Auctioning a discrete public good under incomplete information

Abstract

We study a dynamic auction mechanism in the context of private provision of a discrete public good under incomplete information. The bidders have private valuations, and the cost of the public good is common knowledge. No bidder is willing to provide the good on her own. We show that a natural application of open ascending auctions in such environments fails dramatically: The probability of provision is zero in any equilibrium. The mechanism effectively auctions off the ‘right’ to be the last one to contribute, but intuition suggests that neither player wishes to be the last one to contribute. Since the player who contributes first has the advantage of being able to free ride on the contributions of the other players, no player wants to ‘win’ the auction.

Appendix

Proof of Lemma 4

Consider the values $x$ of bidder $i$ such that $x<y^{\beta _{-i}}=c-b^{\beta _{-i}}.$ That is, $c-x>b^{\beta _{-i}}.$ Suppose that $\beta _{{-i}}(.)$ is strictly increasing. (We will consider the values $x>y^{\beta _{-i}}$ and the case where $\beta _{{-i}}(.)$ is weakly increasing later.) Suppose bidder $i$ bids an amount $b\ge b^{\beta _{-i}}$ when her value is $x.$ There are two cases, $c-x > b\ge b^{\beta _{-i}}$ and $b\ge c-x.$

$(i)$ $c-x>b\ge b^{\beta _{-i}}:$ For any $y$ such that $y<\beta _{-i}^{-1}(b),$ bidder $-i$ drops out first and her drop-out price is less than $c-x.$ Thus, for such values of bidder $-i,$ there is no provision. For values $y$ such that $y>\beta _{-i}^{-1}(b),$ bidder $i$ drops out first, and her drop out price$,$ $b,$ is bigger than $c-y.$ Thus, there is provision, and bidder $i$ gets $x-b$ for such values of bidder $-i.$ So, the payoff for bidder $i$ is

$$\begin{aligned} \pi ^{i}(x,b,\beta _{-i})=\int \limits _{\beta _{-i}^{-1}(b)}^{1}[x-b]f(y)\mathrm{{d}}y=(x-b)[1-F(\beta _{-i}^{-1}(b))]. \end{aligned}$$

As $b$ decreases, both $(x-b)$ and $[1-F(\beta _{-i}^{-1}(b))]$ increase. Thus, in the range $c-x>b\ge b^{\beta _{-i}},$ optimal bid is $b^{\beta _-{i}}.$

$(ii)$ $b\ge c-x>b^{\beta _{-i}}:$ In this case, for values $y,$ such that $y<\beta _{-i}^{-1}(c-x)$ there is no provision since bidder $-i$ drops out first, and her drop-out price is less than $c-x.$ For values $y$ such that $\beta _{-i}^{-1}(c-x)<y<\beta _{-i}^{-1}(b),$ bidder $-i$ is still the first one to drop out, and her drop-out price determines whether bidder $i$ will contribute the rest or not. For values $y$ such that $y>\beta _{-i}^{-1}(b),$ there is provision as explained above. So, the payoff for bidder $i$ is given by

$$\begin{aligned} \pi ^{i}(x,b,\beta _{-i})=\int \limits _{\beta _{-i}^{-1}(c-x)}^{\beta _{-i}^{-1}(b)}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+(x-b)[1-F(\beta _{-i}^{-1}(b))]. \end{aligned}$$

Suppose instead of $b,$ bidder $i$ bids $b-\varepsilon ,$ where $\varepsilon >0.$ Then, the relevant payoff is given by

$$\begin{aligned} \pi ^{i}(x,b-\varepsilon ,\beta _{-i})=\int \limits _{\beta _{-i}^{-1}(c-x)}^{\beta _{-i}^{-1}(b-\varepsilon )}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+(x-b+\varepsilon )[1-F(\beta _{-i}^{-1}(b-\varepsilon ))]. \end{aligned}$$

Then,

$$\begin{aligned}&\pi ^{i}(x,b-\varepsilon ,\beta _{-i})-\pi ^{i}(x,b,\beta _{-i})\\&\quad =(x-b+\varepsilon )[1-F(\beta _{-i}^{-1}(b-\varepsilon ))]-(x-b)[1-F(\beta _{-i}^{-1}(b))] \\&\qquad +\int \limits _{\beta _{-i}^{-1}(c-x)}^{\beta _{-i}^{-1}(b-\varepsilon )}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y-\int \limits _{\beta _{-i}^{-1}(c-x)}^{\beta _{-i}^{-1}(b)}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y \\&\quad \ge (x-b)[F(\beta _{-i}^{-1}(b))-F(\beta _{-i}^{-1}(b-\varepsilon ))] +\varepsilon [1-F(\beta _{-i}^{-1}(b-\varepsilon ))]\\&\qquad -\int \limits _{\beta _{-i}^{-1}(b-\varepsilon )}^{\beta _{-i}^{-1}(b)}[x-c+\beta _{-i}(\beta _{-i}^{-1}(b))]f(y)\mathrm{{d}}y >0.\\ \end{aligned}$$

The weak inequality follows from the fact that $y<\beta _{-i}^{-1}(b)$. By Lemma 1, the strict inequality follows from the fact that $b\le c/2$ and that $\varepsilon >0.$ Thus, the expected profit is strictly decreasing in $b$ over the range $[b^{\beta _{-i}},\overline{\beta }_{^{i}}).$ Thus, whenever bidder $i$ has a value $x<y^{\beta _-i}$, and $\beta _{-i}(.)$ is strictly increasing, she never makes a bid higher than $b^{\beta _{-i}}$. Now consider the case where $\beta _{-i}(y)=b$ for $y\in (y_{1},y_{2})\subseteq [y^{\beta _{-i}},1],$ for some $y_{1},$ $y_{2}.$ ¹⁷ Again we consider two possible cases separately.

