Comparing first and second price auctions with asymmetric bidders

Abstract

We study procurement auctions in which, as is common in practice, a group of sellers (incumbents, qualified bidders) is given an advantage, based, for example, on better reliability, quality, or incumbency status. We show conditions under which for any given first price handicap auction, there is a simple second-price design which dominates it. This generalizes a previous result for the case of an auction with one insider and one outsider (Mares and Swinkels in J Econ Theory, 2013) and sharpens our understanding of the classical result by Maskin and Riley (Rev Econ Stud 67:413–438, 2000).

Appendix

Proof of Lemma 1

Note first that because there are no atoms and $\beta _{I}(c)\le 1,$ a bid of $1-A$ or above gives zero profit to an outsider as does a bid $\beta _{O}(c)=c$. But any bid $b\in (c,1-A)$ makes $b-c>0$ when it wins, and at a minimum wins when $c_{i}\in (b+A,1]$ for all incumbents and $c_{j}\in (b,1]$ for all other outsiders. The fact that $\phi (1-A)=1$ follows directly.

Next, let us show that $\beta _{I}\left( 0\right) \le \beta _{0}\left( 0\right) +A,$ so that $\phi \left( 0\right) \ge 0$. Assume that this is false. It must then be that $n_{O}>1,$ else the (single) outsider is submitting bids strictly below where he is already winning for sure. Let $R$ be the support of bids submitted by $O$ that are below $\beta _{I}\left( 0\right) -A$. Standard arguments tell us that bids are non-atomic on $R,$ that $R$ is an interval, and that the upper bound $\bar{r}$ of $R$ is equal to $\beta _{I}\left( 0\right) -A$ (since otherwise, there is an effective gap in the support of bids). Equivalently, $\beta _{O}$ is continuous and strictly increasing on $\beta _{O}^{-1}\left( R\right)$. For any given $b\in R,$ let $\bar{H}\left( b\right) =\bar{F}\left( \beta _{O}^{-1}\left( b\right) \right)$ be the probability that any given outsider bids above $b,$ and let $h\left( \cdot \right)$ be the associated density. The profits of an outsider with cost $c$ and bid $b$ can be expressed as

$$\begin{aligned} \Pi _{O}\left( c,b\right) =\bar{H}^{n_{O}-1}\left( b\right) \left( b-c\right) , \end{aligned}$$

and so

$$\begin{aligned} \frac{\partial }{\partial b}\Pi _{O}\left( c,b\right) =\left( n_{O}-1\right) \bar{H}^{n_{O}-2}\left( b\right) \left( -h\left( b\right) \right) \left( b-c\right) +\bar{H}^{n_{O}-1}\left( b\right) , \end{aligned}$$

which has the same sign as

$$\begin{aligned} -\left( n_{O}-1\right) h\left( b\right) \left( b-c\right) +\bar{H}\left( b\right) . \end{aligned}$$

Evaluated at $c=\beta _{O}^{-1}\left( b\right) ,$ this equals 0.

But, if we let $\Pi _{I}\left( c,b\right)$ be analogously defined as the profits of an incumbent, we have (for $b\in \,R$) that

$$\begin{aligned} \Pi _{I}\left( 0,b+A\right) =\bar{H}^{n_{O}}\left( b\right) \left( b+A\right) , \end{aligned}$$

and so

$$\begin{aligned} \frac{\partial }{\partial b}\Pi _{O}\left( 0,b+A\right) =n_{O}\bar{H} ^{n_{O}-1}\left( b\right) \left( -h\left( b\right) \right) \left( b+A\right) +\bar{H}^{n_{O}}\left( b\right) , \end{aligned}$$

which has the same sign as

$$\begin{aligned} -n_{O}h\left( b\right) \left( b+A\right) +\bar{H}\left( b\right) . \end{aligned}$$

But then, since $\left( b+A\right) >b>b-\beta _{O}^{-1}\left( b\right) ,$

$$\begin{aligned} \frac{\partial }{\partial b}\Pi _{O}\left( 0,b+A\right) <0. \end{aligned}$$

As this holds for all $b\in R,$ when $I$ has cost 0, his profits strictly increase by lowering his bid a little, and by continuity, this holds for all $c$ sufficiently close to 0 as well. Hence, $\beta _{I}$ is not optimal, a contradiction. But then, note that by (1)

$$\begin{aligned} \beta _{I}\left( \phi \left( 0\right) \right) =\beta _{I}\left( 0\right) +A\ge \beta _{I}\left( 0\right) \end{aligned}$$

and hence $\phi \left( 0\right) \ge 0$ since $\beta _{I}$ is increasing.

Thus, since $\beta _{I}\left( \cdot \right)$ and $\beta _{O}\left( \cdot \right)$ are continuous and increasing, (1) is well defined for $c\in \left[ 0,1-A\right]$. $\square$

Proof of Lemma 2

Let us show first that if $F$ and $G$ are concave positive functions and if $\alpha \ge 0$ and $\beta \ge 0$ satisfy $\alpha +\beta \le 1,$ then $H=F^{\alpha }G^{\beta }$ is concave. The claimed result then follows since by definition, $F^{\rho }$ and $G^{\rho }$ are concave, and so $H^{\rho }$ is concave.

Observe that

$$\begin{aligned} \frac{H^{\prime }}{H}=\underset{\equiv J}{\underbrace{\alpha \frac{F^{\prime }}{F}+\beta \frac{G^{\prime }}{G}}} \end{aligned}$$

and thus, since $H^{\prime }=HJ,$

$$\begin{aligned} H^{\prime \prime }&= \left( HJ\right) ^{\prime } \\&= H^{\prime }J+HJ^{\prime } \\&= HJ^{2}+HJ^{\prime } \end{aligned}$$

which has the same sign as $J^{2}+J^{\prime }$. But

$$\begin{aligned} J^{\prime }=\alpha \left( \frac{F^{\prime \prime }}{F}-\left( \frac{ F^{\prime }}{F}\right) ^{2}\right) +\beta \left( \frac{G^{\prime \prime }}{G} -\left( \frac{G^{\prime }}{G}\right) ^{2}\right) \end{aligned}$$

and thus $J^{2}+J^{\prime }$ is equal to

$$\begin{aligned}&\left( \alpha \frac{F^{\prime }}{F}+\beta \frac{G^{\prime }}{G}\right) ^{2}+\alpha \left( \frac{F^{\prime \prime }}{F}-\left( \frac{F^{\prime }}{F} \right) ^{2}\right) +\beta \left( \frac{G^{\prime \prime }}{G}-\left( \frac{ G^{\prime }}{G}\right) ^{2}\right) \\&\qquad =\alpha \frac{F^{\prime \prime }}{F}+\beta \frac{G^{\prime \prime }}{G} +\alpha ^{2}\left( \frac{F^{\prime }}{F}\right) ^{2}+2\alpha \beta \frac{ F^{\prime }}{F}\frac{G^{\prime }}{G}+\beta ^{2}\left( \frac{G^{\prime }}{G}\right) ^{2} \\&\qquad \quad -\alpha \left( \frac{F^{\prime }}{F}\right) ^{2}-\beta \left( \frac{ G^{\prime }}{G}\right) ^{2} \\&\qquad \le \alpha (\alpha -1)\left( \frac{F^{\prime }}{F}\right) ^{2}+2\alpha \beta \frac{F^{\prime }}{F}\frac{G^{\prime }}{G}+\beta \left( \beta -1\right) \left( \frac{G^{\prime }}{G}\right) ^{2} \end{aligned}$$

since $F^{\prime \prime },G^{\prime \prime }\le 0$.

