The Variance Gamma call price formula is given by:
$$C(0)= \int\gamma(R) e^{-rT} \int f\left(S(0) e^{\theta R+\omega T+\frac12 \sigma^2 R} e^{rT-\frac12 \sigma^2 R+\sqrt{T}\sqrt{R/T} \sigma W}\right)g(W)\,dW\,dR$$
How do I use Newton-Raphson search to obtain the inverse in terms of $\sigma$?
Notes:
- The inner integral is the Black-Scholes price of an option with payoff $f$ with spot equal to $S(0) e^{\theta R+\omega T+\frac12 \sigma^2 R}$ and volatility $\sigma \sqrt{R/T}$. Black-Scholes can be substituted for the inner integral and then the outer integral can be computed numerically
- $\gamma(R)$ is the density of the Gamma distribution
- $g$ is the standard normal density
- $W$ is a random $N(0,1)$ draw
- $R$ is fixed random time
- $r, \theta, \omega, T$ are parameters
Theorem 2 of The Variance Gamma Process and Option Pricing gives a closed-form solution for the call price in terms of $\sigma$, the volatility associated with the underlying Wiener process.
Call this $C\left(\sigma\right)$. If you know $C\left(\sigma\right)=C_0$ for some constant $C_0$ (i.e. the real observed price), you can use a Newton's method in the usual way:
$$\sigma_{n+1}=\sigma_{n}-(C\left(\sigma_{n}\right)-C_0)/C^{\prime}\left(\sigma_{n}\right)$$
with an initial guess $\sigma_0$. Under reasonable conditions, the iteration converges, and the point to which it converges is the volatility you are looking for.
It sounds to me, however, that you should first familiarize yourself with Newton's method.