$(i)$ $c-x>b\ge b^{\beta _{-i}}:$ Over the range $(y_{1},y_{2})$ both bidders have the same bid, so with probability $1/2$ bidder $i$ gets to be the first one to drop out and gets a payoff of $\int _{y_{1}}^{y_{2}}[x-~b]f(y)\mathrm{{d}}y,$ and with probability $1/2$ she gets to be the last one to drop out and gets a payoff of zero since $b<c-x$. Thus, bidder $i$’s payoff when she bids $b$ and her valuation is $x$, is

$$\begin{aligned} \pi ^{i}(x,b,\beta _{-i})=\frac{1}{2}\int \limits _{y_{1}}^{y_{2}}[x-b]f(y)\mathrm{{d}}y+ \int \limits _{y_{2}}^{1}[x-b]f(y)\mathrm{{d}}y. \end{aligned}$$

But, if she bids $b-\varepsilon ,$ for an arbitrarily small $\varepsilon >0,$ then her payoff is

$$\begin{aligned} \pi ^{i}(x,b-\varepsilon ,\beta _{-i})=\int \limits _{\widetilde{y} _{1}}^{y_{2}}[x-b+\varepsilon ]f(y)\mathrm{{d}}y+\int \limits _{y_{2}}^{1}[x-b+\varepsilon ]f(y)\mathrm{{d}}y, \end{aligned}$$

where $\widetilde{y}_{1}$ is such that $\beta ^{-i}(\widetilde{y} _{1})=b-\varepsilon$ if $\beta _{-i}(.)$ is continuous at $y_{1},$ and $\widetilde{y}_{1}=y_{1}$ if $\beta _{-i}(.)$ is not continuous at $y_{1}.$ For $\varepsilon >0$ small enough, $\pi ^{i}(x,b-\varepsilon ,\beta _{-i})$ is strictly bigger than $\pi ^{i}(x,b,\beta _{-i}).$ Since any $b$ such that $b>b^{\beta _{-i}}$ is strictly dominated by $b-\varepsilon ,$ any bid $b\in (c-x,b^{\beta _{-i}})$ cannot be optimal. Thus, bidder $i$ bids $b^{\beta _{-i}}.$

$(ii)$ $b\ge c-x:$ Define $y_{0}$ such that $\beta _{-i}(y_{0})=c-x.$ ¹⁸ Then, bidder $i$’s payoff when she bids $b$, and her valuation is $x$, is

$$\begin{aligned} \pi ^{i}(x,b,\beta _{-i})&=\int \limits _{y_{0}}^{y_{1}}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+ \frac{1}{2}\int \limits _{y_{1}}^{y_{2}}[x-b]f(y)\mathrm{{d}}y\\&\quad +\,\frac{1}{2}\int \limits _{y_{1}}^{y_{2}}[x-c+b]f(y)\mathrm{{d}}y+\int \limits _{y_{2}}^{1}[x-b]f(y)\mathrm{{d}}y \\&=\int \limits _{y_{0}}^{y_{1}}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+\int \limits _{y_{1}}^{y_{2}}[x-c/2]f(y)\mathrm{{d}}y\\&\quad +\,\int \limits _{y_{2}}^{1}[x-b]f(y)\mathrm{{d}}y. \end{aligned}$$

But, the bidder with value $x$ can bid $b-\varepsilon$ and improve her payoff. To see this, let us first write the payoff from bidding $b-\varepsilon$:

$$\begin{aligned} \pi ^{i}(x,b-\varepsilon ,\beta _{-i})&=\int \limits _{y_{0}}^{\widetilde{y} _{1}}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y\\&\quad +\int \limits _{\widetilde{y}_{1}}^{y_{2}}[x-b+ \varepsilon ]f(y)\mathrm{{d}}y+\int \limits _{y_{2}}^{1}[x-b+\varepsilon ]f(y)\mathrm{{d}}y, \end{aligned}$$

where $\widetilde{y}_{1}$ is defined as in Case $(i)$ above. If $\widetilde{y}_{1}=y_{1},$ then $\pi ^{i}(x,b-\varepsilon ,\beta _{-i})>\pi ^{i}(x,b,\beta _{-i})$ since

$$\begin{aligned} \int \limits _{\widetilde{y}_{1}}^{y_{2}}[x-b+\varepsilon ]f(y)\mathrm{{d}}y > \int \limits _{y_{1}}^{y_{2}}[x-b]f(y)\mathrm{{d}}y \ge \int \limits _{y_{1}}^{y_{2}}[x-c/2]f(y)\mathrm{{d}}y, \end{aligned}$$

where the last inequality follows from $b\le c/2.$ If $\widetilde{y}_{1}<y_{1},$ then $\pi ^{i}(x,b-\varepsilon ,\beta _{-i})>\pi ^{i}(x,b,\beta _{-i})$ since

$$\begin{aligned}&\int \limits _{y_{0}}^{\widetilde{y}_{1}}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+\int \limits _{ \widetilde{y}_{1}}^{y_{2}}[x-b+\varepsilon ]f(y)\mathrm{{d}}y \\&\quad >\int \limits _{y_{0}}^{\widetilde{y}_{1}}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+\int \limits _{ \widetilde{y}_{1}}^{y_{2}}[x-c/2]f(y)\mathrm{{d}}y \\&\quad >\int \limits _{y_{0}}^{y_{1}}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y+\int \limits _{y_{1}}^{y_{2}}[x-c/2]f(y)\mathrm{{d}}y, \end{aligned}$$

