Probabilistic procurement auctions

Abstract

We analyse procurement auctions in which sellers are distinguished on the basis of the ratios of quality per unit of money that they offer. Sellers are privately informed on the offered quality of the technology or good. We assume that the procurer cannot perfectly identify the best offer. Thus, with positive and decreasing probability, the second, third, etc. best ratio offered is selected as the winner of the auction. We model this decision process as based on a general noisy ranking of offers. We show that, although the problem seems to be analytically intractable in general, there exists a simple symmetric, pure-strategy equilibrium in which everyone follows the simple heuristic to match the same ‘focal’ price–quality ratio.

Appendix

Proof of Proposition 1

Consider seller $i$’s utility maximisation problem, given that all other sellers $j\ne i$ choose their bids $b_j$ such that they all offer the same constant quality–price ratio $c$, i.e., $b_j = \beta (\theta _j)=\theta _j/c$. Seller $i$ needs to choose a bid $b_i$, given the quality $\theta _i$, in order to maximise

$$\begin{aligned} u_i(\tilde{c}_i) = b_i \int \limits _{\Theta _{-i}} \pi _i(\tilde{c}_i) f_{-i}(\theta _{-i}) d \theta _{-i} \end{aligned}$$

(18)

in which

$$\begin{aligned} {\tilde{c}_i = \left( c_{i1}, c_{i2}, \ldots , c_{ii},\ldots , c_{in}\right) = \left( \frac{c_i}{c}, \frac{c_i}{c}, \ldots , 1, \ldots , \frac{c_i}{c}\right) } \end{aligned}$$

(19)

and $c_i = \theta _i / b_i$. Thus,

$$\begin{aligned} {u_i (\tilde{c}_i) = b_i \pi _i(\tilde{c}_i) = b_i \pi _i\left( \frac{\theta _i}{b_i c}, \frac{\theta _i}{b_i c}, \ldots , 1, \ldots , \frac{\theta _i}{b_i c}\right) .} \end{aligned}$$

(20)

Note that the slope of $c_{ii}$ with respect to $b_i$ is identically zero because $c_{ii}\equiv 1$. Thus, we only need to consider $\tfrac{\partial \pi _i(\tilde{c}_i)}{\partial b_i}$ for $j\ne i$ in the following. The first-order condition is given by

$$\begin{aligned} {\frac{\partial u_i (\tilde{c}_i)}{\partial b_i} = 0 \;\iff \; \pi _i(\tilde{c}_i) + b_i \sum \limits _{j\ne i} \left( \frac{\partial \pi _i(\tilde{c}_i)}{\partial c_{ij}}\frac{\partial c_{ij}}{\partial b_i}\right) = 0.} \end{aligned}$$

(21)

In our symmetric, pure-strategy equilibrium candidate, all players choose the constant ratio $c$, i.e., $c_i = c$ and $\tilde{c}_i = (\tfrac{c}{c}, \tfrac{c}{c}, \ldots , 1, \ldots , \tfrac{c}{c}) = (1,1,\ldots , 1) = \mathbf {1}$, implying $\pi _i(\mathbf {1}) = 1/n$. By A1, the slopes $\tfrac{\partial \pi _i(\tilde{c}_i)}{\partial c_{ij}}$ for $j\ne i$ are equal. Recall that the uniform ratio $c$ implies $c_{ij}=1$ in which case that slope was denoted by $\pi '(\mathbf {1})$. Finally,

$$\begin{aligned} \frac{\partial c_{ij}}{\partial b_i} = - \frac{\theta _i}{b_i^2 c}. \end{aligned}$$

(22)

Evaluated at the equilibrium candidate (in which $\theta _i / b_i = c$), this is $\tfrac{\partial c_{ij}}{\partial b_i} = - \tfrac{1}{b_i}$. Therefore, the first-order condition, evaluated at the equilibrium candidate, can be written as

$$\begin{aligned} {\pi _i(\mathbf {1}) + b_i (n-1) \pi '(\mathbf {1}) \left( -\frac{1}{b_i}\right) = 0 \; \iff \; \pi '(\mathbf {1}) = \frac{1}{n(n-1)}} \end{aligned}$$