Since $\alpha +\beta \le 1,\,\alpha -1\le -\beta ,$ and so

$$\begin{aligned} \alpha (\alpha -1)\le -\alpha \beta , \end{aligned}$$

and similarly,

$$\begin{aligned} \beta \left( \beta -1\right) \le -\alpha \beta . \end{aligned}$$

Hence,

$$\begin{aligned} J^{2}+J^{\prime }&\le -\alpha \beta \left( \left( \frac{F^{\prime }}{F} \right) ^{2}-2\frac{F^{\prime }}{F}\frac{G^{\prime }}{G}+\left( \frac{G^{\prime }}{G}\right) ^{2}\right) \\&= -\alpha \beta \left( \frac{F^{\prime }}{F}-\frac{G^{\prime }}{G}\right) ^{2} \\&\le 0. \end{aligned}$$

Thus, $H^{\prime \prime }\le 0,$ and so $H$ is concave. $\square$

Proof of Proposition 1

Differentiate the second and third expressions of (5), and rearrange to get

$$\begin{aligned} 0=\bar{Q}_{O}\left( \phi \left( c\right) \right) \beta _{O}^{\prime }\left( c\right) -q_{O}\left( \phi \left( c\right) \right) \phi ^{\prime }\left( c\right) \left( \beta _{O}\left( c\right) -c\right) \end{aligned}$$

or

$$\begin{aligned} \beta _{O}^{\prime }\left( c\right) =\frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}\left( \phi \left( c\right) \right) }\phi ^{\prime }\left( c\right) \left( \beta _{O}\left( c\right) -c\right) . \end{aligned}$$

Thus,

$$\begin{aligned} \beta _{O}^{\prime }\left( c\right) =\phi ^{\prime }\left( c\right) S_{O}\left( c\right) \frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q} _{O}^{2}\left( \phi \left( c\right) \right) }. \end{aligned}$$

(12)

Analogously, from (7)

$$\begin{aligned} 0=\bar{Q}_{I}\left( \psi \left( c\right) \right) \beta _{I}^{\prime }\left( c\right) -q_{I}\left( \psi \left( c\right) \right) \psi ^{\prime }\left( c\right) \left( \beta _{I}\left( c\right) -c\right) \end{aligned}$$

or

$$\begin{aligned} \beta _{I}^{\prime }\left( c\right) =\frac{q_{I}\left( \psi \left( c\right) \right) }{\bar{Q}_{I}\left( \psi \left( c\right) \right) }\psi ^{\prime }\left( c\right) \left( \beta _{I}\left( c\right) -c\right) , \end{aligned}$$

and so

$$\begin{aligned} \beta _{I}^{\prime }\left( \phi \left( c\right) \right) =S_{I}\left( \phi \left( c\right) \right) \frac{q_{I}\left( c\right) }{\bar{Q}_{I}^{2}\left( c\right) }\frac{1}{\phi ^{\prime }\left( c\right) }. \end{aligned}$$

(13)

Differentiate (1) to get

$$\begin{aligned} \phi ^{\prime }\left( c\right) =\frac{\beta _{O}^{\prime }\left( c\right) }{ \beta _{I}^{\prime }\left( \phi \left( c\right) \right) }. \end{aligned}$$

Substituting (12) and (13) yields

$$\begin{aligned} \phi ^{\prime }\left( c\right) =\frac{\phi ^{\prime }\left( c\right) S_{O}\left( c\right) }{S_{I}\left( \phi \left( c\right) \right) \frac{1}{ \phi ^{\prime }\left( c\right) }}\left( \frac{\frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}^{2}\left( \phi \left( c\right) \right) }}{ \frac{q_{I}\left( c\right) }{\bar{Q}_{I}^{2}\left( c\right) }}\right) \end{aligned}$$

or,

$$\begin{aligned} \phi ^{\prime }\left( c\right) =\frac{S_{I}\left( \phi \left( c\right) \right) }{S_{O}\left( c\right) }\left( \frac{\frac{q_{I}\left( c\right) }{ \bar{Q}_{I}^{2}\left( c\right) }}{\frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}^{2}\left( \phi \left( c\right) \right) }}\right) . \end{aligned}$$

For the differentiability of $\beta _{I},\beta _{O}$ and $\phi$ follow the same argument as in Mares and Swinkels (2013). That $\phi ^{\prime }\left( c\right) >0$ on $[0,1-A)$ follows from (7) noting that the integrand is strictly positive at $\phi (c)<1$ and thus for some interval to the right of $\phi (c)$. $\square$

Proof of Lemma 3

Note that

$$\begin{aligned} \frac{\frac{q_{I}}{\bar{Q}_{I}}\left( c\right) }{\frac{q_{O}}{\bar{Q}_{O}} \left( \phi \left( c\right) \right) }&= \frac{-\frac{\partial }{\partial c} \log \bar{Q}_{I}\left( c\right) }{\left. -\frac{\partial }{\partial s}\log \bar{Q}_{O}\left( s\right) \right| _{s=\phi \left( c\right) }} \\&= \frac{n_{O}\frac{f}{\bar{F}}\left( c\right) +\left( n_{I}-1\right) \phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }{\left. \left( n_{O}-1\right) \psi ^{\prime }\left( s\right) \frac{f }{\bar{F}}\left( \psi \left( s\right) \right) +n_{I}\frac{f}{\bar{F}} \left( s\right) \right| _{s=\phi \left( c\right) }} \\&= \frac{n_{O}\frac{f}{\bar{F}}\left( c\right) +\left( n_{I}-1\right) \phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }{\left( n_{O}-1\right) \frac{1}{\phi ^{\prime }\left( c\right) } \frac{f}{\bar{F}}\left( c\right) +n_{I}\frac{f}{\bar{F}}\left( \phi \left( c\right) \right) } \\&= \phi ^{\prime }\left( c\right) \left( \frac{n_{O}\frac{f}{\bar{F}}\left( c\right) +\left( n_{I}-1\right) \phi ^{\prime }\left( c\right) \frac{f}{\bar{ F}}\left( \phi \left( c\right) \right) }{\left( n_{O}-1\right) \frac{f}{\bar{ F}}\left( c\right) +n_{I}\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}} \left( \phi \left( c\right) \right) }\right) \\&= \phi ^{\prime }\left( c\right) \left( 1+\frac{\frac{f}{\bar{F}}\left( c\right) -\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }{\left( n_{O}-1\right) \frac{f}{\bar{F}}\left( c\right) +n_{I}\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }\right) . \end{aligned}$$

