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Define $q(t)$ as the log price minus a linear trend

$$ q(t) = \ln P(t) - \mu t $$

Assume the log price process = Equation 1: $$ dq(t) = - \Theta q(t) dt + \sigma dW(t) $$

Can you show that the solution to Equation 1 is: $$ \ln P(t+h) - \ln P(t) = \mu h + (\exp(-h \Theta) - 1) \ln P(t) + \sigma \int_t^{t+h} \exp(-\Theta(t-u))dW_u $$

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  • $\begingroup$ Changed the term $\int^t_{t+h} \exp(-\Theta(t-u)dW_u)$ to $\int_t^{t+h} \exp(-\Theta(t-u))dW_u$, according to the original paper. $\endgroup$ – Gordon Nov 2 '15 at 16:07
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by application of Ito's lemma , we have $$d\left(q(t)e^{\Theta\,t}\right)=\Theta \,q(t)e^{\Theta\,t}dt+e^{\Theta\,t}dq(t)+0$$ then $$d\left(q(t)e^{\Theta\,t}\right)=\sigma e^{\Theta\,t}dW_t$$ in other words $$q(t+h)e^{\Theta\,(t+h)}-q(t)e^{\Theta\,t}=\sigma\int_{t}^{t+h}e^{\Theta\,u}dW_u\Rightarrow$$ $$q(t+h)-q(t)=\left(e^{-h\Theta}-1\right)q(t)+\sigma\int_{t}^{t+h}e^{-\Theta\,(t+h-u)}dW_u$$ By substituting $\ln P(t)-\mu\,t$ to last equation , we have $$ \ln P(t+h) -\ln P(t) =\mu h+\left(e^{-h\Theta}-1\right)(\ln P(t)-\mu t) + \sigma\int_{t}^{t+h}e^{-\Theta\,(t+h-u)}dW_u $$

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  • $\begingroup$ Hi - Why was it necessary to edit the question? I posted lnP(t+h) - lnP(t) not lnP(t) - exp(ho)lnP(t+h). You changed the entire question. I rolled it back. I am looking to see how its derived to the form I posted. $\endgroup$ – joesyc Aug 1 '15 at 20:35
  • $\begingroup$ because it was wrong $\endgroup$ – user16891 Aug 1 '15 at 21:06
  • $\begingroup$ I pulled it from this paper dx.doi.org/10.4236/tel.2012.23050 equation 2 $\endgroup$ – joesyc Aug 1 '15 at 21:11
  • $\begingroup$ You typed equation (2) incorrectly and I think that it is wrong $\endgroup$ – user16891 Aug 1 '15 at 22:04
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    $\begingroup$ From EQ(1) we get $$ \ln P(t) = q(t) + \mu t $$ Hence $$ d \ln P(t) = dq(t) + \mu dt $$ Now put the solution for q into the equation and u'll get the result from the paper. $\endgroup$ – Phun Aug 1 '15 at 22:54

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