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274 9. Mesure de la dépendance extrême entre deux actifs financiers
where ɛ is a random variable independent of Y and α a non random positive coefficient. Assume that Y and
ɛ have a Student’s distribution with density:
Cν
PY (y) =
1 + y2
ν
Pɛ(ɛ) =
Cν
σ 1 + ɛ2
ν σ2 ν+1
2
ν+1
2
, (E.97)
. (E.98)
We first give a general expression for the probability for X to be larger than F −1
X (u) knowing that Y is
(u) :
larger than F −1
Y
LEMMA 1
The probability that X is larger than F −1
X
with
Pr[X > F −1
−1
X (u)|Y > FY (u)] = ¯ Fɛ(η) + α
∞
1 − u F −1
Y (u)
−1
(u) knowing that Y is larger than F (u) is given by :
η = F −1
X
Y
dy ¯ FY (y) · Pɛ[αF −1
Y (u) + η − αy] , (E.99)
−1
(u) − αF (u). (E.100)
The proof of this lemma relies on a simple integration by part and a change of variable, which are detailed
in appendix E.1.
Introducing the notation
we can show that
η = α 1 +
Y
˜Yu = F −1
Y (u) , (E.101)
σ
ν1/ν − 1 ˜Yu + O(
α
˜ Y −1
u ), (E.102)
which allows us to conclude that η goes to infinity as u goes to 1 (see appendix E.2 for the derivation of this
result). Thus, ¯ Fɛ(η) goes to zero as u goes to 1 and
∞ α
λ = lim dy
u→1 1 − u
¯ FY (y) · Pɛ(α ˜ Yu + η − αy) . (E.103)
Now, using the following result :
LEMMA 2
Assuming ν > 0 and x0 > 1,
1
lim
ɛ→0 ɛ
∞
1
˜Yu
dx 1
x ν
Cν
1 + x−x0
ɛ
2 ν+1
2
whose proof is given in appendix E.3, it is straigthforward to show that
The final steps of this calculation are given in appendix E.4.
= 1
xν , (E.104)
0
1
λ =
1 +
σ ν . (E.105)
α
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