$\square$

Proof of Corollary 1

Using (9) and (10),

$$\begin{aligned} 1=\frac{S_{I}\left( \phi \left( c\right) \right) }{S_{O}\left( c\right) } \frac{\bar{Q}_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{I}\left( c\right) }T\left( c\right) . \end{aligned}$$

But, from (7) and (5) we have

$$\begin{aligned} \frac{S_{I}(\phi (c))}{S_{O}(c)}\frac{\bar{Q}_{O}(\phi (c))}{\bar{Q}_{I}(c)}= \frac{\beta _{I}\left( \phi (c)\right) -\phi (c)}{\beta _{O}\left( c\right) -c}. \end{aligned}$$

while by (1)

$$\begin{aligned} \frac{\beta _{I}\left( \phi (c)\right) -\phi (c)}{\beta _{O}\left( c\right) -c}=\frac{\beta _{O}\left( c\right) +A-\phi (c)}{\beta _{O}\left( c\right) -c }=1+\frac{A+c-\phi (c)}{\beta _{O}\left( c\right) -c}. \end{aligned}$$

Combining these three expressions yields the result. $\square$

Proof of Proposition 2

We proceed through a series of lemmas. $\square$

Lemma 7

For any $c<1-A$ and $\phi \left( c\right) =c,\,\phi ^{\prime }\left( c\right) >1$.

Proof

Recall from Corollary 1 that

$$\begin{aligned} 1=\left( 1+\frac{\frac{f}{\bar{F}}\left( c\right) -\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }{\left( n_{O}-1\right) \frac{f}{\bar{F}}\left( c\right) +n_{I}\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }\right) \left( 1+\frac{A+c-\phi (c)}{\beta _{O}\left( c\right) -c}\right) , \end{aligned}$$

which when $\phi \left( c\right) =c$ reduces to

$$\begin{aligned} 1=\left( 1+\frac{1-\phi ^{\prime }\left( c\right) }{n_{O}-1+\left( n_{I}\right) \phi ^{\prime }\left( c\right) }\right) \left( 1+\frac{A}{\beta _{O}\left( c\right) -c}\right) . \end{aligned}$$

Since $\beta _{O}\left( c\right) \ge c$ for all $c,$ the last term is greater than 1. Thus it must be that

$$\begin{aligned} 1+\frac{1-\phi ^{\prime }\left( c\right) }{n_{O}-1+\left( n_{I}\right) \phi ^{\prime }\left( c\right) }<1 \end{aligned}$$

or

$$\begin{aligned} \phi ^{\prime }\left( c\right) -1>0. \end{aligned}$$

$\square$

Corollary 3

$\phi \left( c\right) >c$ for all $c>0$.

This is immediate, since by Lemma 1 $\phi \left( 0\right) \ge 0,$ and by Lemma 7 anytime that $\phi \left( c\right) =c,\,\phi \left( c\right) -c$ is strictly increasing.

Next, let us show that $\phi \left( c\right) <c+A$ for all $c$.

Lemma 8

$\phi \left( 0\right) \le A$.

Proof

Assume $\phi \left( 0\right) >A$. Choose $\hat{c} \in \left( A,\phi \left( 0\right) \right)$. For any $b\in \left( \beta _{I}\left( \hat{c}\right) ,\beta _{O}\left( 0\right) +A\right) ,$ let $\bar{H }\left( b\right) =\Pr \left( I\text { bids above }b\right) =\bar{F}\left( \beta _{I}^{-1}\left( b\right) \right)$. Fix any such $\tilde{b},$ and let $\tilde{c}=\beta _{I}^{-1}\left( \tilde{b}\right) >\hat{c}>A$. Note that for $b\in \left( \beta _{I}\left( \hat{c}\right) ,\beta _{O}\left( 0\right) +A\right) ,$

$$\begin{aligned} \Pi _{I}\left( c,b\right) =\bar{H}^{n_{I}-1}\left( b\right) \left( b-c\right) \end{aligned}$$

and so

$$\begin{aligned} \frac{\partial }{\partial b}\Pi _{I}\left( c,b\right) =\bar{H} ^{n_{I}-2}\left( b\right) \left( -\left( n_{I}-1\right) h\left( b\right) \left( b-c\right) +\bar{H}\left( b\right) \right) . \end{aligned}$$

Thus,

$$\begin{aligned} \left( -\left( n_{I}-1\right) h\left( \tilde{b}\right) \left( \tilde{b}- \tilde{c}\right) +\bar{H}\left( \tilde{b}\right) \right) =0, \end{aligned}$$

and so,

$$\begin{aligned} \left( -n_{I}h\left( \tilde{b}\right) \left( \tilde{b}-\tilde{c}\right) + \bar{H}\left( \tilde{b}\right) \right) <0, \end{aligned}$$

But then, for any $c<\hat{c}-A,\,\tilde{b}-A-c>\tilde{b}-\tilde{c}$ and so

$$\begin{aligned} \left( -n_{I}h\left( \tilde{b}\right) \left( \tilde{b}-A-c\right) +\bar{H} \left( \tilde{b}\right) \right) <0, \end{aligned}$$

and so

$$\begin{aligned} \frac{\partial }{\partial b}\Pi _{O}\left( c,\tilde{b}-A\right) <0. \end{aligned}$$

Since this is true for all $\tilde{b}\in \left( \beta _{I}\left( \hat{c} \right) ,\beta _{O}\left( 0\right) +A\right) ,$ and hence for all $\tilde{b} -A\in \left( \beta _{I}\left( \hat{c}\right) -A,\beta _{O}\left( 0\right) \right)$ But then, any outsider with cost $\hat{c}-A$ is better of with bids strictly below $\beta _{O}\left( 0\right)$ than with $\beta _{O}\left( 0\right) ,$ contradicting the optimality of $\beta _{O}$ for types near 0. $\square$

Lemma 9

For all $c\ge 0,\,\phi \left( c\right) <c+A$.

