statistique, théorie et gestion de portefeuille - Docs at ISFA

100 3. Distributions exponentielles étirées contre distributions régulièrement variables
Now, accounting for the relation E1[xc ] = uc +d c , the first and third term within the brackets cancel out, and
accounting for equation (104) yields
˜C12(2) = − c∗
d∗
1
u
+
c∗ d∗ c∗
· ln u
− E ln
d∗ x
d∗ . (110)
Finally, ˜C12 can be consistently estimated by replacing (c ∗ ,d ∗ ) by (ĉ, ˆ
d) and using the relation
Let us now define
g1(x|c ∗ ,d ∗ ) =
E
ln x
d
∂ln f1(x|c ∗ ,d ∗ )
∂c ∗
∂ln f1(x|c ∗ ,d ∗ )
∂d ∗
g2(x|b ∗ ) = ∂ln f2(x|b ∗ )
∂b ∗
The matrices Ci j are defined as
which can be consistently estimated by
= ln u 1
+
d c e( u d ) c
· Γ 0,
=
1
c∗ + ln u
d∗ − c∗
d∗
u
c . (111)
d
u
d∗ c∗
+ ln x
d∗ − ln x
d∗ 1 + u
d∗ c∗ − x
d∗ c∗
x c∗ d
(112)
= 1
+ lnu − lnx. (113)
b∗
Ci j = E0 gi(x) · g j(x) t , i, j = 1,2, (114)
Ĉi j = 1
T
T
∑
i=1
D.2 Testing the (SE) model against the Pareto model
gi(x) · g j(x) t , i, j = 1,2. (115)
Let us now assume that, beyond a given high threshold u, the true model is the Pareto model, that is, the true
returns distribution is a power law with pdf
This will be our null hypothesis H0.
f0(x|b) = b ub
, x ≥ u. (116)
xb+1 Now consider the maximum likelihood estimators (ĉ, d) ˆ of the model (SE). They are solution of equations
(46-47)
1
c =
d c = uc
T
1
T ∑T xi
i=1 u
1
T ∑T xi
i=1 u
T
xi
∑
i=1 u
Under H0, (ĉ, ˆ
d) converges to the pseudo-true values, solutions of
1
c
x
E0 u =
x
E0 u
c xi ln
u
c −
− 1 1 T
T ∑ ln
i=1
xi
,
u
(117)
c − 1. (118)
c
x ln
u
c − 1 − E0
ln x
, (119)
u
d c = E0 [x c ] − u c , (120)
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