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Ambiguity Increases and Insurance Deductibles

�
� �
if
−1
,
−1 if ,−1 if ,
� � �
�� ��
��
�

if , , �
= �
= � − if = � � − if ,
−
� �
�
� � � � � �
otherwise,0
0 0 oth erw is e , otherwise,
where is a positive scalar that makes satisfy the integration by parts of FOC (8). Then, we is a positive scalar that makes satisfy the integration by parts of FOC (8). Then, we is a positive scalar that makes satisfy the integration by parts of FOC (8). Then, we
��
�� ��
where
where
� � � � � �
find that
find that find that
(1+ )�1− ( ; )��1 + (1 + )�1 − ( ; )�� 1 + (1 + )�1 − ( ; )�� 1 + (1 + )�1 − ( ; )��
∗ ∗
∗
∗ ∗
∗
∗ ∗
∗
∗ ∗
∗
(1+ )�1− ( ; )��
(1+ )�1− ( ; )��
� � � � � �
∗ ∗ ∗
� � �
� �
�
∗ ∗
∗
× �� ( − − ( ))×[ � � ; � − � ; ��� − � ; �� � �
× �� ( −
× �� ( − − ( ))×[ � � ; − ( ))×[ � � ; � − � ; ��
� �
� �
� �
�
� � �
+ (1− )� ( ; ) − ( ; )�] � − ( ; )�] � ) − ( ; )�] �
+ (1− )� ( ; ) + (1− )� ( ;
�
�
� �
� �
� � �
− � � �
∗ ∗
∗
� �
�
∗ ∗
� �
∗
�
= (1 + )�1 − ( ; )��− + −
= (1 + )�1
�
= (1 + )�1 − ( ; )��− + � ( ; )��− + � − � �
− �
�
� �
� �
� � � �
= (1 + )�1 − ( ; )�(− ) >0, − ) >0, ; )�(− ) >0,
∗
∗
∗ ∗
� �
∗ ∗
�
= (1 + )�1 − ( ; )�( + )�1 − (
= (1
� � � � � �
which co
ntradi
.
which contradicts Equation (D1).
on (D1)
s Equati
ct
which contradicts Equation (D1).
� �
�
We now prove that Condition (D2) holds for all ( , 1] and [0, ] if and only if, for , 1] and [0, ] if and only if, for , 1] and [0, ] if and only if, for
∗
∗ ∗
We now prove that Condition (D2) holds for all (
We now prove that Condition (D2) holds for all (
� � � � � �
all [0, ], ], all [0, ],
∗
∗ ∗
all [0,
� � �
� ; � ≥ � ; � , ; � , � � � � ; � ≥ � ; � � , ( D D 3 3 ) ) ( ( D 3 )
� ; � ≥ �
� �
� ( ; )− ( ; )� + [ ( ; ) − ( ; )] ≥0.)− ( ; )� + [ ( ; ) − ( ; )] ≥0.)− ( ; )� + [ ( ; � � ) − ( ; ( )] ≥0. D D 4 4 ) ) ( ( D 4 )
� ( ;
� ( ;
�
� �
� �
� �
�
� �
The if part: Rewrite Condition (D2) as Rewrite Condition (D2) as
The if part: Rewrite Condition (D2) as
:
if

The
part

(2 − 1)� � ; � − � ; �� � − � ; ��
(2 − 1)� �
(2 − 1)� � ; � − � ; �� ;
� � � � � �
.
(
D5)
+ (1 − ) [ ( ; )− ( ; )] + [ ( ; )− ( ; )]} ≥ 0. (D5) ; )]} ≥ 0. (D5)
+
)−
;
≥

)−
;
+
;
(
;
;
)

[
[
)]

+
(

;

(1
−
)]}

[
(

;
0
)]