Proof

Recall from Corollary 1 that

$$\begin{aligned} 1=\left( 1+\frac{\frac{f}{\bar{F}}\left( c\right) -\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }{\left( n_{O}-1\right) \frac{f}{\bar{F}}\left( c\right) +n_{I}\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( \phi \left( c\right) \right) }\right) \left( 1+\frac{A+c-\phi (c)}{\beta _{O}\left( c\right) -c}\right) . \end{aligned}$$

When $\phi \left( c\right) =c+A,$ the second term is 1. Thus,

$$\begin{aligned} 1=1+\frac{\frac{f}{\bar{F}}\left( c\right) -\phi ^{\prime }\left( c\right) \frac{f}{\bar{F}}\left( c+A\right) }{n_{O}\frac{f}{\bar{F}}\left( c\right) +\left( n_{I}-1\right) \frac{f}{\bar{F}}\left( c+A\right) \phi ^{\prime }\left( c\right) } \end{aligned}$$

The denominator of the fraction is positive and so

$$\begin{aligned} \frac{f}{\bar{F}}\left( c\right) -\phi ^{\prime }\left( c\right) \frac{f}{ \bar{F}}\left( c+A\right) =0. \end{aligned}$$

Since $\frac{f}{\bar{F}}$ is increasing, it follows that $\phi ^{\prime }\left( c\right) <1$. Thus $\phi \left( c\right)$ crosses $c+A$ at most once and from above. Since by Lemma 8 $\phi \left( 0\right) \le A$ we have our desired result. $\square$

Proof of Lemma 4

Recall from Lemma 1 that $\phi \left( 0\right) \ge 0$. If $\phi \left( 0\right) =0,$ then (8) reduces to $\phi ^{\prime }=\frac{ S_{I}\left( 0\right) }{S_{O}\left( 0\right) }>1,$ since $S_{I}\left( 0\right) >S_{O}\left( 0\right)$.

So, assume that $\phi \left( 0\right) >0$. The proof will proceed in three steps.

• Step 1 Note that $\beta _{I}$ is differentiable on $[0,\phi \left( 0\right) )$ and on $(\phi \left( 0\right) ,1-A)$. The second part of this statement follows from Proposition 1. The first part is a direct consequence that for cost types below $\phi \left( 0\right)$ only incumbents are competing and thus $\beta _{I}$ follows the construction of a standard symmetric first price auction.

• Step 2 Let

  $$\begin{aligned} P_{I}\left( s\right) \equiv \bar{F}^{n_{I}-1}\left( s\right) \bar{F} ^{n_{O}}\left( \psi \left( s\right) \right) . \end{aligned}$$

  This is the probability that $I$ wins when be bids $\beta _{I}\left( s\right)$. Let

  $$\begin{aligned} \Pi _{I}\left( c,s\right) =P_{I}\left( s\right) \left( \beta _{I}\left( s\right) -c\right) \end{aligned}$$

  be the profit of $I$ with type $c$ when he bids $\beta _{I}\left( s\right)$. Let

  $$\begin{aligned} P_{O}\left( s\right) \equiv \bar{F}^{n_{I}}\left( s\right) \bar{F} ^{n_{O}-1}\left( \psi \left( s\right) \right) . \end{aligned}$$

  This is the probability that $O$ wins with bid $\beta _{I}\left( s\right) -A.$ Let

  $$\begin{aligned} \Pi _{O}\left( c,s\right) =P_{O}\left( s\right) \left( \beta _{I}\left( s\right) -A-c\right) \end{aligned}$$

  be the profit of $I$ with type $c$ when he bids $\beta _{I}\left( s\right) -A$.

• Step 3 Note that it cannot be that $\left[ \Pi _{O}\left( 0,s\right) \right] _{s}<0$ for all $s$ on some interval of the form $[\hat{c} ,\phi \left( 0\right) ),$ since if it was, then we would have

  $$\begin{aligned} \Pi _{O}\left( 0,\hat{c}\right) >\Pi _{O}\left( 0,\phi \left( 0\right) \right) . \end{aligned}$$

  But, noting that $\beta _{I}\left( \phi \left( 0\right) \right) -A=\beta _{O}\left( 0\right) ,$ the RHS is $O$’s equilibrium payoff with $c=0,$ a contradiction.

Form a sequence $\left\{ c^{k}\right\}$ with $c^{k}\rightarrow \phi \left( 0\right) ,$ such that $c^{k}<\phi \left( 0\right)$ and $\left. \left[ \Pi _{O}\left( 0,s\right) \right] _{s}\right| _{s=c^{k}}\ge 0$ for all $k$. Note that

$$\begin{aligned} \left. \left[ \Pi _{O}\left( 0,s\right) \right] _{s}\right| _{s=c^{k}} =\frac{P_{O}^{\prime }}{P_{O}}\left( s\right) +\frac{\beta _{I}^{\prime }\left( s\right) }{\beta _{I}\left( s\right) -A-c} \end{aligned}$$

(14)

which has the same sign as

$$\begin{aligned} \frac{P_{O}^{\prime }}{P_{O}}\left( s\right) \frac{\beta _{I}\left( s\right) -A-c}{\beta _{I}^{\prime }\left( s\right) }+1, \end{aligned}$$

and so, for all $k,$ we have

$$\begin{aligned} \frac{-P_{O}^{\prime }}{P_{O}}\left( c^{k}\right) \frac{\beta _{I}\left( c^{k}\right) -A}{\beta _{I}^{\prime }\left( c^{k}\right) }\le 1. \end{aligned}$$

(15)

Similarly,

$$\begin{aligned} \left[ \log \Pi _{I}\left( c,s\right) \right] _{s}=\frac{ P_{I}^{\prime }}{P_{I}}\left( s\right) +\frac{\beta _{I}^{\prime }\left( s\right) }{\beta _{I}\left( s\right) -c} \end{aligned}$$

(16)

which has the same sign as

$$\begin{aligned} \frac{P_{I}^{\prime }}{P_{I}}\left( s\right) \frac{\beta _{I}\left( s\right) -c}{\beta _{I}^{\prime }\left( s\right) }+1, \end{aligned}$$

and so, for all $k,$

$$\begin{aligned} \frac{-P_{I}^{\prime }}{P_{I}}\left( c^{k}\right) \frac{\beta _{I}\left( c^{k}\right) -c^{k}}{\beta _{I}^{\prime }\left( c^{k}\right) }=1. \end{aligned}$$

(17)

Comparing (15) and (17) ,

$$\begin{aligned} \frac{-P_{O}^{\prime }}{P_{O}}\left( c^{k}\right) \left( \beta _{I}\left( c^{k}\right) -A\right) \le \frac{-P_{I}^{\prime }}{P_{I}}\left( c^{k}\right) \left( \beta _{I}\left( c^{k}\right) -c\right) . \end{aligned}$$

