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?
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$\begingroup$ Are $W_1$ and $W_2$ correlated or independent? $\endgroup$ – Gordon Jun 23 '16 at 15:36
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$\begingroup$ they are independent $\endgroup$ – snowave Jun 23 '16 at 15:41
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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.