(23)

which establishes our claim.$\square$

Proof of Proposition 2

Inserting the equilibrium candidate $b_j= \theta _j/c$, (2) simplifies to

$$\begin{aligned} {u_i(\theta _i,b_i) = b_i \int \limits _{\Theta _{-i}} \frac{(\theta _i/b_i)^r}{(\theta _i/b_i)^r + (n-1)c^r} f_{-i}(\theta _{-i}) d \theta _{-i} = b_i \frac{(\theta _i/b_i)^r}{(\theta _i/b_i)^r + (n-1)c^r}.} \nonumber \\ \end{aligned}$$

(24)

The first-order condition with respect to $b_i$ is

$$\begin{aligned} \frac{\left( \frac{\theta _i}{b_i}\right) ^r \left( \left( \frac{\theta _i}{b_i}\right) ^r-(n-1) (r-1)c^r\right) }{\left( \left( \frac{\theta _i}{b_i}\right) ^r+(n-1)c^r\right) ^2} = 0. \end{aligned}$$

(25)

Note that only the big parenthesis in the numerator is of importance, given that all players’ ratios are positive. Inserting the symmetric candidate $b_i = \theta _i /c$, simplifying, and solving for $r$ we get

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

(26)

In order to demonstrate existence, we evaluate player $i$’s incentives to deviate from the equilibrium candidate quality–price ratio $c$. To simplify the exposition, we denote $i$’s deviations from quality–price ratio $c$ by $k c$ with $k> 0$, i.e., we consider bids $b_i = \theta _i / (kc)>0$ which implies that $i$ offers a quality–price ratio of $k c$, with $k=1$ in equilibrium. Thus, the payoff from ‘deviation $k$’ is computed by inserting $b_i = \theta _i / (kc)$ in (24),

$$\begin{aligned} {u_i^{dev} (k) = \frac{\theta _i}{k c} \frac{(k c)^r}{(k c)^r + (n-1)c^r}= \frac{\theta _i}{c} \frac{k^{r-1}}{k^r+n-1},\quad r=\frac{n}{n-1}.} \end{aligned}$$

(27)

In the equilibrium candidate, $k=1$,

$$\begin{aligned} u_i^{dev} (1) = \frac{\theta _i}{cn}. \end{aligned}$$

(28)

We claim that $k=1$ is the most profitable deviation, confirming the equilibrium candidate. Thus, we want to show that

$$\begin{aligned} {u_i^{dev}(1) > u_i^{dev}(k), \quad \forall k\ne 1, \quad r=\frac{n}{n-1}.} \end{aligned}$$

(29)

This is equivalent to

$$\begin{aligned} {\frac{1}{n} > \frac{k^{\frac{1}{n-1}}}{k^{\frac{n}{n-1}}+n\!-\!1} \iff k^{\frac{n}{n-1}}+n\!-\!1 > n k^{\frac{1}{n-1}}\iff n-1 > (n-k)k^{\frac{1}{n-1}}.}\qquad \end{aligned}$$

(30)

The latter inequality is obviously satisfied for $k\ge n$ (then the right-hand side is nonpositive). Thus, it remains to consider $k<n$. Note that the left-hand side of the inequality is constant, whereas the right-hand side depends on $k$. In the following, we maximise the right-hand side. The first-order condition for a maximum is

$$\begin{aligned} {\frac{\partial }{\partial k} \left( (n-k)k^{\frac{1}{n-1}}\right) = 0 \iff \cdots \iff k^{\frac{1}{n-1}} (k-1) = 0.} \end{aligned}$$

(31)

Since $k>0$, the only solution is $k=1$. Note that the second derivative is negative:

$$\begin{aligned} {\frac{\partial ^2}{(\partial k)^2} \left( (n-k)k^{\frac{1}{n-1}}\right) = \frac{n}{(n-1)^2}k^{\frac{3-2n}{n-1}}(2-n-k) < 0.} \end{aligned}$$

(32)

Thus, the right-hand side of our condition is uniquely maximised at $k=1$. The value of the right-hand side at $k=1$ is $n-1$ which implies that the condition holds for all $k\ne 1$. Finally, since (28) is positive, $b_i = 0$ can be ruled out as a best response. This completes the proof.$\square$