(18)

But,

$$\begin{aligned} \frac{-P_{O}^{\prime }}{P_{O}}\left( s\right) =n_{I}\frac{f}{\bar{F}}\left( s\right) +\left( n_{O}-1\right) \frac{f}{\bar{F}}\left( s\right) \left( \psi \left( s\right) \right) \psi ^{\prime }\left( s\right) \end{aligned}$$

(19)

and

$$\begin{aligned} \frac{-P_{I}^{\prime }}{P_{I}}\left( s\right) =\left( n_{I}-1\right) \frac{f }{\bar{F}}\left( s\right) +n_{O}\frac{f}{\bar{F}}\left( s\right) \left( \psi \left( s\right) \right) \psi ^{\prime }\left( s\right) . \end{aligned}$$

(20)

Since $c^{k}<\phi \left( 0\right) ,\,\psi ^{\prime }\left( c^{k}\right) =0,$ and (18) reduces to

$$\begin{aligned} n_{I}\frac{f}{\bar{F}}\left( c^{k}\right) \left( \beta _{I}\left( c^{k}\right) -A\right) \le \left( n_{I}-1\right) \frac{f}{\bar{F}}\left( c^{k}\right) \left( \beta _{I}\left( c^{k}\right) -c^{k}\right) \end{aligned}$$

or

$$\begin{aligned} \frac{\beta _{I}\left( c^{k}\right) -c^{k}}{\beta _{I}\left( c^{k}\right) -A} \ge \frac{n_{I}}{n_{I}-1}. \end{aligned}$$

(21)

As this is true for all $k,$

$$\begin{aligned} \frac{\beta _{I}\left( \phi \left( 0\right) \right) -\phi \left( 0\right) }{ \beta _{I}\left( \phi \left( 0\right) \right) -A}\ge \frac{n_{I}}{n_{I}-1}. \end{aligned}$$

(22)

Now, form a sequence $\left\{ c^{k}\right\}$ with $c^{k}\rightarrow \phi \left( 0\right) ,$ such that $c^{k}>\phi \left( 0\right)$ for all $k$. At each $k,$ we have

$$\begin{aligned} \frac{-P_{O}^{\prime }}{P_{O}}\left( c^{k}\right) \frac{\beta _{I}\left( c^{k}\right) -A-c^{k}}{\beta _{I}^{\prime }\left( c^{k}\right) }=1 \end{aligned}$$

and

$$\begin{aligned} \frac{-P_{I}^{\prime }}{P_{I}}\left( c^{k}\right) \frac{\beta _{I}\left( c^{k}\right) -c^{k}}{\beta _{I}^{\prime }\left( c^{k}\right) }=1, \end{aligned}$$

and hence

$$\begin{aligned} \frac{-P_{O}^{\prime }}{P_{O}}\left( c^{k}\right) \left( \beta _{I}\left( c^{k}\right) -A-c^{k}\right) =\frac{-P_{I}^{\prime }}{P_{I}}\left( c^{k}\right) \left( \beta _{I}\left( c^{k}\right) -c\right) , \end{aligned}$$

and so

$$\begin{aligned} \frac{\frac{-P_{O}^{\prime }}{P_{O}}\left( c^{k}\right) }{\frac{ -P_{I}^{\prime }}{P_{I}}\left( c^{k}\right) }=\frac{\beta _{I}\left( c^{k}\right) -c^{k}}{\beta _{I}\left( c^{k}\right) -A-c^{k}}. \end{aligned}$$

The RHS converges to at least

$$\begin{aligned} \frac{n_{I}}{n_{I}-1} \end{aligned}$$

by (22). Consider the $LHS$. By (19) and (20) , for any $k,$ it is given by

$$\begin{aligned} \frac{\frac{-P_{O}^{\prime }}{P_{O}}\left( c^{k}\right) }{\frac{ -P_{I}^{\prime }}{P_{I}}\left( c^{k}\right) }&= \frac{n_{I}\frac{f}{\bar{F}} \left( c^{k}\right) +\left( n_{O}-1\right) \frac{f}{\bar{F}}\left( c^{k}\right) \left( \psi \left( c^{k}\right) \right) \psi ^{\prime }\left( c^{k}\right) }{\left( n_{I}-1\right) \frac{f}{\bar{F}}\left( c^{k}\right) +n_{O}\frac{f}{\bar{F}}\left( c^{k}\right) \left( \psi \left( c^{k}\right) \right) \psi ^{\prime }\left( c^{k}\right) } \\&= \frac{n_{I}}{n_{I}-1}\frac{\frac{f}{\bar{F}}\left( c^{k}\right) +\frac{ n_{O}-1}{n_{I}}\frac{f}{\bar{F}}\left( c^{k}\right) \left( \psi \left( c^{k}\right) \right) \psi ^{\prime }\left( c^{k}\right) }{\frac{f}{\bar{F}} \left( c^{k}\right) +\frac{n_{O}}{n_{I}-1}\frac{f}{\bar{F}}\left( c^{k}\right) \left( \psi \left( c^{k}\right) \right) \psi ^{\prime }\left( c^{k}\right) }. \end{aligned}$$

But then,

$$\begin{aligned} \lim \inf \frac{\frac{f}{\bar{F}}\left( c^{k}\right) +\frac{n_{O}-1}{n_{I}} \frac{f}{\bar{F}}\left( c^{k}\right) \left( \psi \left( c^{k}\right) \right) \psi ^{\prime }\left( c^{k}\right) }{\frac{f}{\bar{F}}\left( c^{k}\right) + \frac{n_{O}}{n_{I}-1}\frac{f}{\bar{F}}\left( c^{k}\right) \left( \psi \left( c^{k}\right) \right) \psi ^{\prime }\left( c^{k}\right) }\ge 1. \end{aligned}$$

(23)

Now, note that

$$\begin{aligned} \frac{n_{O}-1}{n_{I}}<\frac{n_{O}}{n_{I}-1} \end{aligned}$$

(cross multiply), and so (23) can only hold if $\lim \sup \psi ^{\prime }\left( c^{k}\right) =0$. But then, $\phi ^{\prime }\left( 0\right) =\infty ,$ which completes the proof that $\phi ^{\prime }\left( 0\right) >1$.