where the first inequality follows from the fact that $x-b+\varepsilon >x-c/2,$ and the second inequality follows from the fact that $x-c/2\ge x-c+\beta _{-i}(y)$ and $\widetilde{y}_{1}<y_{1}.$ Thus, any bid $b\ge c-x$ is strictly dominated by $b-\varepsilon .$ Specifically $c-x$ is dominated by $c-x-\varepsilon .$ Thus, we are back in Case $(i)$, and the optimal bid is $b^{\beta _{-i}}.$ Now consider the values $x>y^{\beta _{-i}}.$ But in this case, there is only one region to check, which is $b>b^{\beta _{-i}},$ since $c-x<b^{\beta _{-i}}.$ By the same argument in Case $(ii)$ above, we can conclude that $\pi ^{i}(x,b-\varepsilon ,\beta _{-i})>\pi ^{i}(x,b,\beta _{-i}),$ and hence any $b>b^{\beta _{-i}}$ is dominated by $b^{\beta _{-i}}.$ $\square$

Proof of Lemma 5

First note that it is straightforward from Lemma 4 that $b^{i}=b^{\beta _{-i}}$. Thus, we have $\beta _{^{i}}(x)\le b^{\beta _{-i}}$ for all $x\in [0,1].$ Since $\beta _{^{i}}(.)$ is non-decreasing, definition of $b^{\beta _{-i}}$ implies that $\beta _{^{i}}(x)=b^{\beta _{-i}}$ for any $x>c-b^{\beta _{-i}}=y^{\beta _{-i}}.$ So, we have shown that $\beta _{{i}}(x)=\beta _{-i}(y)=b$ for $x,y\in [c-b,1],$ in an equilibrium with positive probability of provision with non-decreasing bidding functions (In fact, we have not shown that $\beta _{{i}}(c-b)=\beta _{-i}(c-b)=b.$ But, if $\beta _{{i}}(c-b)<b$ and $\beta _{-i}(c-b)<b,$ then one of the bidders deviates to $b-\varepsilon ,$ which will be clear below). We argue that this cannot be the case in an equilibrium with positive probability of provision , simply because bidder $i$ would deviate to $b-\varepsilon$ for some $x$ where $\varepsilon >0$ is arbitrarily small. To see this, first note that bidder $i$ with value $x$, when she bids $b,$ gets a payoff of $x-b$ with probability $1/2$ and a payoff of $x-c+b$ with probability $1/2.$ Note also that $x-b\ge x-c+b$ since $b\le c/2.$ But, dropping out at $b-\varepsilon$ generates a payoff of $x-b+\varepsilon$ over the range $[c-b+\varepsilon ,1],$ and zero over $[c-b,c-b+\varepsilon ).$ So, over the range $[c-b,1],$ the net gain from deviating from $b$ to $b-\varepsilon$ is equal to

$$\begin{aligned}&\int \limits _{c-b+\varepsilon }^{1}[x-b+\varepsilon ]f(y)\mathrm{{d}}y\!-\!\left\{ 1/2\int \limits _{c-b}^{1}[x\!-\!b]f(y)\mathrm{{d}}y\!+\!1/2\int \limits _{c\!-\!b}^{1}[x\!-\!c\!+\!b]f(y)\mathrm{{d}}y\right\} \\&\quad =[x-b][1-F(c-b+\varepsilon )]-[x-c/2][1-F(c-b)]\!+\!\varepsilon [1-F(c-b+\varepsilon )]. \end{aligned}$$

This expression is positive for sufficiently small $\varepsilon >0$ since $x-b\ge x-c/2$.¹⁹ There is, however, a potential loss over the range $(\widetilde{y},c-b)$ where $\widetilde{y}$ is defined to be the solution to $\beta _{-i}(y)=b-\varepsilon$ if there is a solution, and if there is no solution, $\widetilde{y}$ is such that $\beta _{-i}(y)<b-\varepsilon$ for all $y<\widetilde{y},$ and $\beta _{-i}(y)>b-\varepsilon$ for all $c-b>y> \widetilde{y}.$ This loss, however, depends on $\beta _{-i}(y)$ for $y<c-b,$ and there are three possible cases: (1) $\beta _{-i}(.)$ is not continuous at $c-b.$ Then, $\beta _{-i}(y)<b$ for all $y<c-b,$ which, together with discontinuity, implies $\widetilde{y}=c-b.$ Thus, the loss is zero. Thus, the deviation to $b-\varepsilon$ is profitable for any $x\in [c-b,1]$. (2) $\beta _{-i}(.)$ is continuous at $c-b$, and $\beta _{-i}(y)<b$ for all $y<c-b.$ Then, the loss from deviation is equal to $\int _{ \widetilde{y}}^{c-b}[x-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y.$ But, for $x=c-b,$ this expression is negative since $\beta _{-i}(y)<b$ for $y<c-b.$ Thus, the deviation to $b-\varepsilon$ is profitable for $x=c-b$. (3) $\beta _{-i}(.)$ is continuous at $c-b,$ and $\beta _{-i}(y)=b$ for $y\in (\widehat{y},c-b),$ for some $\widehat{y}\in [0,c-b)$. Then, the loss is equal to $1/2\int _{\widehat{y}}^{c-b}[x-c+b]f(y)\mathrm{{d}}y,$ which is zero when $x=c-b.$ Thus, the deviation to $b-\varepsilon$ is profitable for $x=c-b.$ $\square$

