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Pricing a Lognormal Forward Option with Time-Varying Volatility

Article Quant Q&A · Author: snowave
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Summary

The document derives a European call value when a forward price follows a two-factor diffusion with volatility components that vary over time. It applies the exponential solution for the forward process and combines the factor variances into an integrated variance over the option’s life. Dividing that integrated variance by the option maturity defines an equivalent constant volatility for use in the pricing formula.

Under the stated dynamics, the forward price at option expiry is lognormally distributed. The resulting payoff value is given by a Black-style expression using the forward price, strike, discounting, and normal cumulative probabilities. The derivation assumes deterministic volatility functions, independent standard Brownian drivers, and the specified risk-neutral setup. It addresses a European payoff at the option maturity; it does not treat early exercise or model additional market features such as stochastic rates.

Key ideas

  • Integrate the squared volatility components over the option life to obtain total variance.
  • Convert integrated variance to an equivalent constant volatility by dividing by option maturity.
  • The forward price is lognormally distributed under the stated deterministic volatility dynamics.
  • Use a Black-style call formula with the equivalent volatility and discount to option expiry.
  • The derivation covers a European payoff and relies on its stated model assumptions.

Tags

OptionsDerivatives pricingVolatility
Full text
# Option price derivation with these dynamics


# Option price derivation with these dynamics












If my underlying follows a dynamics of the form \begin{align*} dF(t,T)/F(t,T)=\sigma_1(t,T)dW_1(t)+\sigma_2(t,T)dW_2(t), \end{align*} where $\sigma_1(t,T)=h_1e^{-\lambda(T-t)}+h_0$, and $\sigma_2(t,T)=h_2e^{-\lambda(T-t)}$. How to derive an option price?

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/27766

You can proceed similarly to this question.

For $0 < T_0\le T$, consider the option with payoff, at the option maturity $T_0$, of the form \begin{align*} \max(F_{T_0, T}-K, \, 0).\tag{1} \end{align*} Note that \begin{align*} F_{T_0, T} &= F_{0, T}\exp\Bigg(-\frac{1}{2}\int_0^{T_0} \left[\left(h_1e^{-\lambda (T-t)}+h_0\right)^2 + h_2^2e^{-2\lambda (T-t)} \right] dt\\ &\qquad\qquad\qquad +\int_0^{T_0}\left[\left(h_1e^{-\lambda (T-t)}+h_0\right)dW_t^1 + h_2e^{-\lambda (T-t)}dW_t^2\right]\Bigg). \end{align*} Let \begin{align*} \hat{\sigma}^2 &= \frac{1}{T_0}\int_0^{T_0} \left[\left(h_1e^{-\lambda (T-t)}+h_0\right)^2 + h_2^2e^{-2\lambda (T-t)} \right] dt\\ &=\frac{e^{-2\lambda T}(h_1^2+h_2^2)}{2\lambda T_0}\left(e^{2\lambda T_0} -1\right)+\frac{2e^{-\lambda T}h_0h_1}{\lambda T_0}\left(e^{\lambda T_0} -1\right) + h_0^2. \end{align*} Then, in distribution, \begin{align*} F_{T_0, T} = F_{0, T}\exp\left(-\frac{\hat{\sigma}^2}{2} T_0 + \hat{\sigma} \sqrt{T_0} Z\right), \end{align*} where $Z$ is a standard normal random variable. The value of Payoff $(1)$ is now given by \begin{align*} e^{-r T_0}\Big[F_{0, T}\Phi(d_1) - K\Phi(d_2) \Big], \end{align*} where \begin{align*} d_1 &= \frac{\ln \frac{F_{0, T}}{K} + \frac{\hat{\sigma}^2}{2} T_0}{\hat{\sigma} \sqrt{T_0}},\\ d_2 &= d_1 - \hat{\sigma} \sqrt{T_0}, \end{align*} and $\Phi$ is the cumulative distribution function of a standard normal random variable.

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