Proof of Proposition 3

The proof is very similar to that of Proposition 1. Hence, we concentrate mainly on the differences. Here, seller $i$ needs to choose a bid $b_i$, given the quality $\theta _i$, in order to maximise

$$\begin{aligned} {u_i (\tilde{c}_i) = (b_i-\gamma \theta _i) \pi _i(\tilde{c}_i) = (b_i-\gamma \theta _i) \pi _i\left( \frac{\theta _i}{b_i c}, \frac{\theta _i}{b_i c}, \ldots , 1, \ldots , \frac{\theta _i}{b_i c}\right) .} \end{aligned}$$

(33)

The corresponding first-order condition is given by

$$\begin{aligned} {\frac{\partial u_i (\tilde{c}_i)}{\partial b_i} = 0 \;\iff \; \pi _i(\tilde{c}_i) + (b_i-\gamma \theta _i) \sum \limits _{j\ne i} \left( \frac{\partial \pi _i(\tilde{c}_i)}{\partial c_{ij}}\frac{\partial c_{ij}}{\partial b_i}\right) = 0.} \end{aligned}$$

(34)

Evaluated at the equilibrium candidate $c=\theta _i/b_i$, i.e., inserting $\tilde{c}_i = (\tfrac{c}{c}, \tfrac{c}{c}, \ldots , 1, \ldots , \tfrac{c}{c}) = (1,1,\ldots , 1) = \mathbf {1}$, $\frac{\partial c_{ij}}{\partial b_i}$ $=-\tfrac{1}{b_i}$, and, due to $c_{ij}=1$, denoting $\pi '(\mathbf {1})$, the first-order condition can be written as

$$\begin{aligned} \begin{array}{c} \pi _i(\mathbf {1}) + (b_i-\gamma \theta _i) (n-1) \pi '(\mathbf {1}) \left( -\dfrac{1}{b_i}\right) = 0 \\ \iff \dfrac{1}{n} - (1-\gamma c) (n-1) \pi '(\mathbf {1}) = 0 \\ \iff c = \dfrac{1}{\gamma }\left( 1- \dfrac{1}{n(n-1)\pi '(\mathbf {1})}\right) \end{array}\end{aligned}$$

(35)

which establishes our claim.

Proof of Proposition 4

Consider Eq. (4). First, we determine the equilibrium value of $c$. The derivative of (4) w.r.t. $b_i$ can be written as

$$\begin{aligned} \theta _i \frac{\left( \frac{\theta _i}{b_i}\right) ^r \left( b_i \left( \frac{\theta _i}{b_i}\right) ^r+c^r (n-1)(b_i(1- r)+r \gamma \theta _i)\right) }{b_i \left( c^r(n-1)+\left( \frac{\theta _i}{b_i}\right) ^r\right) ^2}. \end{aligned}$$

(36)

Inserting the candidate $b_i = \theta _i/c$, this can be simplified to

$$\begin{aligned} \frac{n+(n-1)r(c \gamma -1)}{n^2}. \end{aligned}$$

(37)

The first-order condition (i.e., setting the above equal to zero) delivers the equilibrium value of $c$,

$$\begin{aligned} c = \frac{(n-1) r-n}{(n-1) r \gamma }. \end{aligned}$$

(38)

Second, we evaluate player $i$’s deviation incentives if all $n-1$ rivals bid the above ratio. We do this by expressing, w.l.o.g., $i$’s deviations from ratio $c$ by $k c$ in which $k>0$. Thus, $k=1$ represents ‘no deviation.’ We insert (38) as well as $b_i = \theta _i/(k c)$ into (4). Then seller $i$’s objective is to maximise

$$\begin{aligned} \frac{k^{r-1} (k n-(k-1) (n-1) r) \gamma \theta _i}{\left( k^r+n-1\right) (n r-n-r)}. \end{aligned}$$

(39)

The derivative w.r.t. deviation $k$ of the above is

$$\begin{aligned} \frac{k^{r-2} (n-1) r \overbrace{\left( \left( 1-k^r\right) +(1-k)(n r-n-r)\right) }^{=(A)} \gamma \theta i}{\left( k^r+n-1\right) ^2 (n r-n-r)} \end{aligned}$$

(40)