Proof of Lemma 6

Suppose that for bidder $i$ we have, $x_{1}<x_{2}$ and $\beta _{^{i}}(x_{1})>\beta _{^{i}}(x_{2}),$ and bidder $-i$ bids according to $\beta _{-i}(y).$ We show that the bidding function $\beta _{^{i}}(x)$ is not incentive compatible. More precisely, we show that $\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})\ge \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})$ implies $\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})$ , and hence we will conclude that such $\beta _{^{i}}(x)$ cannot be decreasing over some set of values. To see this, first define

$$\begin{aligned} Y_{H}&= \{y|\beta _{-i}(y)\ge \beta _{^{i}}(x_{1})\}, \\ Y_{M}&= \{y|\beta _{^{i}}(x_{1})>\beta _{-i}(y)\ge \beta _{^{i}}(x_{2})\}, \\ Y_{L}&= \{y|\beta _{^{i}}(x_{2})>\beta _{-i}(y)\}. \end{aligned}$$

There are five cases defined in terms of the relative magnitudes between $\beta _{^{i}}(x_{k})$ and $c-x_{k},$ where $k=1,2.$

$(i)\,c-x_{1}>c-x_{2}>\beta _{^{i}}(x_{2}),$ $c-x_{1}>\beta _{^{i}}(x_{1}):$ The relevant payoffs are given as follows:

$$\begin{aligned} \pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})&=\int \limits _{Y_{H}\cap \{y|y\ge c-\beta _{^{i}}(x_{1})\}}[x_{1}-\beta _{^{i}}(x_{1})]f(y)\mathrm{{d}}y=[x_{1}-\beta _{^{i}}(x_{1})]P_{1}, \\ \pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})&\!=\!\int \limits _{(Y_{H}\cup Y_{M})\cap \{y|y\ge c\!-\!\beta _{^{i}}(x_{2})\}}[x_{2}\!-\!\beta _{^{i}}(x_{2})]f(y)\mathrm{{d}}y\!=\![x_{2}-\beta _{^{i}}(x_{2})]P_{2}, \\ \pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})&\!=\!\int \limits _{(Y_{H}\cup Y_{M})\cap \{y|y\ge c\!-\!\beta _{^{i}}(x_{2})\}}[x_{1}\!-\!\beta _{^{i}}(x_{2})]f(y)\mathrm{{d}}y\!=\![x_{1}-\beta _{^{i}}(x_{2})]P_{2}, \\ \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})&=\int \limits _{Y_{H}\cap \{y|y\ge c-\beta _{^{i}}(x_{1})\}}[x_{2}-\beta _{^{i}}(x_{1})]f(y)\mathrm{{d}}y\\&\quad +\int \limits _{Y_{M}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y \\&=[x_{2}\!-\!\beta _{^{i}}(x_{1})]P_{1}\,+\,\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c\!+\!\beta _{-i}(y)]f(y)\mathrm{{d}}y, \end{aligned}$$

where $P_{1}=\int _{Y_{H}\cap \{y|y\ge c-\beta _{^{i}}(x_{1})\}}f(y)\mathrm{{d}}y$ and $P_{2}=\int _{(Y_{H}\cup Y_{M})\cap \{y|y\ge c-\beta _{^{i}}(x_{2})\}}f(y)\mathrm{{d}}y.$ ? Now, $\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})\ge \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})$ implies

$$\begin{aligned}{}[x_{2}-\beta _{i}(x_{2})]P_{2}\ge [x_{2}-\beta _{i}(x_{1})]P_{1}+\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \end{aligned}$$

which implies $[x_{2}-\beta _{^{i}}(x_{2})]P_{2}\ge [x_{2}-\beta _{^{i}}(x_{1})]P_{1},$ or $P_{2}\ge \frac{[x_{2}-\beta _{^{i}}(x_{1})]}{[x_{2}-\beta _{^{i}}(x_{2})]}P_{1}.$ ²⁰ Since $x_{1}<x_{2},$ we have $P_{2}>\frac{ [x_{1}-\beta _{i}(x_{1})]}{[x_{1}-\beta _{i}(x_{2})]}P_{1}.$ This follows from the fact that

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}x}\left( \frac{[x-\beta _{^{i}}(x_{1})]}{[x-\beta _{^{i}}(x_{2})]}\right) = \frac{[x-\beta _{^{i}}(x_{2})-x+\beta _{^{i}}(x_{1})]}{[x-\beta _{^{i}}(x_{2})]}=\frac{[\beta _{^{i}}(x_{1})-\beta _{^{i}}(x_{2})]}{[x-\beta _{^{i}}(x_{2})]^{2}}>0. \end{aligned}$$

$P_{2}>\frac{[x_{1}-\beta _{^{i}}(x_{1})]}{[x_{1}-\beta _{^{i}}(x_{2})]} P_{1}$ implies $\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i}).$

Note that, we implicitly assumed that $P_{1}>0.$ If $P_{1}=0,$ then $\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})=0$. Thus, the probability of provision is zero at $x_{1}$. Now, at this equilibrium with positive probability of provision, $(\beta _{i},\beta _{-i})$, where the probability of provision is strictly positive, define $x_{i}^{*}$ to be the infimum over the set of values of bidder $i$ such that the probability of provision in this equilibrium is positive. Then, the probability of provision is strictly positive for all $x\in (x_{i}^{*},1]$ and positive only for those values. Thus, $x_{1}<x^{*}$. Also, $\beta _{i}$ must be non-decreasing in $(x^{*},1]$, otherwise we could apply the same argument in the interval $(x^{*},1]$ for those values exhibiting decreasing bidding behaviour, and get a higher $x^{*}$ contradicting the definition of $x^{*}$. Since the probability of provision is zero for all $x\in [0,x_{i}^{*}],$ proofs of Lemma 4 and Lemma 5 can be used to show that this cannot be an equilibrium with positive probability of provision.