For the characterization of $\phi ^{\prime }$ close to $1-A,$ recall that by Lemma 9,

$$\begin{aligned} A+c-\phi \left( c\right) >0 \end{aligned}$$

for any $c<1-A$. Since $\phi \left( 1-A\right) =1,$

$$\begin{aligned} A+1-A-\phi \left( 1-A\right) =0, \end{aligned}$$

and thus

$$\begin{aligned} A+c-\phi \left( c\right) -\left( A+1-A-\phi \left( 1-A\right) \right) >0, \end{aligned}$$

or

$$\begin{aligned} c-\left( 1-A\right) -\left( \phi \left( c\right) -\phi \left( 1-A\right) \right) >0, \end{aligned}$$

or

$$\begin{aligned} 1-\frac{\phi \left( 1-A\right) -\phi \left( c\right) }{\left( 1-A\right) -c} <0, \end{aligned}$$

which is to say

$$\begin{aligned} \frac{\phi \left( 1-A\right) -\phi \left( c\right) }{\left( 1-A\right) -c}>1. \end{aligned}$$

Since this is true for all $c,$ it cannot be that on any interval of the form $[\hat{c},1)\,\phi ^{\prime }\left( c\right) <1$ everywhere. But then, if there is any point where $\phi ^{\prime }<1,$ then there is a point to the right of it where $\phi ^{\prime }\ge 1$. $\square$

Proof of Lemma 3

Let $\tilde{c}$ be any point in $[0,1-A)$ with $\phi ^{\prime }\left( \tilde{c}\right) \le 1,$ and $\phi ^{\prime \prime }\left( \tilde{c}\right) =0$. Let

$$\begin{aligned} \tau \equiv \sup \left\{ c|\phi \left( c\right) \le \phi \left( \tilde{c} \right) +\phi ^{\prime }\left( \tilde{c}\right) \left( c-\tilde{c}\right) \right\} . \end{aligned}$$

Such a $\tau$ exists since $\phi$ is continuous on $[0,1-A]$, and $\tau \in [\tilde{c},1-A)$ since $\phi ^{\prime }\left( \tilde{c}\right) \le 1,\,\phi \left( \tilde{c}\right) <\tilde{c}+A,$ and $\phi \left( 1-A\right) =1$. On $\left[ \tilde{c},\tau \right] ,\,\phi ^{\prime }$ is continuously differentiable, and so reaches a minimum at some $r\in \left[ \tilde{c},\tau \right]$. If

$$\begin{aligned} \tilde{c}\in \arg \min _{c\in \left[ \tilde{c},\tau \right] }\phi ^{\prime }\left( c\right) , \end{aligned}$$

take $r=\tilde{c}$. Otherwise,

$$\begin{aligned} \phi ^{\prime }\left( r\right) <\phi ^{\prime }\left( \tilde{c}\right) \le \phi ^{\prime }\left( \tau \right) , \end{aligned}$$

and so $r$ is interior. In either case, $\phi ^{\prime \prime }\left( r\right) =0,$ and, of course,

$$\begin{aligned} \phi ^{\prime }\left( r\right) \le \phi ^{\prime }\left( \tilde{c}\right) \le 1. \end{aligned}$$

It remains to show that for $s\ge r,\,\phi \left( s\right) \ge l\left( s\right)$. But, for $s\in \left[ r,\tau \right] ,$

$$\begin{aligned} \phi \left( s\right)&= \phi \left( r\right) +\int \limits _{r}^{s}\phi ^{\prime }\left( t\right) dt \\&\ge \phi \left( r\right) +\int \limits _{r}^{s}\phi ^{\prime }\left( r\right) dt \\&= \phi \left( r\right) +\phi ^{\prime }\left( r\right) \left( s-r\right) \\&= l\left( s\right) . \end{aligned}$$

So, $\phi \left( s\right) \ge l\left( s\right)$ for all $s\in \left[ r,\tau \right] ,$ and, in particular, $\phi \left( \tau \right) \ge l\left( \tau \right)$. But, by construction,

$$\begin{aligned} \phi \left( \tau \right) =\phi \left( \tilde{c}\right) +\phi ^{\prime }\left( \tilde{c}\right) \left( \tau -\tilde{c}\right) , \end{aligned}$$

and hence, for all $s\in (\tau ,1-A],$

$$\begin{aligned} \phi \left( s\right)&> \phi \left( \tilde{c}\right) +\phi ^{\prime }\left( \tilde{c}\right) \left( s-\tilde{c}\right) \\&= \phi \left( \tilde{c}\right) +\phi ^{\prime }\left( \tilde{c}\right) \left( \tau -\tilde{c}\right) +\phi ^{\prime }\left( \tilde{c}\right) \left( s-\tau \right) \\&= \phi \left( \tau \right) +\phi ^{\prime }\left( \tilde{c}\right) \left( s-\tau \right) \\&\ge l\left( \tau \right) +\phi ^{\prime }\left( r\right) \left( s-\tau \right) \\&= l\left( s\right) . \end{aligned}$$

$\square$

Proof of Proposition 4

Taking logs on both sides of (9) and differentiating, we have

$$\begin{aligned} \frac{\phi ^{\prime \prime }\left( c\right) }{\phi ^{\prime }\left( c\right) }=\frac{\left( S_{I}\left( \phi \left( c\right) \right) \right) ^{\prime }}{ S_{I}\left( \phi \left( c\right) \right) }-\frac{\left( S_{O}\left( c\right) \right) ^{\prime }}{S_{O}\left( c\right) }+\frac{\partial }{\partial c}\log \frac{\bar{Q}_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{I}\left( c\right) }+\frac{\partial }{\partial c}\log \left( \frac{\frac{q_{I}\left( c\right) }{\bar{Q}_{I}\left( c\right) }}{\frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}\left( \phi \left( c\right) \right) }}\right) . \end{aligned}$$

But, using (5) and (7) ,

$$\begin{aligned} \frac{\left( S_{I}\left( \phi \left( c\right) \right) \right) ^{\prime }}{ S_{I}\left( \phi \left( c\right) \right) }-\frac{\left( S_{O}\left( c\right) \right) ^{\prime }}{S_{O}\left( c\right) }&= -\phi ^{\prime }\left( c\right) \frac{\bar{Q}_{I}\left( c\right) }{S_{I}\left( \phi \left( c\right) \right) }+\frac{\bar{Q}_{O}\left( \phi \left( c\right) \right) }{S_{O}\left( c\right) } \\&= -\phi ^{\prime }\left( c\right) \frac{\frac{q_{I}\left( c\right) }{\bar{Q} _{I}\left( c\right) }}{S_{I}\left( \phi \left( c\right) \right) \frac{ q_{I}\left( c\right) }{\bar{Q}_{I}^{2}\left( c\right) }}+\frac{\frac{ q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}\left( \phi \left( c\right) \right) }}{S_{O}\left( c\right) \frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}^{2}\left( \phi \left( c\right) \right) }} \\&= \frac{-\phi ^{\prime }\left( c\right) \frac{q_{I}\left( c\right) }{\bar{Q} _{I}\left( c\right) }+\phi ^{\prime }\left( c\right) \frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}\left( \phi \left( c\right) \right) }}{ S_{I}\left( \phi \left( c\right) \right) \frac{q_{I}\left( c\right) }{\bar{Q} _{I}^{2}\left( c\right) }}\,\hbox { using } (8)\\&\ge \frac{-\frac{q_{I}\left( c\right) }{\bar{Q}_{I}\left( c\right) }+\phi ^{\prime }\left( c\right) \frac{q_{O}\left( \phi \left( c\right) \right) }{ \bar{Q}_{O}\left( \phi \left( c\right) \right) }}{S_{I}\left( \phi \left( c\right) \right) \frac{q_{I}\left( c\right) }{\bar{Q}_{I}^{2}\left( c\right) }}\,\, (\hbox {since } \phi ^{\prime }\left( c\right) \le 1) \\&= \frac{\frac{\partial }{\partial c}\log \frac{\bar{Q}_{I}\left( c\right) }{ \bar{Q}_{O}\left( \phi \left( c\right) \right) }}{S_{I}\left( \phi \left( c\right) \right) \frac{q_{I}\left( c\right) }{\bar{Q}_{I}^{2}\left( c\right) }}. \end{aligned}$$

