Handbook of Corporate Finance Empirical Corporate Finance Volume 1

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92 S. Dasgupta and R.G. Hansen


so that bidderj’s bid isb(vj).Ifb(v)represents a (symmetric) bidding equilibrium,
then bidderi’s optimal decision will be to reportv=vi, so that her bid isb(vi).
In the situation where the otherN−1 bidders are both usingb(v)and reporting
honestly, we can re-write(3)as


E(πi)=Pr(win)(vi−b)
=G(v) (4)

(


vi−b(v)

)


,


where we assume bidderiis usingb(v)but not requiringv=vi. Note that bidderiwins
if all otherN−1 values are less than thevthat bidderireports, hence the conversion
of Pr(win)intoG(v), the distribution for the highest value among the remainingN− 1
values.
Now we simply require that bidderi’s optimum decision is also honest reporting.
Taking the first derivative of(4)with respect tov,wehave


(5)


dE(πi)
dv

=g(v)

(


vi−b(v)

)


−G(v)

db(v)
dv

= 0.


The first term of(5)shows the marginal benefit of bidding higher while the second
term shows the marginal cost. Re-arranging, we have


G(v) (6)

db(v)
dv

=g(v)

(


vi−b(v)

)


.


For equilibrium, we require that(6)hold atv=vi. Hence we get

G(v) (7)

db(v)
dv

=g(v)

(


v−b(v)

)


.


Equation(7)is a standard first-order differential equation that can be solved via
integration-by-parts.^4 Doing this yields


b(v)= (8)

1


G(v)

∫v

0

yg(y) dy.

Equation(8)can be easily interpreted. AsG(x)is the distribution for the highest
value among the remainingN−1 values,g(x)/G(v)is the density of that value condi-
tional on it being lower thanv. Equation(8)tells a bidder to calculate the expected value
of the highest value among the remainingN−1 bidders, conditional on that value be-
ing less than bidderi’s own, and to bid that amount. This is about as far as intuition can
take us: the expected value of the second-highest value is in some sense bidderi’s real
competition, and equilibrium bidding calls for her to just meet that competition. (One
other intuitive approach involves marginal revenue; we will turn to this view below.)


(^4) Rewrite(7)asG(y) db+bdG=ydGord(G(y)b(y))=ydG. Integrating, and using the fact that
G( 0 )=0, we getb(v)=
∫v
0 ydG/G(v)=
∫v
0 yg(y) dy/G(v).

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