statistique, théorie et gestion de portefeuille - Docs at ISFA

where E0[·] denotes the expectation with respect to the power-law distribution f0. We have
x
c E0
u
E0 ln
=
b
, and c < b,
b − c
(121)
x
u
= 1
x
c E0 ln
u
,
b
(122)
x
u
=
b
.
(b − c) 2 (123)
Thus, we easily obtain that the unique solution of (119) is c = 0, and equation (120) does not make sense
any more. So, under H0, ĉ goes to zero and dˆ is not well defined. Thus, Wald test cannot be performed under
such a null hypothesis. We must find another way to test (SE) against H0.
In this goal, we remark that the quantity
ˆηT = ĉ
ĉ
u
+ 1
dˆ
101
(124)
is still well defined. Using (120), it is easy to show that, as T goes to infinity, this quantity goes to b,
whatever ĉ being positive or equal to zero. For positive c, this is obvious from (120) and (121), while for
c = 0, expanding (118) around ĉ = 0 yields
ĉ
ĉ u
dˆ
=
1
T
T
∑
i=1
ln xi
u
−1
(125)
= ˆb → b. (126)
In order to test the descriptive power of the (SE) model against the null hypothesis that the true model is the
Pareto model, we can consider the statistic
ˆηT
ζT = T − 1 , (127)
ˆb
which asymptoticaly follows a χ 2 -distribution with one degree of freedom. Indeed expanding the quantity
(xi/u) ĉ in power series around c = 0 gives
which allows us to get
where
xi
ĉ
∼= 1 + ĉ · log(
u
xi
u
) + ĉ2
2 · log2
xi
+
u
ĉ3
6 · log3
xi
+ ··· , as ĉ → 0, (128)
u
1
T ∑
xi
ĉ
∼= 1 + ĉ · S1 +
u
ĉ2
2 · S2 + ĉ3
3 S3, (129)
1
T ∑
xi
c
xi
log ∼= S1 + ĉ · S2 +
u u
ĉ2
2 S3, (130)
S1 = 1
T
S2 = 1
T
S3 = 1
T
T
∑
i=1
T
∑
i=1
T
∑
i=1
log
xi
u
log 2 xi
u
log 3 xi
u
37
, (131)
, (132)
. (133)
(134)