Observe that the term $(nr-n-r)$ occurs twice in the above, and is positive iff $r > n/(n-1)$. Thus, our assumption of $r>2$ is sufficient to make this term positive. Therefore, the sign of (40) is determined by the sign of term $(A)$. It is easy to see that this term is positive if $k<1$ and negative for $k>1$, confirming existence of the equilibrium.$\square$

Proof of Proposition 5

By assumption, qualities are fixed and observable by the buyer. Therefore, a seller’s choice variable is the financial bid, $\beta (\theta )$, and each bid implies a unique quality–price ratio. W.l.o.g., we express strategies in terms of quality–price ratios in which a ratio is denoted by $\tilde{\beta }(\theta ) = \tfrac{\theta }{\beta (\theta )}$. We conjecture that there is a symmetric, pure-strategy equilibrium, where every seller bids a monotonically increasing (and, thus, invertible) quality–price ratio, i.e., $\tilde{\beta }'(\theta )>0$. Denote the c.d.f. of the largest quality among $n-1$ sellers by $G(\theta )=F^{n-1}(\theta )$ and its density by $g(\theta )$. Suppose seller $i$ submits the financial bid $b_i$. We derive the first-order condition of seller $i$’s expected profit maximization given that the other $j\ne i$ sellers bid according to the ratio $\tilde{\beta }(\theta _j)$.

$$\begin{aligned} \pi _i(\theta _i)&= \Pr \left\{ \frac{\theta _i}{b_i} > \max _{j\ne i} \tilde{\beta }(\theta _j) \right\} b_i = \Pr \left\{ \tilde{\beta }^{-1}\left( \frac{\theta _i}{b_i}\right) > \max _{j\ne i} \theta _j\right\} b_i\nonumber \\&= G\left( \tilde{\beta }^{-1}\left( \frac{\theta _i}{b_i}\right) \right) b_i. \nonumber \\ \pi _i'(\theta _i)&= \left( - \frac{g\left( \tilde{\beta }^{-1}\left( \frac{\theta _i}{b_i}\right) \right) }{\tilde{\beta }'\left( \tilde{\beta }^{-1}\left( \frac{\theta _i}{b_i}\right) \right) }\frac{\theta _i}{b_i^2}\right) b_i + G\left( \tilde{\beta }^{-1}\left( \frac{\theta _i}{b_i}\right) \right) =0. \end{aligned}$$

(41)

In symmetric equilibrium, $\tilde{\beta }(\theta _i) = \frac{\theta _i}{b_i}$. Thus, the first-order condition simplifies to the differential equation

$$\begin{aligned} {- \frac{g(\theta _i)}{\tilde{\beta }'(\theta _i)} \tilde{\beta }(\theta _i) + G(\theta _i) = 0 \iff \frac{G(\theta _i)}{g(\theta _i)} = \frac{\tilde{\beta }(\theta _i)}{\tilde{\beta }'(\theta _i)}} \end{aligned}$$

(42)

The solution of this differential equation is

$$\begin{aligned} \tilde{\beta }(\theta ) = k G(\theta ), \end{aligned}$$

(43)

in which $k>0$ is a constant. Thus, the ratio is indeed monotonically increasing in quality. The associated financial bid $\beta (\theta )$ is determined using $\tilde{\beta }(\theta ) = \theta / \beta (\theta )$, i.e., $\beta (\theta ) = \theta / (k G(\theta ))$. In order to see that the above candidate is an equilibrium, we insert it into $i$’s expected profit from (41).

$$\begin{aligned} \pi _i(\theta _i)&= \Pr \left\{ \dfrac{\theta _i}{b_i} > \max _{j\ne i} k G(\theta _j) \right\} b_i = \Pr \left\{ \dfrac{\theta _i}{b_i} > k G\left( \max _{j\ne i} \theta _j\right) \right\} b_i\nonumber \\&= \Pr \left\{ G^{-1}\left( \dfrac{\theta _i}{k b_i}\right) > \max _{j\ne i} \theta _j \right\} b_i= G\left( G^{-1}\left( \dfrac{\theta _i}{k b_i}\right) \right) b_i = \dfrac{\theta _i}{k}.\quad \end{aligned}$$

(44)

Thus, $i$’s expected profit is constant, implying that the ratio $\tilde{\beta }(\theta _i)$ is a best response to the other sellers’ ratios $\tilde{\beta }(\theta _j)$, $j\ne i$.$\square$