \frac{1}{2\varepsilon } dT\right)$ is the actual expected cost of type $\theta$ (incurred in stages 2 and 3). p(z) is the price quoted by type $\theta$ when it is mimicking type z.

$$\begin{aligned} p\left( z\right)= & {} \int _{q^{*}\left( z\right) -\varepsilon }^{k}\frac{1 }{\left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{T}-1\right) \frac{1}{2\varepsilon }dT+c(q^{*}\left( z\right) ,z)+I(z) \\ \text {where }I(z)= & {} \int _{z}^{{\overline{\theta }}}c_{\theta }\left( q^{*}\left( u\right) ,u\right) \frac{(1-F(u))^{n-1}}{(1-F(z))^{n-1}}du. \end{aligned}$$

Also note the following.

$$\begin{aligned} c(q^{*}\left( \theta \right) ,\theta )+\int _{q^{*}\left( \theta \right) -\varepsilon }^{k}\frac{1}{\left( \alpha +\beta \varepsilon \right) } \left( \frac{k}{T}-1\right) \frac{1}{2\varepsilon }dT=p\left( \theta \right) -I\left( \theta \right) \end{aligned}$$

Hence,

$$\begin{aligned} \Pi _{1}(\theta ,z)=(1-F(z))^{n-1}\left[ p\left( z\right) -\left( p\left( \theta \right) -I\left( \theta \right) \right) \right] \end{aligned}$$

Now,

$$\begin{aligned} \frac{\partial }{\partial z}\left[ \Pi _{1}(\theta ,z)\right] =\left[ \begin{array}{c} -(n-1)(1-F(z))^{n-2}f(z)\left[ p\left( z\right) -\left( p\left( \theta \right) -I\left( \theta \right) \right) \right] \\ +\left( 1-F\left( z\right) \right) ^{n-1}p^{\prime }\left( z\right) \end{array} \right] \end{aligned}$$

(20)

Now, using (18) we get

$$\begin{aligned} p^{\prime }(z)=\left( n-1\right) \frac{f\left( z\right) }{1-F\left( z\right) }I\left( z\right) \end{aligned}$$

Using the above in (20) we have

$$\begin{aligned} \frac{\partial }{\partial z}\left[ \Pi _{1}(\theta ,z)\right]= & {} \left[ \begin{array}{c} -(n-1)(1-F(z))^{n-2}f(z)\left[ p\left( z\right) -\left( p\left( \theta \right) -I\left( \theta \right) \right) \right] \\ +(n-1)(1-F(z))^{n-2}f(z)I\left( z\right) \end{array} \right] \nonumber \\= & {} (n-1)(1-F(z))^{n-2}f(z)\left[ \begin{array}{c} \left\{ p\left( \theta \right) -I\left( \theta \right) \right\} \\ -\left\{ p\left( z\right) -I\left( z\right) \right\} \end{array} \right] \end{aligned}$$

(21)

From (21) we know that since $(n-1)(1-F(z))^{n-2}f(z)>0$, $\frac{\partial }{ \partial z}\left[ \Pi _{1}(\theta ,z)\right]$ has the same sign as $\left[ \left\{ p\left( \theta \right) -I\left( \theta \right) \right\} -\left\{ p\left( z\right) -I\left( z\right) \right\} \right]$. Let

$$\begin{aligned} X\left( \theta ,z\right) =\left[ p\left( \theta \right) -I\left( \theta \right) \right] -\left[ p\left( z\right) -I\left( z\right) \right] \end{aligned}$$

(22)

For any given $\theta$, we have

$$\begin{aligned} \frac{\partial X\left( .\right) }{\partial z}=-\left[ p^{\prime }\left( z\right) -I^{\prime }\left( z\right) \right] \end{aligned}$$

(23)

Using (16) and (18) we have

$$\begin{aligned} p^{\prime }\left( z\right) -I^{\prime }\left( z\right) =c_{\theta }(q^{*}\left( z\right) ,z)>0 \end{aligned}$$

(24)

