7
$\begingroup$

Can anyone give me a detailed explanation of how below equations (3) and (4) are derived from (1) and (2)? \begin{align*} \frac{dF_{t,T}}{F_{t,T}} &=\sigma e^{-\lambda(T-t)}dB_t, \tag{1}\\ \ln(F_{t,T})&=\ln(F_{0,T})-1/2\int_{0}^{t}\sigma^2 e^{-2\lambda(T-s)}ds+\int_{0}^{t}\sigma e^{-\lambda(T-s)}dB_s.\tag{2} \end{align*} Given $\ln(S_t)=\ln(F_{t,t})$, we have: \begin{align*} \frac{dS_t}{S_t}=(\mu_t-\lambda \ln(S_t))dt+\sigma dB_t,\tag{3} \end{align*} where \begin{align*} \mu_t=\frac{\partial \ln(F_{0,t})}{\partial t} +\lambda \ln(F_{0,t})+\frac{1}{4}\sigma^2(1-e^{-2\lambda t}). \tag{4} \end{align*} Or anything related to them will be helpful.

$\endgroup$
10
$\begingroup$

From $(2)$, \begin{align*} \ln S_t &=\ln F_{t, t} \\ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\\ &=\ln F_{0, t}-\frac{\sigma^2}{4\lambda} \left(1-e^{-2\lambda t}\right)+e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s. \end{align*} Then, \begin{align*} \lambda e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s = \lambda \ln S_t - \lambda \ln F_{0, t} + \frac{\sigma^2}{4} \left(1-e^{-2\lambda t}\right). \end{align*} Therefore, \begin{align*} d\ln S_t &= \left(\frac{\partial \ln F_{0, t}}{\partial t}-\frac{\sigma^2}{2}e^{-2\lambda t} - \lambda e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s\right)dt +\sigma dB_t\\ &=\left[\frac{\partial \ln F_{0, t}}{\partial t}-\frac{\sigma^2}{2}e^{-2\lambda t}+\lambda \ln F_{0, t} - \frac{\sigma^2}{4} \left(1-e^{-2\lambda t}\right) -\lambda \ln S_t\right]dt + \sigma dB_t\\ &=\left(\frac{\partial \ln F_{0, t}}{\partial t}+\lambda \ln F_{0, t} -\frac{\sigma^2}{4} - \frac{\sigma^2}{4} e^{-2\lambda t} -\lambda \ln S_t\right)dt + \sigma dB_t. \end{align*} Note that \begin{align*} d\langle \ln S, \ln S \rangle_t= \sigma^2 dt. \end{align*} By Ito's lemma, \begin{align*} dS_t &= de^{\ln S_t}\\ &= e^{\ln S_t} d \ln S_t + \frac{1}{2}e^{\ln S_t}d\langle \ln S, \ln S \rangle_t\\ &=S_t d \ln S_t + \frac{1}{2} \sigma^2 S_t dt\\ &= S_t\left[\left(\frac{\partial \ln F_{0, t}}{\partial t}+\lambda \ln F_{0, t} -\frac{\sigma^2}{4} - \frac{\sigma^2}{4} e^{-2\lambda t} -\lambda \ln S_t\right)dt + \sigma dB_t + \frac{\sigma^2}{2} dt \right]\\ &=S_t\big[\left(\mu_t - \lambda \ln S_t\right)dt + \sigma dB_t\big], \end{align*} where \begin{align*} \mu_t = \frac{\partial \ln F_{0, t}}{\partial t}+\lambda \ln F_{0, t} +\frac{\sigma^2}{4}\left(1- e^{-2\lambda t}\right). \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.