C. Lef`evre, S. Utev Insurance: Mathematics and Economics 28 2001 21–30 27
Proof. We begin by observing that by 3.16,
r
k
:= EX
k
P
m j =1
EX
j
= EY
k
P
m j =1
EY
j
, k = 1, . . . , m.
3.18 Now, by 3.15 and the scaling property 2.5 of the convex order,
r
k
X
k
EX
k cx
r
k
Y
k
EY
k
, k = 1, . . . , m.
Using the convolution property 2.4 of the convex order, we then get
m
X
k=1
r
k
X
k
EX
k cx
m
X
k=1
r
k
Y
k
EY
k
, k = 1, . . . , m,
which can be rewritten, by 3.18, as P
m k=1
X
k
P
m k=1
EX
k cx
P
m k=1
Y
k
P
m k=1
EY
k
, i.e., 3.17.
Counterexample 2. We show that without 3.16, the order
l
does not verify property 3.1 for any random variable X. Thus, the additional condition 3.16 is necessary for obtaining 3.17. Indeed, assume that 3.1 is satis-
fied by X with EX = 1. Fix Y with a non-degenerated distribution and EY = 1. We see that for Z, U ∈ G,
Z =
l
U if and only if ZEZ =
d
UEU, 3.19
and moreover, aZ =
l
bZ for any a, b 0.
3.20 Thus, 3.1 with Y
1
=
d
aY and Y
2
=
d
bY would imply X + aY
l
X + bY for any a, b 0.
3.21 Note that 3.16 is not satisfied. Since a and b can be permuted, the sign
l
in 3.21 is in fact a sign =
l
, so that by 3.19 and taking a = 1, we would have
X + Y =
d
X + bY 2
1 + b for any b 0.
But this equality holds if and only if either X = Y = 1 a.s., or X and Y have a positive Cauchy distribution, which has no finite mean, hence the contradiction.
4. Preservation under mixture
In this part, we are going to show that both orders
hamr
and
l
do not satisfy the mixture property 2.3 in general. Furthermore, we will prove that, surprisingly, the property holds true when condition 3.16 is added to the
hypotheses.
28 C. Lef`evre, S. Utev Insurance: Mathematics and Economics 28 2001 21–30
4.1. Mixture and hamr-order
Property 4.1. The mixture property is satisfied for
hamr
under the additional condition 3.16.
Proof. By definition, 2.3 for
hamr
means that
m
X
k,i=1
p
i
p
k
EX
k
− u
+
EY
i
≤
m
X
k,i=1
p
i
p
k
EY
k
− u
+
EX
i
. 4.1
Now, by 3.16 and since X
k hamr
Y
k
, we find that EX
k
− u
+
EY
i
= EX
k
− u
+
EX
k
EX
k
EY
i
≤ EY
k
− u
+
EY
k
EX
k
EY
i
= EY
k
− u
+
EX
i
EX
k
EY
k
. EX
i
EY
i
= EY
k
− u
+
EX
i
, which implies 4.1.
Counterexample 3. We show that the order
hamr
does not satisfy the mixture property 2.3 in general, and thus that the additional condition 3.16 is necessary. Fix positive a and u such that 1 1 + a u
3 2
. In the context of property 2.3, we take
m = 2, p
1
= p
2
= 0.5
and X
1
= 1, X
2
= 4, Y
1
= 1 + a, Y
2
= 4 + a.
First, we note that X
1 hamr
Y
1
and X
2 hamr
Y
2
, and 3.16 is not satisfied. Now, let L and R be the l.h.s. and r.h.s. of 4.1 calculated for these r.v.’s and at point u.
Since 1 + a u 4, we get L =
1 4
EX
2
− u
+
EY
1
+ EY
2
=
1 4
4 − u5 + 2a, and
R =
1 4
EY
2
− u
+
EX
1
+ EX
2
=
1 4
4 + a − u5. Thus L R when u
3 2
, i.e., the mixture property 2.3 does not hold. 4.2. Mixture and Lorenz order
Property 4.2. The mixture property is satisfied for
l
under the additional condition 3.16.
Proof. We proceed as for Property 3.4. By 3.16, we have that
s
k
:= EX
k
P
m j =1
EX
j
p
j
= EY
k
P
m j =1
EY
j
p
j
, k = 1, . . . , m.
4.2 Now, by 2.5 and 3.15, we get
s
k
X
k
EX
k cx
s
k
Y
k
EY
k
, k = 1, . . . , m.
C. Lef`evre, S. Utev Insurance: Mathematics and Economics 28 2001 21–30 29
Thus, for any convex function f ,
m
X
k=1
p
k
Ef [s
k
X
k
EX
k
] ≤
m
X
k=1
p
k
Ef [s
k
Y
k
EY
k
], i.e.,
m
X
k=1
p
k
Ef X
k
P
m j =1
EX
j
p
j
≤
m
X
k=1
p
k
Ef Y
k
P
m j =1
EY
j
p
j
, which is equivalent to 2.3.
Counterexample 4. We show that the order
l
does not satisfy the mixture property 2.3 in general, meaning that the extra condition 3.16 is necessary. Let X, Z be two independent r.v.’s with a negative exponential distribution
of parameter 1. Take X
1
=
d
X, X
2
=
d
X, Y
1
=
d
X, Y
2
=
d
Xa. Then,
X
1 l
Y
1
and X
2 l
Y
2
. Now, let us suppose that the mixture property is satisfied with m = 2 and 0 p
1
= 1 − p
2
1. Arguing as before in Counterexample 2, we get
F
X
p
1
+ F
X
p
2
=
d
F
Xb
p
1
+ F
Xab
p
2
for any a, b 0, which yields
e
− x
= e
− xb
p
1
+ e
− xba
p
2
for any x ≥ 0. But this is possible only when b = 1 = a, i.e., for the mixture of identically distributed r.v.’s.
5. Main result