$(ii)\,c-x_{1}>\beta _{^{i}}(x_{1})>\beta _{^{i}}(x_{2})>c-x_{2}:$ The relevant payoffs are

$$\begin{aligned} \pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})&= [x_{1}-\beta _{^{i}}(x_{1})]P_{1}, \\ \pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})&= [x_{2}-\beta _{i}(x_{2})]P_{2}\\&\quad +\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \\ \pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})&= [x_{1}-\beta _{i}(x_{2})]P_{2}\\&\quad +\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \\ \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})&= [x_{2}-\beta _{^{i}}(x_{1})]P_{1}\\&\quad +\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\})}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

The inequality $\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})\ge \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})$ implies

$$\begin{aligned}&[x_{2}-\beta _{^{i}}(x_{2})]P_{2}+\int \limits _{Y_{L}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y\\&\quad \ge [x_{2}-\beta _{^{i}}(x_{1})]P_{1}+\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\})}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \end{aligned}$$

which implies $[x_{2}-\beta _{^{i}}(x_{2})]P_{2}\ge [x_{2}-\beta _{^{i}}(x_{1})]P_{1}+\int _{Y_{M}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y.$ Thus, $[x_{2}-\beta _{^{i}}(x_{2})]P_{2}\ge [x_{2}-\beta _{^{i}}(x_{1})]P_{1}.$ By a similar argument in Case $(i)$ above, we get $[x_{1}-\beta _{^{i}}(x_{2})]P_{2}>[x_{1}-\beta _{^{i}}(x_{1})]P_{1},$ which in turn implies

$$\begin{aligned}{}[x_{1}-\beta _{^{i}}(x_{2})]P_{2}+\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y>[x_{1}-\beta _{^{i}}(x_{1})]P_{1}, \end{aligned}$$

that is, $\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i}).$ ²¹

$(iii)\,\beta _{^{i}}(x_{1})>c-x_{1}>c-x_{2}>\beta _{^{i}}(x_{2}):$ The relevant payoffs are

$$\begin{aligned} \pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})&= [x_{1}-\beta _{^{i}}(x_{1})]P_{1}\\&\quad +\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \\ \pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})&= [x_{2}-\beta _{^{i}}(x_{2})]P_{2}, \\ \pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})&= [x_{1}-\beta _{^{i}}(x_{2})]P_{2}, \\ \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})&= [x_{2}-\beta _{^{i}}(x_{1})]P_{1}\\&\quad +\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

$\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})\ge \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})$ implies

$$\begin{aligned}{}[x_{2}-\beta _{^{i}}(x_{2})]P_{2}\ge [x_{2}-\beta _{^{i}}(x_{1})]P_{1}+\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \end{aligned}$$

which in turn implies

$$\begin{aligned} P_{2}\ge \frac{[x_{2}-\beta _{^{i}}(x_{1})]}{ [x_{2}-\beta _{^{i}}(x_{2})]}P_{1}+\frac{1}{[x_{2}-\beta _{^{i}}(x_{2})]} \int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

Since $x_{2}>x_{1}$, the above inequality implies

$$\begin{aligned} P_{2}>\frac{[x_{1}-\beta _{^{i}}(x_{1})]}{[x_{1}-\beta _{^{i}}(x_{2})]}P_{1}+\frac{1}{[x_{2}-\beta _{^{i}}(x_{2})]}\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

Arranging terms and adding and subtracting $\int _{Y_{M}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y$, we get,

$$\begin{aligned}&\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}))>\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}))\\&\quad +\,\frac{[x_{1}-\beta _{^{i}}(x_{2})]}{[x_{2}-\beta _{^{i}}(x_{2})]}\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y \\&\quad -\,\int \limits _{Y_{M}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

We are done if we can show

$$\begin{aligned}&\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}\frac{[x_{2}-c+\beta _{-i}(y)]}{[x_{2}-\beta _{^{i}}(x_{2})]}f(y)\mathrm{{d}}y\\&\qquad \ge \int \limits _{Y_{M}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{1}\}}\frac{[x_{1}-c+\beta _{-i}(y)]}{[x_{1}-\beta _{^{i}}(x_{2})]}f(y)\mathrm{{d}}y. \end{aligned}$$

First, we show that for any set of values $Y\subseteq Y_{M},$ we have

$$\begin{aligned} \int \limits _{Y}\frac{[x_{2}-c+\beta _{-i}(y)]}{[x_{2}-\beta _{^{i}}(x_{2})]} f(y)\mathrm{{d}}y>\int \limits _{Y}\frac{[x_{1}-c+\beta _{-i}(y)]}{[x_{1}-\beta _{^{i}}(x_{2})]} f(y)\mathrm{{d}}y. \end{aligned}$$

To see this,

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}x}\int \limits _{Y}\frac{[x-c+\beta _{-i}(y)]}{[x-\beta _{^{i}}(x_{2})]} f(y)\mathrm{{d}}y&= \int \limits _{Y}\frac{\mathrm{{d}}}{\mathrm{{d}}x}\frac{[x-c+\beta _{-i}(y)]}{[x-\beta _{^{i}}(x_{2})]}f(y)\mathrm{{d}}y \\&= \int \limits _{Y}\frac{[c-\beta _{^{i}}(x_{2})-\beta _{-i}(y)]}{[x-\beta _{^{i}}(x_{2})]^{2}}f(y)\mathrm{{d}}y >0. \end{aligned}$$

