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102 3. Distributions exponentielles étirées contre distributions régulièrement variables
Thus, we get from equation (117):
and from equations (124) and (118):
ĉ
1
2 S2 − S 2 1
1
2S1S2 − 1
3S3 ˆηT
ˆb − 1 = ĉ · S2 1
1 − 2S2 + ĉ
2 [S1S2 − 1
S1 + ĉ
2S2 + ĉ2
6 S3
, (135)
3 S3]
. (136)
Now, accounting for the fact that the variables ξ1 = S1 − b−1 , ξ2 = S2 − 2b−2 and ξ3 = S3 − 6b−3 are
asymptoticaly Gaussian random variables with zero mean and variance of order T −1/2 , we obtain, at the
lowest order in T −1/2 :
ĉ = b 2
2ξ1 − b
2 ξ2
, (137)
and
which shows that
S2 1
1 − 2S2 + ĉ
2 [S1S2 − 1
3S3] S1 + ĉ
2 S2 + ĉ2
6 S3
ˆηT
ˆb
− 1 = b2
= 2ξ1 − b
2 ξ2
, (138)
2ξ1 − b
2 ξ2
2
. (139)
We now use the fact that ξ1, ξ2 are asymptotically Gaussian random variables with zero mean. We find their
variances:
Var(ξ1) = 1
T b2 , Var(ξ2) = 20
T b4 , and Cov(ξ1,ξ2) = 4
.
T b3 (140)
Using equation (139), the random variable b(2ξ1 − bξ2/2) is in the limit of large T a Gaussian rv with zero
mean and variance 1/T , i.e., ζT is asymptoticaly distributed according to a χ 2 distribution with one degree
of freedom.
Thus, the test consists in accepting H0 if ζT ≤ χ2 1−ε (1) and rejecting it otherwise. (ε denotes the asymptotic
level of the test).
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