Online Appendix
”Core Deviation Minimizing Auctions”
by Isa E. Hafalir and Hadi Yektaş
April 23, 2014
Proposition 3 The core deviation minimization problem can be written as a standard calculus of variations
problem of the form
Z 2
A (r (v) , r0 (v) , v) dv.
min
(1)
r
0
Proof of Proposition 3 Denote
Z
2
min{v,1}
Z
Z
!
min{v−v1 ,1}
(v1 + v2 − r (v1 + v2 ))dF (v2 ) dF (v1 ) dG (r (v))
0
0
0
by O1 and
2
Z
Z
1
Z
!
1
max{r (v) − max{v − v1 , 0} − max{v − v2 , 0}, 0}dF (v2 ) dF (v1 ) dG (r (v))
0
max{v−1,0}
max{v−v1 ,0}
by O2 .
Denote the expectation of F by L,
v
Z
v 0 f (v 0 ) dv 0 .
L (v) =
(2)
0
Moreover, denote v1 + v2 by t and the distribution of t by H. That is
min{t,1}
Z
min{t−v1 ,1}
Z
H (t) =
dF (v2 ) dF (v1 ) .
0
(3)
0
Also denote its density by h, h (t) = H 0 (t) , and expectation by K,
t
Z
t0 h (t0 ) dt0 .
K (t) =
(4)
0
Lastly, denote v1 + v2 conditional on v1 , v2 < v by tv and its distribution by Hv . That is,
Z
min{tv ,v}
Z
min{t−v1 ,v}
Hv (tv ) =
dF (v2 ) dF (v1 ) .
0
Also denote its density by hv , hv (t) =
(5)
0
d
dt Hv
(t) , and expectation by Kv ,
Z
t
t0 hv (t0 ) dt0 .
Kv (t) =
0
1
(6)
Then, we can write
Z
2
Z
2
v
O1 =
0
0
Z
(t − r (t))dH (t) dG (r (v))
Z
K (v) −
=
0
v
r (t) h (t) dt dG (r (v))
0
2
Z
Z
2
Z
2
Z
v
K (v) dG (r (v)) −
=
0
0
0
2
Z
Z
K (v) dG (r (v)) −
=
2
Z
2
Z
0
dG (r (v)) r (t) h (t) dt
K (v) dG (r (v)) −
=
2
t
0
0
r (t) h (t) dt dG (r (v))
(1 − G (r (t))) r (t) h (t) dt
0
2
Z
(K (v) g (r (v)) r0 (v) − (1 − G (r (v))) r (v) h (v)) dv
=
0
where, to achieve the fourth line, we change order of integration.
On the other hand,
Z 1 Z 1 Z 1
r (v) dF (v2 ) dF (v1 ) dG (r (v))
O2 =
Z
0
1
Z
v
1
v
v
Z
+
0
v
1
Z
Z
max{r (v) − v + v2 , 0}dF (v2 ) dF (v1 ) dG (r (v))
0
v
1
Z
max{r (v) − v + v1 , 0}dF (v2 ) dF (v1 ) dG (r (v))
+
0
0
1 Z
Z
v
vZ
v
max{r (v) − 2v + v1 + v2 , 0}dF (v2 ) dF (v1 ) dG (r (v))
+
0
0
2
Z
Z
v−v1
Z 1
1
+
1
v−1
max{r (v) − 2v + v1 + v2 , 0}dF (v2 ) dF (v1 ) dG (r (v)) .
v−v1
Let us denote the first, second, third, fourth and fifth summands by O2,1 , O2,2 , O2,3 , O2,4 and O2,5
respectively.
We can write
Z
1
2
r (v) (1 − F (v)) dG (r (v))
O2,1 =
0
and
1
Z
1
Z
Z
!
v
(r (v) − v + v2 ) dF (v2 ) dF (v1 ) dG (r (v))
O2,2 =
0
v
1
Z
v−r(v)
1
Z
((r (v) − v) (F (v) − F (v − r (v))) + L (v) − L (v − r (v))) dF (v1 ) dG (r (v))
=
0
v
1
Z
((r (v) − v) (F (v) − F (v − r (v))) + L (v) − L (v − r (v))) (1 − F (v)) dG (r (v))
=
0
and
Z
1
Z
v
1
Z
O2,3 =
0
Z
0
v
max{r (v) − v + v1 , 0} (1 − F (v)) dF (v1 ) dG (r (v))
!
(r (v) − v + v1 ) dF (v1 ) (1 − F (v)) dG (r (v))
=
0
Z
v−r(v)
1
((r (v) − v) (F (v) − F (v − r (v))) + L (v) − L (v − r (v))) (1 − F (v)) dG (r (v))
=
0
= O2,2
2
and
Z
2v
1
Z
1
Z
O2,4 =
v
0
Z
2v
max{r (v) − 2v + tv , 0}dHv (tv ) dG (r (v))
!
(r (v) − 2v + tv ) dHv (tv ) dG (r (v))
=
2v−r(v)
0
Z
1
((r (v) − 2v) (Hv (2v) − Hv (2v − r (v))) + Kv (2v) − Kv (2v − r (v))) dG (r (v))
=
0
and lastly,
Z
2
2
Z
O2,5 =
v
1
Z
2
Z
2
max{r (v) − 2v + t, 0}dH (t) dG (r (v))
!
(r (v) − 2v + t) dH (t) dG (r (v))
=
2v−r(v)
1
Z
2
((r (v) − 2v) (1 − H (2v − r (v))) + K (2) − K (2v − r (v))) dG (r (v)) .
=
1
Adding all the terms up, we obtain
Z
1
(K (v) g (r (v)) r0 (v) − (1 − G (r (v))) r (v) h (v)) dv
O=
0
Z 1
+
2
r (v) (1 − F (v)) g (r (v)) r0 (v) dv
0
Z
1
2 ((r (v) − v) (F (v) − F (v − r (v))) + L (v) − L (v − r (v))) (1 − F (v)) g (r (v)) r0 (v) dv
+
0
Z
1
((r (v) − 2v) (Hv (2v) − Hv (2v − r (v))) + Kv (2v) − Kv (2v − r (v))) g (r (v)) r0 (v) dv
+
0
Z
2
(K (v) g (r (v)) r0 (v) − (1 − G (r (v))) r (v) h (v)) dv
+
1
Z
+
2
((r (v) − 2v) (1 − H (2v − r (v))) + K (2) − K (2v − r (v))) g (r (v)) r0 (v) dv.
1
This problem is now of the form (1)
3
(7)