To see the last inequality, note that for any $y\in Y_{M}$ we have $\beta _{-i}(y)<\beta _{^{i}}(x_{1})\le x_{1}$ and $\beta _{^{i}}(x_{2})\le c-x_{2}<c-x_{1}.$ Thus, for any $y\in Y_{M},$ we have $\beta _{^{i}}(x_{2})+\beta _{-i}(y)<c.$ To conclude,

$$\begin{aligned}&\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}\frac{[x_{2}-c+\beta _{-i}(y)]}{[x_{2}-\beta _{^{i}}(x_{2})]}f(y)\mathrm{{d}}y\\&\quad >\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}\frac{[x_{1}-c+\beta _{-i}(y)]}{[x_{1}-\beta _{^{i}}(x_{2})]}f(y)\mathrm{{d}}y \\&\quad \ge \int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}}\frac{[x_{1}-c+\beta _{-i}(y)]}{[x_{1}-\beta _{^{i}}(x_{2})]}f(y)\mathrm{{d}}y. \end{aligned}$$

The last inequality above follows from $Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}\subseteq Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}$.²² $(iv)\,\beta _{^{i}}(x_{1})>c-x_{1}>\beta _{^{i}}(x_{2})>c-x_{2}:$ The relevant payoffs are

$$\begin{aligned}&\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i}) \!=\![x_{1}-\beta _{^{i}}(x_{1})]P_{1}\!+\!\int \limits _{Y_{M}\cap \{y|\beta _{-i}(y)\!\ge \! c\!-\!x_{1}\}}[x_{1}-c\!+\!\beta _{-i}(y)]f(y)\mathrm{{d}}y, \\&\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i}) \!=\![x_{2}\!-\!\beta _{^{i}}(x_{2})]P_{2}\!+\!\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\!\ge \! c-x_{2}\}}[x_{2}-c\!+\!\beta _{-i}(y)]f(y)\mathrm{{d}}y, \\&\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i}) =[x_{1}-\beta _{^{i}}(x_{2})]P_{2}, \\&\pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i}) =[x_{2}-\beta _{^{i}}(x_{1})]P_{1}\\&\qquad \qquad \qquad \qquad \qquad \qquad +\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{2}\})}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

The inequality $\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})\ge \pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})$ implies

$$\begin{aligned}&[x_{2}-\beta _{^{i}}(x_{2})]P_{2}+\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y \\&\qquad \ge [x_{2}-\beta _{^{i}}(x_{1})]P_{1}+\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\})}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

That is,

$$\begin{aligned}&P_{2}\ge \frac{[x_{2}-\beta _{^{i}}(x_{1})]}{ [x_{2}-\beta _{^{i}}(x_{2})]}P_{1}+\frac{1}{[x_{2}-\beta _{^{i}}(x_{2})]} \int \limits _{Y_{M}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

Since $x_{2}>x_{1}$, we get

$$\begin{aligned} P_{2}>\frac{[x_{1}-\beta _{^{i}}(x_{1})]}{[x_{1}-\beta _{^{i}}(x_{2})]}P_{1}+\frac{1}{[x_{2}-\beta _{^{i}}(x_{2})]} \int \limits _{Y_{M}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

After arranging terms, we add and subtract $\int _{Y_{M}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y$ to get

$$\begin{aligned}&\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})\\&\quad +\frac{[x_{1}-\beta _{^{i}}(x_{2})]}{[x_{2}-\beta _{^{i}}(x_{2})]}\int \limits _{Y_{M}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y-\int \limits _{Y_{M}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

Since $\beta _{^{i}}(x_{2})<c-x_{1},$ for any $y\in Y_{M},$ we have $\beta _{i}(x_{2})+\beta _{-i}(y)<c.$ So, for any $Y\subseteq Y_{M},$

$$\begin{aligned} \int \limits _{Y}\frac{[x_{2}-c+\beta _{-i}(y)]}{[x_{2}-\beta _{^{i}}(x_{2})]} f(y)\mathrm{{d}}y>\int \limits _{Y}\frac{[x_{1}-c+\beta _{-i}(y)]}{[x_{1}-\beta _{^{i}}(x_{2})]} f(y)\mathrm{{d}}y. \end{aligned}$$

Thus, $\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})$.²³

$(v)\,\beta _{i}(x_{1})>\beta _{i}(x_{2})>c-x_{1}>c-x_{2}:$ Since $\overline{ \beta }_{i}\le c/2$ and $\beta _{-i}(y)<\beta _{i}(x_{1})$ for any $y\in Y_{M}$, we have $\beta _{i}(x_{2})+\beta _{-i}(y)<c$ for any $y\in Y_{M}.$ Then, the argument given above in Case $(iv)$ proves that $\pi ^{i}(x_{1},\beta _{i}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{i}(x_{1}),\beta _{-i}).$ To see this,

$$\begin{aligned}&\pi ^{i}(x_{1},\beta _{^{i}}(x_{1}),\beta _{-i})=[x_{1}-\beta _{^{i}}(x_{1})]P_{1}+\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\})}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y,\\&\pi ^{i}(x_{2},\beta _{^{i}}(x_{2}),\beta _{-i})=[x_{2}-\beta _{^{i}}(x_{2})]P_{2}+\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y,\\&\pi ^{i}(x_{1},\beta _{^{i}}(x_{2}),\beta _{-i})=[x_{1}-\beta _{^{i}}(x_{2})]P_{2}+\int \limits _{Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y,\\&\pi ^{i}(x_{2},\beta _{^{i}}(x_{1}),\beta _{-i})=[x_{2}-\beta _{^{i}}(x_{1})]P_{1}+\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{2}\})}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y, \end{aligned}$$

