Advances in Risk Management

140 A COMPARATIVE ANALYSIS OF DEPENDENCE LEVELS

If both intensities are perfectly correlated:λA=λB=λ, then:

pAB= 2 p+E

(

exp

(

− 2

∫T

0

λ(s)ds

))

−1, wherep=pA=pB

The correlation between the two default events is then:

ρ

def

=

AB

pAB−pApB

√

pA(1−pA)pB(1−pB)

(7.7)

=

2 p+E(exp(− 2

∫T

0 λds))−^1 −p

2

p(1−p)

=

E(exp(− 2

∫T

0 λds))−(1−p)

2

p(1−p)

=

Var(exp(−

∫T

0 λds))

p(1−p)

(7.8)

If we assume that the variance of the survival probability is at most of order
p^2 , then the correlation is of orderp. Nonetheless, we argue that this is far
from being satisfied usually.
To justify his assumption, Schönbucher (2003) suggested a normally dis-
tributedintegratedintensity, forwhichweassumethattheintegratedhazard
function between 0 andTis following a normal law N (μ,σ^2 ).
Note that such an assumption does not generate a “true” intensity pro-
cess because some values of the integrated intensity may be negative.
Nonetheless, forgetting such a detail, we get:

E

(

exp

(

−

∫T

0

λ(s)ds

))

= 1 −p=exp

(

−μ+

1

2

σ^2

)

E

(

exp

(

− 2

∫T

0

λ(s)ds

))

=exp(− 2 μ+ 2 σ^2 )

and we deduce:

ρ=(e−^2 μ+^2 σ

2

−e−^2 μ+σ

2

)/(p−p^2 )≈(1−p)(eσ

2

−1)/p (7.9)

Ifσ≈λT, we get thatρandpare of the same order with this normal inten-
sities specification. Clearly, it is a very crude approximation. A more careful
approximation provides:

exp(σ^2 )− 1 ≈σ^2 ≈2(μ−p),

because 1−p=exp(−μ+σ^2 /2)≈ 1 −μ+σ^2 /2. Thus, we getρ≈2(μ−p)/p,
but we have no ideas (a priori) concerning the size of the latter ratio. To