But then,

$$\begin{aligned} \frac{\phi ^{\prime \prime }\left( c\right) }{\phi ^{\prime }\left( c\right) }&\ge \frac{\frac{\partial }{\partial c}\log \frac{\bar{Q}_{I}\left( c\right) }{\bar{Q}_{O}\left( \phi \left( c\right) \right) }}{S_{I}\left( \phi \left( c\right) \right) \frac{q_{I}\left( c\right) }{\bar{Q} _{I}^{2}\left( c\right) }}+\frac{\partial }{\partial c}\log \frac{\bar{Q} _{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{I}\left( c\right) }+\frac{ \partial }{\partial c}\log \left( \frac{\frac{q_{I}\left( c\right) }{\bar{Q} _{I}\left( c\right) }}{\frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{ Q}_{O}\left( \phi \left( c\right) \right) }}\right) \\&= \left( \frac{1}{S_{I}\left( \phi \left( c\right) \right) \frac{ q_{I}\left( c\right) }{\bar{Q}_{I}^{2}\left( c\right) }}-1\right) \frac{ \partial }{\partial c}\log \frac{\bar{Q}_{I}\left( c\right) }{\bar{Q} _{O}\left( \phi \left( c\right) \right) }+\frac{\partial }{\partial c}\log \left( \frac{\frac{q_{I}\left( c\right) }{\bar{Q}_{I}\left( c\right) }}{ \frac{q_{O}\left( \phi \left( c\right) \right) }{\bar{Q}_{O}\left( \phi \left( c\right) \right) }}\right) . \end{aligned}$$

The result then follows by (11). $\square$

Proof of Lemma 5

Note first that

$$\begin{aligned} S_{O}\left( \phi \left( r\right) \right)&= \int \limits _{r}^{1-A}\bar{F} ^{n_{I}}\left( \phi \left( s\right) \right) \bar{F}^{n_{O}-1}\left( s\right) ds \nonumber \\&< \int \limits _{r}^{1-A}\bar{F}^{n_{I}}\left( l\left( s\right) \right) \bar{F}^{n_{O}-1}\left( s\right) ds \nonumber \\&= \int \limits _{r}^{1-A}K\left( s\right) ds \nonumber \\&\le \int \limits _{r}^{1}K\left( s\right) ds. \end{aligned}$$

(24)

where $K\left( s\right) \equiv 0$ if $l\left( s\right) >1,$ and where the inequality is strict since $\phi \left( s\right) >l\left( s\right)$ on $\left[ \tau ,1-A\right]$.

Now, since $\bar{Q}_{O}\left( \phi \left( s\right) \right) =\bar{F} ^{n_{I}}\left( \phi \left( s\right) \right) \bar{F}^{n_{O}-1}\left( s\right) ,$ and since $l\left( r\right) =\phi \left( r\right) , \,\bar{Q}_{O}\left( \phi \left( r\right) \right) =K\left( r\right)$. Further,

$$\begin{aligned} \bar{Q}_{O}^{\prime }\left( \phi \left( s\right) \right) \phi ^{\prime }\left( s\right) =\left[ \bar{F}^{n_{I}}\left( \phi \left( s\right) \right) \bar{F}^{n_{O}-1}\left( s\right) \right] _{s}, \end{aligned}$$

and so, since $l^{\prime }\left( r\right) =\phi ^{\prime }\left( r\right) ,$

$$\begin{aligned} \bar{Q}_{O}^{\prime }\left( \phi \left( r\right) \right) \phi ^{\prime }\left( r\right) =K^{\prime }\left( r\right) \end{aligned}$$

and thus

$$\begin{aligned} q_{O}\left( \phi \left( r\right) \right) =\frac{-K^{\prime }\left( r\right) }{\phi ^{\prime }\left( r\right) } \end{aligned}$$

using that $\phi ^{\prime \prime }\left( r\right) =0$. Thus,

$$\begin{aligned} S_{O}\left( \phi \left( r\right) \right) \frac{q_{O}\left( \phi \left( r\right) \right) }{\bar{Q}_{O}^{2}\left( \phi \left( r\right) \right) }&< \frac{1}{\phi ^{\prime }\left( r\right) }\int \limits _{r}^{1}K\left( s\right) ds \frac{-K^{\prime }(r)}{\left[ K\left( r\right) \right] ^{2}} \\&= \frac{1}{\phi ^{\prime }\left( r\right) }\left( 1-\rho _{\int K\left( s\right) }\left( r\right) \right) , \end{aligned}$$

and we are done since by (8),

$$\begin{aligned} S_{I}\left( \phi \left( r\right) \right) \frac{q_{I}\left( r\right) }{\bar{Q} _{I}^{2}\left( r\right) }=\phi ^{\prime }\left( r\right) S_{O}\left( \phi \left( r\right) \right) \frac{q_{O}\left( \phi \left( r\right) \right) }{ \bar{Q}_{O}^{2}\left( \phi \left( r\right) \right) }. \end{aligned}$$

$\square$

Proof of Lemma 6

We begin with a technical result. $\square$

Lemma 10

Let $t\ge 1$. Then,

$$\begin{aligned} \frac{1}{\frac{n_{O}}{n_{I}-1}\frac{1}{t}+1}- \frac{1}{\frac{n_{O}-1}{n_{I}} \frac{1}{t}+1}>\frac{-1}{n-1}. \end{aligned}$$

Proof

Since

$$\begin{aligned} \frac{-1}{n-1}=\frac{1}{\frac{n_{O}}{n_{I}-1}+1}- \frac{1}{\frac{n_{O}-1}{ n_{I}}+1} \end{aligned}$$

it would be enough to show that

$$\begin{aligned} Y\left( t\right) =\frac{1}{\frac{n_{O}}{n_{I}-1}\frac{1}{t}+1}-\frac{1}{ \frac{n_{O}-1}{n_{I}}\frac{1}{t}+1} \end{aligned}$$

is increasing in $t$.

