Separating equilibria in auctions with two types of bidders

Abstract

We consider two simultaneous, private value, second-price auctions with identical objects for sale and two types of bidders, strong and weak, in the sense of hazard rate stochastic dominance. We show that if the strong bidders are not too strong with respect to the weak bidders, then, by setting an appropriate reserve price, a separating equilibrium exists in which strong bidders will participate in the auction with only strong bidders, and weak bidders will participate in the auction with only weak bidders.

Appendices

Appendix A: Proof of Corollary 1

Using the proposition, it is sufficient to prove that $\lambda _{w}\ge \lambda _{s}$ and the condition $\frac{\lambda _{w}}{\lambda _{s}}\le \frac{1}{F_{s}}$ is satisfied for $F_{s}(x)=x^{\alpha }$ and $F_{w}(x)=x^{\beta \text { }}$where $\alpha >\beta .$

Let us first prove the first part, namely, $1\le \frac{\lambda _{w}}{ \lambda _{s}}.$

Substituting the distributions, we need to prove that the following inequality holds for all x

$$\begin{aligned} 1\le \frac{\beta x^{\beta -1\text { }}(1-x^{\alpha })}{\alpha x^{\alpha -1}(1-x^{\beta })}. \end{aligned}$$

or, alternatively, we need to show that

$$\begin{aligned} \alpha x^{\alpha -\beta }(1-x^{\beta })-\beta (1-x^{\alpha })\le 0 \end{aligned}$$

Note that in the case of $x=0$, we get $-\beta$, and when $x=1$, we get zero. Thus, what we have left to prove is that the function is increasing. The derivative of the LHS is

$$\begin{aligned} \frac{\partial \left( \alpha x^{\alpha -\beta }(1-x^{\beta })-\beta (1-x^{\alpha })\right) }{\partial x}=(\alpha -\beta )\alpha x^{\alpha -\beta -1}-(\alpha -\beta )\alpha x^{\alpha -1}\ge 0 \end{aligned}$$

providing the proof of the first part.

Now consider the proof of the following inequality $\frac{\lambda _{w}}{ \lambda _{s}}\le \frac{1}{F_{s}}.$ In other words,

$$\begin{aligned} \frac{\beta x^{\beta -1\text { }}(1-x^{\alpha })}{\alpha x^{\alpha -1}(1-x^{\beta })}\le x^{-\alpha } \end{aligned}$$

or, alternatively, we need to show that

$$\begin{aligned} \beta x^{\beta \text { }}(1-x^{\alpha })-\alpha (1-x^{\beta })\le 0 \end{aligned}$$

Note that when $x=0$, we get $-\alpha$, and if $x=1$, we get zero. Then, if the function is increasing, we will complete the proof. Differentiating the LHS gives

$$\begin{aligned} \frac{\partial \left( \beta x^{\beta \text { }}(1-x^{\alpha })-\alpha (1-x^{\beta })\right) }{\partial x}=(\alpha +\beta )\beta x^{\beta -1}-(\alpha +\beta )\beta x^{\alpha +\beta -1}\ge 0. \end{aligned}$$

$\square$

Appendix B: Proof of Corollary 2

First, we show that $F_{s}=x^{\alpha }$ stochastically dominates $F_{w}(v)= \frac{1-e^{-\lambda x}}{1-e^{-\lambda }}$ in the sense of the hazard rate. Namely, $\lambda _{w}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda x}-e^{-\lambda }}\ge \lambda _{s}=\frac{\alpha x^{\alpha -1}}{1-x^{\alpha }}$ or $\frac{\lambda e^{-\lambda x}}{e^{-\lambda x}-e^{-\lambda }}-\frac{ \alpha x^{\alpha -1}}{1-x^{\alpha }}\ge 0$. Alternatively, for $x<1$ and $x>0$ we should prove that $z(x)=\lambda e^{-\lambda x}\left( 1-x^{\alpha }\right) -\alpha x^{\alpha -1}\left( e^{-\lambda x}-e^{-\lambda }\right) \ge 0$. Observe that in the bounds, $z(0)=\lambda$ and that $z(1)=0$ thus, if $z^{\prime }(x)<0$ we find that $z(x)\ge 0$. Differentiating z(x) gives

$$\begin{aligned} z^{\prime }(x)=-\lambda ^{2}e^{-\lambda x}(1-x^{\alpha })-\alpha (\alpha -1)x^{\alpha -2}\left( e^{-\lambda x}-e^{-\lambda }\right) . \end{aligned}$$

Observe that $e^{-\lambda x}-e^{-\lambda }>0$ then, $z^{\prime }(x)<0$ if $\alpha \ge 1$ and thus $\lambda _{w}\ge \lambda _{s}$. To prove that $\frac{\lambda _{w}}{\lambda _{s}}\le \frac{1}{F_{s}}$ we show that $\lambda _{w}F_{s}\le \lambda _{s}$ namely, $\frac{\lambda e^{-\lambda x}}{ e^{-\lambda x}-e^{-\lambda }}x^{\alpha }\le \frac{\alpha x^{\alpha -1}}{ 1-x^{\alpha }}$ or $\frac{\alpha }{1-x^{\alpha }}-\frac{\lambda xe^{-\lambda x}}{e^{-\lambda x}-e^{-\lambda }}\ge 0$. Again, for $x<1$ we may show that $T(x)=\alpha \left( e^{-\lambda x}-e^{-\lambda }\right) -\lambda xe^{-\lambda x}\left( 1-x^{\alpha }\right) >0$. In the bounds $T(0)=\alpha \left( 1-e^{-\lambda }\right) >0$ and $T(1)=0$ thus, if $T^{\prime }(x)<0$ we complete the proof. Differentiating T(x) gives

$$\begin{aligned} T^{\prime }(x)=-\lambda e^{-\lambda x}\left( 1-x^{\alpha }\right) (\alpha +1-\lambda x). \end{aligned}$$

If $\alpha +1-\lambda x>0$ we have $T^{\prime }(x)<0$ and this condetion is satidfied if $\lambda <\alpha +1$. $\square$