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Assuming my underline asset price follows the process:

$$d\ln (F_{t,T})=-(1/2)\sigma ^2e^{-2\lambda(T-t)}dt+\sigma e^{-\lambda(T-t)}dB_t $$

How should I derive an option price formula?

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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\left(-\frac{\sigma^2}{2}\int_0^{T_0} e^{-2\lambda (T-t)} dt+\sigma \int_0^{T_0}e^{-\lambda (T-t)} dB_t\right). \end{align*} Let \begin{align*} \hat{\sigma}^2 &= \frac{\sigma^2}{T_0}\int_0^{T_0} e^{-2\lambda (T-t)} dt\\ &=\frac{e^{-2\lambda T}\sigma^2}{2\lambda T_0}\left(e^{2\lambda T_0} -1\right). \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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  • $\begingroup$ That's pretty cool. I'm learning the fundamentals of option pricing, and I'll change my underline process to derive this by myself. Thanks a lot. $\endgroup$ – snowave Jun 16 '16 at 15:21
  • $\begingroup$ @snowave: To deriive it by yourself is a good learning process. $\endgroup$ – Gordon Jun 16 '16 at 15:25
  • $\begingroup$ is the option delta = N(d1)? I assume it is. $\endgroup$ – snowave Jun 20 '16 at 18:12
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    $\begingroup$ @snowave: The delta with respect to $F_{0, T}$ is $e^{-rT_0} \Phi(d_1)$. $\endgroup$ – Gordon Jun 20 '16 at 18:15
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    $\begingroup$ @snowave: I do not know the specific format of the two factor model. Basically, you can proceed similarly. $\endgroup$ – Gordon Jun 23 '16 at 12:58

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