But,

$$\begin{aligned} Y^{\prime }\left( t\right) =\frac{-\frac{n_{O}}{n_{I}-1}\left( \frac{-1}{t^{2}}\right) }{\left( \frac{n_{O}}{n_{I}-1}\frac{1}{t}+1\right) ^{2}}-\frac{-\frac{n_{O}-1}{n_{I}}\left( \frac{-1}{t^{2}}\right) }{\left( \frac{n_{O}-1}{n_{I}}\frac{1}{t}+1\right) ^{2}} \end{aligned}$$

which has the same sign as

$$\begin{aligned} \frac{\frac{n_{O}}{n_{I}-1}}{\left( \frac{n_{O}}{n_{I}-1}\frac{1}{t} +1\right) ^{2}}-\frac{\frac{n_{O}-1}{n_{I}}}{\left( \frac{n_{O}-1}{n_{I}} \frac{1}{t}+1\right) ^{2}}. \end{aligned}$$

Cross multiplying, we have that the sign of $Y^{\prime }\left( t\right)$ is the same as that of

$$\begin{aligned} \frac{n_{O}}{n_{I}-1}\left( \frac{n_{O}-1}{n_{I}t}+1\right) ^{2}-\frac{ n_{O}-1}{n_{I}}\left( \frac{n_{O}}{\left( n_{I}-1\right) t}+1\right) ^{2} \end{aligned}$$

which in turn, after bringing everything to the same denominator, is the same as that of

$$\begin{aligned} \left( n_{O}-1+n_{I}t\right) ^{2}-\frac{n_{I}}{n_{I}-1}\frac{n_{O}-1}{n_{O}} \left( n_{O}+\left( n_{I}-1\right) t\right) ^{2}. \end{aligned}$$

So, since $n_{O}\le n_{I},\,Y^{\prime }\left( t\right)$ is positive whenever

$$\begin{aligned} \left( n_{O}-1+n_{I}t\right) ^{2}-\left( n_{O}+\left( n_{I}-1\right) t\right) ^{2} \end{aligned}$$

is non-negative. Observe that since the last expression is a difference of squares, and since $t\ge 1,$ it has the same sign as

$$\begin{aligned} Y^{\prime }\left( t\right) \underset{s}{\ge }\left[ t-1\right] \left[ \left( n_{O}-1+n_{I}t\right) +\left( n_{O}+\left( n_{I}-1\right) t\right) \right] \ge 0. \end{aligned}$$

We thus conclude that $Y^{\prime }\left( t\right) \ge 0$.

This in hand, note that by Lemma 3, and since $\phi ^{\prime \prime }\left( r\right) =0,$

$$\begin{aligned} \frac{\partial }{\partial r}\log \left( \frac{\frac{q_{I}\left( r\right) }{ \bar{Q}_{I}\left( r\right) }}{\frac{q_{O}\left( \phi \left( r\right) \right) }{\bar{Q}_{O}\left( \phi \left( r\right) \right) }}\right)&= \frac{\partial }{\partial r}\log \left( \phi ^{\prime }\left( r\right) T\left( r\right) \right) \\&= \frac{\partial }{\partial r}\log \left( \frac{n_{O}\frac{f}{\bar{F}} \left( r\right) +\left( n_{I}-1\right) \phi ^{\prime }\left( r\right) \frac{f }{\bar{F}}\left( \phi \left( r\right) \right) }{\left( n_{O}-1\right) \frac{f }{\bar{F}}\left( r\right) +n_{I}\phi ^{\prime }\left( r\right) \frac{f}{\bar{ F}}\left( \phi \left( r\right) \right) }\right) \\&= \frac{\partial }{\partial r}\log \left( \frac{n_{O}+\left( n_{I}-1\right) \phi ^{\prime }\left( r\right) \frac{\frac{f}{\bar{F}}\left( \phi \left( r\right) \right) }{\frac{f}{\bar{F}}\left( r\right) }}{\left( n_{O}-1\right) +n_{I}\phi ^{\prime }\left( r\right) \frac{\frac{f}{\bar{F}}\left( \phi \left( r\right) \right) }{\frac{f}{\bar{F}}\left( r\right) }}\right) \\&= \frac{\partial }{\partial r}\log \left( \frac{n_{O}+\left( n_{I}-1\right) D\left( r\right) }{\left( n_{O}-1\right) +n_{I}D\left( r\right) }\right) , \end{aligned}$$

where

$$\begin{aligned} D\left( r\right) \equiv \phi ^{\prime }\left( r\right) \frac{\bar{F}\left( r\right) }{\bar{F}\left( \phi \left( r\right) \right) }\frac{f\left( \phi \left( r\right) \right) }{f\left( r\right) }>1 \end{aligned}$$

by Corollary 2. Thus,

$$\begin{aligned} \frac{\partial }{\partial r}\log \left( \frac{\frac{q_{I}\left( r\right) }{ \bar{Q}_{I}\left( r\right) }}{\frac{q_{O}\left( \phi \left( r\right) \right) }{\bar{Q}_{O}\left( \phi \left( r\right) \right) }}\right)&= \frac{\left( n_{I}-1\right) D^{\prime }\left( r\right) }{n_{O}+\left( n_{I}-1\right) D\left( r\right) }-\frac{n_{I}D^{\prime }\left( r\right) }{\left( n_{O}-1\right) +n_{I}D\left( r\right) } \\&= \left( \frac{1}{\frac{n_{O}}{\left( n_{I}-1\right) D\left( r\right) }+1}- \frac{1}{\frac{n_{O}-1}{n_{I}D\left( r\right) }+1}\right) \frac{D^{\prime }\left( r\right) }{D\left( r\right) }\,\,(\hbox {using }n_{I}>1) \\&\ge -\frac{1}{n-1}\frac{D^{\prime }\left( r\right) }{D\left( r\right) }, \end{aligned}$$

by Lemma 10. But, since $\phi ^{\prime \prime }\left( r\right) =0,$ we are done, as

$$\begin{aligned} \frac{D^{\prime }\left( r\right) }{D\left( r\right) }=\frac{\partial }{ \partial r}\log \frac{\bar{F}\left( r\right) }{\bar{F}\left( \phi \left( r\right) \right) }+\frac{\partial }{\partial r}\log \frac{f\left( \phi \left( r\right) \right) }{f\left( r\right) }. \end{aligned}$$

$\square$