Classic SABR Model: Exactly Solvable Cases

Abstract

In this chapter, we analyze exactly solvable cases for the SABR model; free Normal SABR, Log-normal SABR, Normal SABR with positive rate and zero correlation between underlying and volatility, and a general zero correlation case.

3.8 Appendix A. Integrals with Square Root \(\sqrt{\cosh s-\cosh u}\)

On several occasions we face contour integrals of the type

$$\begin{aligned} J=\frac{1}{2\mathrm{i}}\int _{C_{u}}\frac{F(u)}{\sqrt{\cosh s-\cosh u}}\, \mathrm{d}u \qquad s>0 \, ,\end{aligned}$$

(3.131)

where the path of integration $C_{u}$ in the complex plane u is defined like in (2.58), recall, it consists of three sides of the infinite rectangle with vertices at $\infty -\pi \mathrm{i},\,-\pi \mathrm{i},\,\pi \mathrm{i},\,\infty +\pi \mathrm{i}$, run in this order. It is supposed that function F(u) is analytic in the half-strip $\Pi$, surrounded by the contour $C_{u}$, and is decreasing fast at $\mathfrak {R}u\rightarrow \infty$ (typically $F(u)\propto \mathrm{e}^{-\frac{u^{2}}{2t}}$). Function

$$\begin{aligned} f(u)=\left( \cosh s - \cosh u\right) ^{1/2} \, , \end{aligned}$$

(3.132)

defined originally on contour $C_{u}$, can be continued analytically into the half-strip $\Pi$, equipped with the cut along the real axis from the branching point $u=s$ to $+\infty$. The cut allows to select a single valued analytic branch of f(u), fixed by the condition of f(u) being positive on $C_{u}$, as discussed in Sect. 3.2.

According to the Cauchy theorem, the value of integral J of the analytic function F(u) / f(u) depends only on the ends of the path $C_{u}$. Even more, the ends themselves may be shifted along the vertical line $\mathfrak {R}u=\infty$ due to decreasing F(u). This means that without changing the value of integral J we may choose any path C that sustains the global properties of the original contour $C_{u}$; it starts at $\mathfrak {R}u=+\infty$ on or below the lower bank of the cut, terminates at $\mathfrak {R}u=+\infty$ on or above the upper bank, and goes around the cut in the negative direction (crossing cut is not allowed). This said, we squeeze contour $C_{u}$, placing integration directly onto the banks of the cut, plus an infinitesimal circumference surrounding point $u=s$. Contribution from the latter vanishes as its radius tends to zero, because the singularity $1/f(u) \propto (\delta u)^{-1/2}$ is integrable. This leaves only two integrals, taken in opposite directions along the banks.

We need now values of function f(u) on the cut banks. On either bank, $u= \mathfrak {R}u>s$ so that f(u) (3.132) is pure imaginary. Proper signs can be determined by moving from contour $C_{u}$ to the cut banks. Take u on the far right end of the upper leg of $C_{u}$, $u=R+\mathrm{i}\pi , \quad R\gg s$, where asymptotics of f(u) (3.132) is $\sqrt{2} f(u) \approx \mathrm{e}^{R/2} = -\mathrm{i}\mathrm{e}^{u/2}$. By continuity, this asymptotics remains valid, when moving down off $C_{u}$ and reaching the upper bank at $u=R$ with $\sqrt{2} f(u) \approx -\mathrm{i}\mathrm{e}^{R/2}$. We conclude that on the upper bank

$$ f(u)=-\mathrm{i}\left| f(u)\right| =-\mathrm{i}\sqrt{\cosh u -\cosh s} $$

In the same manner, moving upward from the far right end of the lower leg of $C_{u}$, we find that on the lower bank

$$ f(u)=+\mathrm{i}\sqrt{\cosh u -\cosh s} $$

(We write $\sqrt{(\cdot )}$, rather than $(\cdot )^{1/2}$, to underline that the expression under square root is positive and that the positive value of the square root is taken).

Signs of f(u) and directions of integration are opposite on the two banks, meaning that contributions from the two are equal, and the whole integral J is twice that along the upper bank,

$$\begin{aligned} J=\frac{1}{2\mathrm{i}}\int _{C_{u}}\frac{F(u)}{\sqrt{\cosh s-\cosh u}}\, \mathrm{d}u =\int _{s}^{\infty }\frac{F(u)}{\sqrt{\cosh u-\cosh s}}\, \mathrm{d}u \end{aligned}$$

(3.133)

We use this transformation when computing PDFs of normal, log-normal, and general SABR with zero correlation; the moment generating function of the inverse random time $\tau$; and option values. In the case of Log-normal SABR the integral to be transformed is slightly different from (3.131), but the consideration follows the same lines. The original and transformed integrals look like follows

$$\begin{aligned} \tilde{J}&=\frac{1}{2\mathrm{i}}\int _{C_{u}}\frac{F(u)\, E^{-\lambda \sqrt{\cosh s-\cosh u}}}{\sqrt{\cosh s-\cosh u}} \, \mathrm{d}u \nonumber \\&= \int _{s}^{\infty }\frac{F(u)\cos (\lambda \sqrt{\cosh u-\cosh s})}{\sqrt{\cosh u-\cosh s}} \, \mathrm{d}u \, .\end{aligned}$$

(3.134)