Calibration of local volatility model with stochastic interest rates by efficient numerical PDE methods

Abstract

Long-maturity options or a wide class of hybrid products are evaluated using a local volatility-type modelling for the asset price S(t) with a stochastic interest rate r(t). The calibration of the local volatility function is challenging and time-consuming because of the multi-dimensional nature of the problem. A key requirement of any equity hybrid derivatives pricing model is the ability to rapidly and accurately calibrate to vanilla option prices. In this paper, we develop a calibration technique based on a partial differential equation (PDE) approach which allows an accurate calibration and provides an efficient implementation algorithm. The essential idea is based on solving the derived forward equation satisfied by $P(t, S, r) \mathcal {Z}(t, S, r)$, where P(t, S, r) represents the risk-neutral probability density of (S(t), r(t)) and $\mathcal {Z}(t, S, r)$ the projection of the stochastic discounting factor in the state variables (S(t), r(t)). The solution provides effective and sufficient information for the calibration and pricing. The PDE solver is constructed by using ADI (alternative direction implicit) method based on an extension of the Peaceman–Rachford scheme. Furthermore, an efficient algorithm to compute all the corrective terms in the local volatility function due to the stochastic interest rates is proposed by using the PDE solutions and grid points. It reduces by one order the computations costs and then allows to speed up significantly the calibration procedure. Different numerical experiments are examined and compared to demonstrate the results of our theoretical analysis.

7 Appendix

For the proof of Corollary 2, we provide the following useful lemma

Lemma 1

Using the definitions in Proposition 3 and Corollary 2, we have

$$\begin{aligned} S_0n(d_1) = K {\textit{Z}C}(0, T) n(d_2) \end{aligned}$$

(88)

Proof

$$\begin{aligned} d_2^2 - d_1^2&= (d_2-d_1)(d_2+d_1) \end{aligned}$$

(89)

$$\begin{aligned}&= -\sqrt{g(T)} (d_2+d_1) \end{aligned}$$

(90)

$$\begin{aligned}&= -\sqrt{g(T)} \left( 2d_1 - \sqrt{g(T)} \right) \end{aligned}$$

(91)

$$\begin{aligned}&= -2 \left( \log \left( \frac{S_0}{K} \right) - \log {\textit{Z}C}(0,T) \right) \end{aligned}$$

(92)

$$\begin{aligned} \log \left( \frac{n(d_1)}{n(d_2)} \right)&= \log \left( \frac{K\log {\textit{Z}C}(0,T)}{S_0} \right)&\end{aligned}$$

(93)

From the last expression, we deduce directly (88). $\square$

For expression (67), we write

$$\begin{aligned} C_T(T, K)&= S_0 n(d_1) d_{1, T} - K \left[ {\textit{Z}C}_\mathrm{T}(0,T) N(d_2) + {\textit{Z}C}(0,T) n(d_2) d_{2,\mathrm{T}} \right] \end{aligned}$$

(94)

$$\begin{aligned}&= S_0 n(d_1)( d_{1, \mathrm{T}}-d_{2, \mathrm{T}} ) - K {\textit{Z}C}_T(0,T) N(d_2) \end{aligned}$$

(95)

$$\begin{aligned}&= \frac{S_0 n(d_1)}{2} \frac{{\hat{\sigma }}^2(T)}{\sqrt{g(T)}} + K {\textit{Z}C}(0, T)f(0, T) N(d_2) \end{aligned}$$

(96)

where we have used (88) in the second equality, $d_{1, \mathrm{T}}-d_{2, \mathrm{T}} = \frac{1}{2}\frac{{\hat{\sigma }}^2(T)}{\sqrt{g(T)}}$ and ${\textit{Z}C}_T(0,T) = -{\textit{Z}C}(0,T) f(0,T)$ to obtain the third expression.

$$\begin{aligned} C_\mathrm{K}(T, K) = S_0 n(d_1) d_{1, \mathrm{K}} - {\textit{Z}C}(0,T) \left[ N(d_2) + K n(d_2) d_{2, \mathrm{K}} \right] \end{aligned}$$

(97)

Using $d_{1, K} = d_{2, \mathrm{K}}$ and result of lemma (88), we get expression (68). Finally, we obtain formula (69) by deriving (68) w.r.t. K and using $d_{1, \mathrm{K}} = -\frac{1}{K \sqrt{g(T)}}$.