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This brings us back to the problem of calculating the same type of integrals as in the uncorrelated case.
Using the expressions of α and β, and taking into account the parity of n and p, we obtain:
γq1q2 (2n, 2p) = χ1 2p χ2 2n−2p (1 − ρ2 1
q1p+q2(n−p)+ ) 2
π
×Γ q2(n − p) + s + 1
2
γq1q2 (2n, 2p + 1) = χ1 2p+1 χ2 2n−2p−1 (1 − ρ2 ) q1p+q2(n−p)+ q1−q2 +1
2
×Γ
q1p + s + 1 + q1
2
Γ
+∞
s=0
(2ρ) 2s
(2s)! Γ
q1p + s + 1
×
2
461
, (150)
+∞
(2ρ)
π
s=0
2s+1
(2s + 1)! ×
q2(n − p) + s + 1 − q2
. (151)
2
Using the definition of the hypergeometric functions 2F1 (Abramovitz and Stegun 1972), and the relation
(9.131) of (Gradshteyn and Ryzhik 1965), we finally obtain
γq1q2 (2n, 2p) = χ1 2p χ2 2n−2p Γ q1p + 1
2 Γ q2(n − p) + 1
2
2F1 −q1p, −q2(n − p);
π
1
; ρ2(152)
,
2
γq1q2 (2n, 2p + 1) = χ1 2p+1 χ2 2n−2p−1 2Γ q1p + 1 + q1
2 Γ q2(n − p) + 1 − q2
2 ρ ×
π
× 2F1 −q1p − q1 − 1
, −q2(n − p) +
2
q2 + 1
;
2
3
; ρ2 . (153)
2
In the asymmetric case, a similar calculation follows, with the sole difference that the results involves four
terms in the integral (148) instead of two.
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