where $P_{1}=\int _{Y_{H}\cap \{y|y\ge c-\beta _{^{i}}(x_{1})\}}f(y)\mathrm{{d}}y$ and $P_{2}=\int _{Y_{M}\cap \{y|y\ge c-\beta _{^{i}}(x_{2})\}}f(y)\mathrm{{d}}y.$ The inequality $\pi ^{i}(x_{2},\beta _{i}(x_{2}),\beta _{-i})\ge \pi ^{i}(x_{2},\beta _{i}(x_{1}),\beta _{-i})$ implies

$$\begin{aligned}&[x_{2}-\beta _{i}(x_{2})]P_{2}+\int \limits _{Y_{L}\cap \{y|\beta _{^{-i}}(y)\ge c-x_{2}\}}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y \\&\qquad \ge [x_{2}-\beta _{i}(x_{1})]P_{1}+\int \limits _{Y_{M}\cup (Y_{L}\cap \{y|\beta _{-i}(y)\ge c-x_{2}\})}[x_{2}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

That is,

$$\begin{aligned} P_{2}\ge \frac{[x_{2}-\beta _{i}(x_{1})]}{[x_{2}-\beta _{i}(x_{2})]}P_{1}+\frac{1}{[x_{2}-\beta _{i}(x_{2})]}\int \limits _{Y_{M}}[x_{2}-c+ \beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

Since $x_{2}>x_{1}$, we get

$$\begin{aligned} P_{2}>\frac{[x_{1}-\beta _{i}(x_{1})]}{[x_{1}-\beta _{i}(x_{2})]}P_{1}+\frac{1}{[x_{2}-\beta _{i}(x_{2})]}\int \limits _{Y_{M}}[x_{2}-c+ \beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

After arranging terms, we add and subtract $\int _{Y_{M}\cap \{y|\beta _{-i}(y)\ge c-x_{1}\}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y$ to get

$$\begin{aligned}&\pi (x_{1},\beta (x_{2}),\beta _{-i})\ge [x_{1}-\beta (x_{2})]P_{2}\\&\quad >\pi (x_{1},\beta (x_{1}),\beta _{-i})+\frac{[x_{1}-\beta (x_{2})]}{[x_{2}-\beta (x_{2})]}\int \limits _{Y_{M}}[x_{2}-c+ \beta _{-i}(y)]f(y)\mathrm{{d}}y\\&\quad -\int \limits _{Y_{M}}[x_{1}-c+\beta _{-i}(y)]f(y)\mathrm{{d}}y. \end{aligned}$$

Since $\overline{\beta }_{^{i}}\le c/2$ and $\beta _{-i}(y)<\beta _{i}(x_{1})$ for any $y\in Y_{M}$, we have $\beta _{i}(x_{2})+\beta _{-i}(y)<c$ for any $y\in Y_{M}.$ Thus, for any $Y\subseteq Y_{M},$ we have

$$\begin{aligned} \int \limits _{Y}\frac{[x_{2}-c+\beta _{-i}(y)]}{[x_{2}-\beta _{i}(x_{2})]} f(y)\mathrm{{d}}y>\int \limits _{Y}\frac{[x_{1}-c+\beta _{-i}(y)]}{[x_{1}-\beta _{i}(x_{2})]} f(y)\mathrm{{d}}y. \end{aligned}$$

Thus, as in Case $(iv)$ we have $\pi ^{i}(x_{1},\beta _{i}(x_{2}),\beta _{-i})>\pi ^{i}(x_{1},\beta _{i}(x_{1}),\beta _{-i}).$ $\square$

Proof of Proposition 1

Pick a non-decreasing and symmetric equilibrium of the endogenous order mechanism. Pick an arbitrary pair of valuations $x$ and $y$ such that $x+y>c.$ ²⁴ Denote the contribution function of the first contributor in the random order mechanism with $\tau (\cdot ),$ and the contribution function of the winner of the auction in the endogenous order mechanism with $\kappa (\cdot ),$ dropping $b_{1}$ and $b_{2}$ to save space.

Case 1 There is no provision in the random order mechanism regardless of the order Then, clearly endogenous order mechanism cannot do worse, simply because the probability of provision cannot be negative.

Case 2 There is provision in the random order mechanism regardless of the order Then there is also provision in the endogenous order mechanism. To see this, consider three subcases regarding possible information revelation in the endogenous order mechanism.

Case 2.1 The winner infers nothing about the valuation of the loser Then, the winner after paying $b_{1}$ (the bid of the loser) will contribute $\max \{\tau (x)-b_{1},0\}$ in the second stage; that is, his total contribution will be at least $\tau (x).$ So, if there is provision in the random order mechanism$,$ there is also provision in the equilibrium of the endogenous order mechanism.

Case 2.2 The winner learns the exact value of the loser Then the winner will contribute just as enough to make the loser pay the rest in the second stage. So, if there is provision in the random order mechanism$,$ there is also provision in the equilibrium of the endogenous order mechanism.

Case 2.3 The winner infers that the valuation of the loser lies in a subinterval $[a,d]$ of $[0,1]$ where $0\le a<d\le 1$. That is, the equilibrium bidding function of the loser is constant over the interval $[a,d].$ Since we are looking at the equilibrium path, Bayes’ rule implies that the posterior is

$$\begin{aligned} \widehat{f}=\left\{ \begin{array}{ll} zf(s) &{}\quad \text {if }a\le s\le d \\ 0 &{}\quad \text {otherwise} \end{array} \right. \end{aligned}$$

where $z=1/\int _{a}^{d}f(s)ds>1.$ Consider three subcases.

