A 1 2 f a 1 2 f a f a 1 2 thus a median f is the

( a ) = 0
⇒ 1 - 2 F ( a ) = 1 - 2 F ( a ) = 0
⇒ F ( a ) =
1
2
Thus, a
*
= median F is the unique critical point ( i.e. , satisfier of the first order necessary
condition). To be complete, check the second order sufficient condition. Recalling that
the first derivative of F exists and is denoted by f ,
d 2
d a 2
EV
F
( a ) =
d
d a
1 - 2 F ( a )
= - 2 f ( a )
Answer. The expected value of a choice a is
EV
F
( a ) =
∞
-∞
- a - x d F
=
a
-∞
- a - x d F +
∞
a
- a - x d F
=
a
-∞
- ( a - x ) d F +
∞
a
- ( x - a ) d F
Taking derivatives (in order to optimize with respect to the choice a )
d
d a
EV
F
( a ) =
d
d a
a
-∞
- ( a - x ) d F +
d
d a
∞
a
- ( x - a ) d F
=
a
-∞
d ( - ( a - x ))
d a
d F +
d a
d a
( - ( a - a )) +
∞
a
d ( - ( x - a ))
d a
d F +
d a
d a
( - ( a - a ))
=
a
-∞
d ( x - a )
d a
d F +
d a
d a
( - ( a - a )) +
∞
a
d ( a - x )
d a
d F +
d a
d a
( - ( a - a ))
=
a
-∞
( - 1 ) d F + ( 1 )( 0 ) +
∞
a
( 1 ) d F + ( 1 )( 0 )
=
a
-∞
( - 1 ) d F +
∞
a
( 1 ) d F
= - F ( a ) + ( 1 - F ( a ))
= 1 - 2 F ( a )
2

3