Using (24) in (23) we get that

$$\begin{aligned} \frac{\partial X\left( .\right) }{\partial z}=-c_{\theta }(q^{*}\left( z\right) ,z)<0 \end{aligned}$$

(25)

That is, for any given $\theta$, $X\left( .\right)$ is strictly decreasing in z. From (22) we get that

$$\begin{aligned} z=\theta \Rightarrow X=0 \end{aligned}$$

(26)

(25) and (26) together mean that

$$\begin{aligned} z<\theta \Rightarrow X>0\text { and }z>\theta \Rightarrow X<0\text {.} \end{aligned}$$

Hence, since $\frac{\partial }{\partial z}\left[ \Pi _{1}(\theta ,z)\right]$ has the same sign as X we have

$$\begin{aligned} \frac{\partial }{\partial z}\left[ \Pi _{1}(\theta ,z)\right] \ \left\{ \begin{array}{ll}>0&{}\quad \text {if }z<\theta \\ =0&{}\quad \text {if }z=\theta \\ <0&{}\quad \text {if }z>\theta \end{array} \right. \end{aligned}$$

Thus, $\Pi _{1}(\theta ,z)$ achieves its maximum value at $z=\theta$. That is, $\Pi _{1}(\theta ,\theta )>\Pi _{1}(\theta ,z)$ for all $z\ne \theta$. This means that when all other firms choose $p\left( .\right)$, for firm 1 whose type is $\theta$, is better off by quoting a price $p(\theta )$ than quoting any other price p(z) where $z\ne \theta$. Consequently, the given strategy profile is a Bayesian Nash equilibrium.$\square$

Proof of Proposition 3

From (7) we get that for the winner with type $\theta$, the optimal quality, $q^{*}\left( \theta ,\alpha ,\beta ,k\right)$ chosen is s.t

$$\begin{aligned} c_{q}\left( q^{*},\theta \right) =\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{q^{*}-\varepsilon }-1\right) \end{aligned}$$

Now note that

$$\begin{aligned}{} & {} v_{q}\left( q^{*}\right) -c_{q}\left( q^{*},\theta \right) \nonumber \\{} & {} \quad =v_{q}\left( q^{*}\right) -\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left( \frac{k}{q^{*}-\varepsilon }-1\right) \nonumber \\{} & {} \quad =v_{q}\left( q^{*}\right) +\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left( \frac{q^{*}-\varepsilon -k}{q^{*}-\varepsilon }\right) \end{aligned}$$

(27)

Note that $q^{*}\in \left( {\underline{q}},k+\varepsilon \right]$ (see remark 6 in section 3). Also note that $\left( \frac{q^{*}-\varepsilon -k }{q^{*}-\varepsilon }\right)$ is strictly increasing in $q^{*}$. So we get that

$$\begin{aligned}{} & {} v_{q}\left( q^{*}\right) +\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left( \frac{q^{*}-\varepsilon -k}{q^{*}-\varepsilon }\right) \nonumber \\{} & {} \quad >v_{q}\left( q^{*}\right) +\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left( \frac{{\underline{q}}-\varepsilon -k}{{\underline{q}} -\varepsilon }\right) \nonumber \\{} & {} \quad =v_{q}\left( q^{*}\right) -\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left[ \frac{k}{{\underline{q}}-\varepsilon }-1\right] \end{aligned}$$

(28)

Since $v_{qq}\le 0$ (assumption 2)and $q^{*}\in \left( {\underline{q}},k+\varepsilon \right)$ (see remark 6 in section 3) we have the following.

$$\begin{aligned} v_{q}\left( k+\varepsilon \right) \ge \frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left[ \frac{k}{{\underline{q}}-\varepsilon }-1 \right] \Rightarrow v_{q}\left( q^{*}\right) -\frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left[ \frac{k}{{\underline{q}} -\varepsilon }-1\right] \ge 0 \end{aligned}$$

(29)

Using (27), (28) and (29) we get the following.