Case 2.3.1 Suppose that $c-\tau (x)\ge d.$ Then, the probability of provision in the random order mechanism is zero. The probability of provision in the equilibrium of the endogenous order is positive since $c-\kappa (x)\le d.$

Case 2.3.2 Suppose that $a<c-\tau (x)<d.$ That is, $c-d<\tau (x)<c-a.$ Note that $\tau (x)$ satisfies the first order condition

$$\begin{aligned} (x-\tau (x))f(c-\tau (x))-\int \limits _{c-\tau (x)}^{1}f(s)ds=0 \end{aligned}$$

Hence,

$$\begin{aligned} (x-\tau (x))\widehat{f}(c-\tau (x))-\int \limits _{c-\tau (x)}^{d}\widehat{f}(s)ds&=z\{(x-\tau (x))f(c-\tau (x))\\&\quad -\int \limits _{c-\tau (x)}^{d}f(s)ds\} \\&=z\int \limits _{d}^{1}f(s)ds \ge 0 \end{aligned}$$

Then, the optimal $\kappa (x)$ is at least as big as $\tau (x).$ Thus, if there is provision in the random order mechanism, then there is also provision in the equilibrium of the endogenous order mechanism.

Case 2.3.3 Suppose that $c-\tau (x)\le a.$ That is, $c-d<c-a\le \tau (x).$ Then for any $\kappa \in [c-d,c-a],$ we have $\kappa \le \tau (x),$ so the strict concavity of the ex ante objective function implies

$$\begin{aligned} (x-\kappa )f(c-\kappa )-\int \limits _{c-\kappa }^{1}f(s)ds>0 \end{aligned}$$

Hence,

$$\begin{aligned} (x-\kappa )\widehat{f}(c-\kappa )-\int \limits _{c-\kappa }^{d}\widehat{f}(s)ds&= z\{(x-\kappa )f(c-\kappa )-\int \limits _{c-\kappa }^{d}f(s)ds\} \\&> z\int \limits _{d}^{1}f(s)ds \ge 0 \end{aligned}$$

Thus, the optimal $\kappa (x)$ is a corner solution, which is $\kappa (x)=c-a.$ That is, $c-\kappa (x)=a.$ Thus, there is provision in the endogenous order mechanism, because $y\in [a,d]$ implies $y\ge c-\kappa (x).$

Case 3 There is provision in the random order mechanism for one order but not the other Without loss of generality, suppose there is provision when the bidder with valuation $x$ contributes first but not when the bidder with valuation $y$ contributes first. Thus, the probability of provision in the random order mechanism given $(x,y)$ is $1/2.$ We first prove that $x>y.$ Let $\tau (z)$ be the contribution of the first player in the random order mechanism, characterized by

$$\begin{aligned} (z-\tau (z))f(c-\tau (z))=1-F(c-\tau (z)) \end{aligned}$$

Define $G(z,\tau )=(z-\tau )f(c-\tau )-[1-F(c-\tau )]=0.$ Then, by implicit function theorem

$$\begin{aligned} \frac{\mathrm{{d}}\tau }{\mathrm{{d}}z}=-\frac{\partial G(z,\tau )/\partial z}{\partial G(z,\tau )/\partial \tau }=-\frac{f(c-\tau )}{\partial G(z,\tau )/\partial \tau } \end{aligned}$$

Note that $\partial G(z,\tau )/\partial \tau <0$, because it is the second derivative of the ex ante objective function which is strictly concave by the assumption $\frac{2f}{|f^{\prime }|}>2-c$. If $|\partial G(z,\tau )/\partial \tau |<f(c-\tau ),$ then $\frac{\mathrm{{d}}\tau }{\mathrm{{d}}z}>1.$ That is, $|-f^{\prime }(c-\tau )(z-\tau )-2f(c-\tau )|<f(c-\tau )$ implies $\frac{ \mathrm{{d}}\tau }{\mathrm{{d}}z}>1.$ Since $\partial G(z,\tau )/\partial \tau <0,$ we have $\frac{ \mathrm{{d}}\tau }{\mathrm{{d}}z}>1$ if $f^{\prime }(c-\tau )(z-\tau )+f(c-\tau )<0$; that is, $\frac{f(c-\tau )}{|f^{\prime }(c-\tau )|}<z-\tau$ since $f^{\prime }<0$. Since $2-c>$ $\frac{f}{|f^{\prime }|}$ implies $\frac{f(c-\tau )}{|f^{\prime }(c-\tau )|}<z-\tau ,$ we get $\frac{\mathrm{{d}}\tau }{\mathrm{{d}}z}>1.$ Since there is no provision when bidder with valuation $y$ moves first we have $x+\tau (y)<c.$ Also since there is provision when bidder with valuation $x$ moves first, we have $y+\tau (x)\ge c.$ Thus, $y+\tau (x)>x+\tau (y).$ That is, $\tau (x)-x>\tau (y)-y$ which implies $x>y$ since $\tau (z)-z$ is strictly increasing.

Since $x>y$ and the equilibrium is non-decreasing and symmetric, $\beta (x)\ge \beta (y).$ If $\beta (x)>\beta (y),$ then bidder with valuation $x$ wins, and argument in Case 2 applies; thus, there is provision. If $\beta (x)=\beta (y),$ then the bidder with valuation $x$ wins with probability $1/2.$ There is provision if she wins the auction; that is, the provision probability in the endogenous order mechanism given $(x,y)$ is $1/2.$

So we have shown that in all cases the probability of provision given $(x,y)$ in the endogenous order mechanism is at least as big as the probability of provision in the random order mechanism, given this $(x,y)$.$\square$