ABSTRACT
FIGURE 6.5: Implied volatility for the SABR model τ = 10Y. α = 0.2, ρ = −0.7, ν22 τ = 5 and β = 1.
For the SABR model, the connection A and the function Q are given by
A = 1 2 (1− ρ2)
( −d ln(C) + ρ
ν ∂fCda
) (6.53)
Q = a2
( C∂2fC −
(∂fC) 2
2 (1− ρ2)
) (6.54)
For β = 0, the function Q and the potential A vanish. p satisfies a heat kernel equation where the differential operator D reduces to the Laplace-Beltrami operator on H2:
∂p
∂τ ′ = ∆H2p (6.55)
≡ y2(∂2x + ∂2y)p (6.56)
2 (6.41,6.42) (with
f − f0). Therefore solving the Kolmogorov equation for the normal SABR model (i.e., β = 0) is equivalent to solving this (Laplacian) heat kernel on H2. Surprisingly, there is an analytical solution for the heat kernel on H2 (6.55) found by McKean [116]. It is connected to the Selberg trace formula [21]. The exact conditional probability density p depends on the hyperbolic distance d(z, z′) and is given by
p(τ ′, d) = √
(cosh b− cosh d(z, z′)) 12 db
The conditional probability in the old coordinates [a, f ] is
p(τ ′, f, a)dfda = ν√
1− ρ2 dfda
a2
√ 2e−
(cosh b− cosh d(z, z′)) 12 db
where the term ν√ 1−ρ2a2 corresponds to the invariant measure
√ g on H2. We
have compared this exact solution (Fig. 6.6) with a numerical solution (PDE) of the normal SABR model and found an agreement. A similar result was obtained independently in [101] for the conditional probability. The value of a European option is
C(T,K|f0) = max(f0 −K, 0) + 12 ∫ T
dτ
da a2p(τ, x1 = K, a|α)
This expression can be obtained by applying the Itoˆ-Tanaka lemma on the payoff max(ft −K, 0). In order to integrate over a, we use a small trick: we interchange the order of integration over b and a. The half space b ≥ d with a arbitrary is then mapped to the half-strip amin ≤ a ≤ amax and b ≥ lmin where6
(amax − amin)2 4(1− ρ2) = −(K − f0)
2ν2 − α (α+ 2νρ(K − f0) + αρ2) + α cosh b
( 2ρ ((K − f0)ν + αρ) + α(1− ρ2) cosh b
) (6.57)
amin + amax 2
= ρ (ν(K − f0) + αρ) + α(1− ρ2) cosh b (6.58)
cosh lmin = −ρ ((K − f0)ν + αρ) +
√ α2 + 2ανρ(K − f0) + ν2(K − f0)2 α(1− ρ2)
(6.59)
