148 A COMPARATIVE ANALYSIS OF DEPENDENCE LEVELS

in Chambers, Mallows and Stuck (1996). LetW be a random variable
exponentially distributed with parameterλ=1, andUa random variable
uniformly distributed on [−π 2 ,π 2 ] and letξ=arctan(βtan(πα/2))/αand the
random variable:

Z=



















sin (α(ξ+U))

√αcos (αξ) cos (U)

(

cos (αξ+(α−1)U)

W

)^1 −α

α ifα= 1

2

π







(π

2

+βU

)

tan (U)−βln







π

2 WcosU

π

2

+βU











 ifα=^1

(7.14)

ThenZ∼S(α,β). To getS(α, β, γ, δ), we invoke the linear transform of
Definition 6.

7.5.4 α-stable intensity-based model

To simulate more heavy tailed random intensities, we are going to replace
the gamma frailty random variable in (7.3) by anα-stable distributed frailty.
As an intensity process is always positive and according to (7.12), we impose
thatα<1,β=1 andδ=0 in order that the support of the frailty is [0,+∞].
We keep the same simple specification as in our first intensity model: for
every obligoriand every timet,

λi(t)=λi=Zλ0,i
Therefore,Z∼S(α,1,γ,0) whereα∈[0, 1]. Indeed, as the frailty variable has
a multiplicative effect on the intensity, its baseline hazard function plays
the role of a scale parameter. Thus, the parameterγis unnecessary. In fact,
we identifyλ0,iby using the Laplace transform of theα-stable distribution
(7.13), which leads to the one-year default probability:

pi= 1 −exp

(

−γαsec

(πα

2

)

γαλα0,i

)

This implies:

λ 0 =

1

γ





ln

(

1

1 −pi

)

sec

(πα

2

)





(^1) α
Hence
λ
d
=λ 0 Z

1
γ


ln
(
1
1 −pi
)
sec
(πα
2
)


(^1) α
γS(α, 1) (7.15)



ln
(
1
1 −pi
)
sec(πα 2 )


1
α
S(α,1)