Advances in Risk Management

(Michael S) #1
JEAN-PAUL PAQUIN, ANNICK LAMBERT AND ALAIN CHARBONNEAU 299

and


̃X=^1
n(1−ρ)
[u ̃ 1 (1−ρn)+u ̃ 2 (1−ρn−^1 )+u ̃ 3 (1−ρn−^2 )+···

+u ̃n− 2 (1−ρ^3 )+u ̃n− 1 (1−ρ^2 )+u ̃n(1−ρ)]

By settingwt= 1 −ρn−t+^1 the mean random variable becomes equal to:


̃X=^1
n(1−ρ)

∑n

t= 1

wtu ̃t (A.5)

Let us now demonstrate that the weighted sum of random variablesu ̃tobeys the CLT.
Given the initial assumptions governing the randomu ̃t’s, we obtain:


E( ̃X)=^1
n(1−ρ)

∑n

t= 1

wtE(u ̃t)= 0

and


V(X ̃)=
1
n^2 (1−ρ)^2

∑n
t= 1

w^2 tV(u ̃t)=
1
n^2 (1−ρ)^2

∑n
t= 1

w^2 t

The CLT will be established once we show that:


nlim→∞

∑n
t= 1

wtu ̃t

∑n
t= 1

w^2 t

→N(0, 1)

Given that theu ̃t’s are independent in probability, then the characteristic function is
equal to:


φ∑n
t= 1
wt ̃ut
√∑n
t= 1
w^2 t

(h)=E

(
e

ih√∑∑wt ̃ut
w^2 t

)
=

∏n

t= 1

e

√ihw∑t ̃ut
w^2 t =
∏n

t= 1

φu ̃


√wth

w^2 t




The logarithm of the characteristic function then becomes:


√∑wtu ̃t
∑w 2
t

=

∑n

t= 1

logφu ̃t




wth
√∑
w^2 t




=

∑n

t= 1

log


 1 +i√wt

w^2 t

hμ 1 −
1
2


√wt

w^2 t




2
h^2 μ 2


i
3!


√wt

w^2 t




3
h^3 μ 3 +
1
4!


√wt

wt^2




4
h^4 μ 4 +...



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