$$\begin{aligned} \forall \theta \in \left[ {\underline{\theta }},{\bar{\theta }}\right] ,\ v_{q}\left( q^{*}\right) -c_{q}\left( q^{*},\theta \right) >0 \end{aligned}$$

(30)

Thus,

$$\begin{aligned} \int _{{\underline{\theta }}}^{{\bar{\theta }}}\left[ v_{q}(q^{*})-c_{q}(q^{*},\theta )\right] f_{1}(\theta )d\theta >0 \end{aligned}$$

(31)

Using (9) and (31) we get

$$\begin{aligned} EW_{q}\left( q^{*}\right) >0 \end{aligned}$$

(32)

From (11) we know that

$$\begin{aligned} EW_{q}\left( q_{w}^{*}\right) =\int _{{\underline{\theta }}}^{{\bar{\theta }}} \left[ v_{q}(q_{w}^{*})-c_{q}(q_{w}^{*},\theta )\right] f_{1}(\theta )d\theta =0 \end{aligned}$$

(33)

Since $EW_{q}\left( q^{*}\right) >0$ (see 32), $EW_{q}\left( q_{w}^{*}\right) =0$ (see (33)) and $EW_{qq}<0$ (see 10)), we have $\forall \theta \in \left[ {\underline{\theta }},{\bar{\theta }}\right]$, $q^{*}<q_{w}^{*}$. $\square$

Proof of Proposition 4

Note that the expected equilibrium welfare is given as follows (from (12)):

$$\begin{aligned} EW\left( q^{*}\right) =\int _{{\underline{\theta }}}^{{\bar{\theta }} }[v(q^{*})-c(q^{*},\theta )]f_{1}(\theta )d\theta \end{aligned}$$

(i) Now

$$\begin{aligned} \frac{\partial EW\left( q^{*}\right) }{\partial k}=\int _{\underline{ \theta }}^{{\overline{\theta }}}[\left\{ (v_{q}(q^{*})-c_{q}(q^{*},\theta )\right\} q_{k}^{*}]f_{1}(\theta )d\theta \end{aligned}$$

(34)

Note that we have

$$\begin{aligned} v_{q}\left( k+\varepsilon \right) \ge \frac{1}{2\varepsilon \left( \alpha +\beta \varepsilon \right) }\left[ \frac{k}{{\underline{q}}-\varepsilon }-1 \right] \text {.} \end{aligned}$$

Using (30) we have $\forall \theta \in \left[ {\underline{\theta }},\bar{\theta }\right] ,\ v_{q}\left( q^{*}\right) -c_{q}\left( q^{*},\theta \right) >0$. From proposition 1(ii) we know that $q_{k}^{*}>0$. Hence, using (34) we get $\frac{\partial EW\left( q^{*}\right) }{\partial k}>0$.

(ii) Note that from Proposition 1(i) we also know that $q_{\alpha }^{*}<0$ and $q_{\beta }^{*}<0$. Using a similar logic as above we get that $\frac{\partial EW\left( q^{*}\right) }{\partial \alpha }<0$ and $\frac{ \partial EW\left( q^{*}\right) }{\partial \beta }<0.$

(iii) Now

$$\begin{aligned} \frac{\partial EW(q^{*})}{\partial \varepsilon }=\int _{\underline{\theta }}^{{\overline{\theta }}}\left[ \left\{ (v_{q}(q^{*})-c_{q}(q^{*},\theta )\right\} q_{\varepsilon }^{*}\right] f_{1}(\theta )d\theta \end{aligned}$$

(35)

From Proposition 1 (iii) we know that if k is small enough, then $q_{\varepsilon }^{*}>0$. Using a logic similar as above, and using (30) and (35) we get that if k is small enough, $\frac{\partial EW\left( q^{*}\right) }{\partial \varepsilon }>0$.

(iv) From Proposition 1(iv) we know that if $\varepsilon$ is small enough and k is large enough, then $q_{\varepsilon }^{*}<0$. Again, using a similar logic as above, we get that if $\varepsilon$ small enough and k is large enough, then $\frac{\partial EW\left( q^{*}\right) }{\partial \varepsilon }<0$. $\square$