(

(
(
(

)
−
[
)−

(1
+
)−

(
�
�
�
�
� �
� �
� �
� �
e
i

d
e
b
Note
a
n
n
h
3

t
d
)
.
f

Note
g
g

o
.
C
e
D
io
n
Note that the term [ ( ; ) − ( ; )] can be negative. Thus, if Conditions (D3) and (D4) hold s (D3) and (D4) hold
e

t
a

h
(
b
T

a
h
that
c
o
s
f
the
,

s
iv

d
term
i
u

t
n
iv
i
t

n
the
,

l
o
a
D
u
)
n
n
e
that
i
n
4
(
C
io

e

T

c
s

a
d
n
)][
)]
term [
(
;

;

−
)

(
;

(

;
)
(

−
�
�
� �
� �
for all [0, ], then Condition (D5) holds for all ( , 1] and all [0, ]. ], then Condition (D5) holds for all ( , 1] and all [0, ]. ], then Condition (D5) holds for all ( , 1] and all [0, ].
�
� �
∗ ∗
∗
∗ ∗
∗
for all [0,
for all [0,
� � � � � �
� � �
e
v
h
f

pa
o

The
t
n
o

y
i

l
a
W
:

e
t
,
t
f
pr
The only if part: We prove that, if there exists [0, ] such that ( ; )< ( ; ) or ] such that ( ; )< ( ; ) or ] such that ( ; )< ( ; ) or
r
i

∗
∗ ∗
The only if part: We prove that, if there exists [0, there exists [0,
� �
�
�
� � �
� �
�
� ( ; ) − ( ; )� + [ ( ; )− ( ; )] <0, then there exists [ , 1] such that the ) − ( ; )� + [ ( ; )− ( ; )] <0, then there exists [ , 1] such that the ) − ( ; )� + [ ( ; )− ( ; )] <0, then there exists [ , 1] such that the
� �
� ( ;
� ( ;
�
�
� �
�
� �
� �
� �
�
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c
e
lo
th
Con
ases.
lo
l
.
f

v
wo
wing
f

Case
o

e
l
o
s
ider
t
t
a

Conditi
f
o
si
Conditi
D5)
negati
v
ev
is
on
(

D5)
(

o
f
left-hand

on
d
e

aluated
si
e
d
ev
a
ider
t
Con
left-hand
s
th
e
negati

left-hand side of Condition (D5) evaluated at is negative. Consider the following two cases. Case wing two cases. Case
is
e
aluated
.

1:
S
upp
∗ ∗
1: Suppose that there exists ose that there exists � � � � �� ∗ � � � � � � � �
1: Suppose that there exists [0, ] such that ( ; )< ( ; ). When evaluated at = [0, ] such that ( ; )< ( ; ). When evaluated at = [0, ] such that ( ; )< ( ; ). When evaluated at =
� �
� �
negative.
iti
on
ase
,
(D
sid
the
f
e
l
-hand
2:
t
C
p
i

f
u
s
S
5)
1, the left-hand side of Condition (D5) is negative. Case 2: Suppose that there exists [0, ] [0, ]
1, the left-hand side of Condition (D5) is negative. Case 2: Suppose that there exists pose that there exists [0, ]
Cond

o

e
∗ ∗
∗
1
� �
�
�
� �
such that � ( ; )− ( ; )� + [ ( ; ) − ( ; )] <0. We can find an = + , ; )− ( ; )� + [ ( ; ) − ( ; )] <0. We can find an = + , ; )− ( ; )� + [ ( ; ) − ( ; )] <0. We can find an = + ,
� �
�
such that � (
such that � (
� �
� � �
�
� �
� �
�
�
�
�
� �
�
�
� �
� �
� �
� � �
where is positive and sufficiently small so that the left-hand side of Condition (D5) is negative.
where is positive and sufficiently small so that the left-hand side of Condition (D5) is negative. Condition (D5) is negative.
tly s
m
an
f
ficien
al
hat th
e
t
l so
left-h
is po
sitiv
of
wher
e
de
d
and su
e
si

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