Derivation using Ito's Lemma of price process

Question

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: $$ \ln P(t+h) - \ln P(t) = \mu h + (\exp(-h \Theta) - 1) \ln P(t) + \sigma \int_t^{t+h} \exp(-\Theta(t-u))dW_u $$

Answer 1

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 $$ \ln P(t+h) -\ln P(t) =\mu h+\left(e^{-h\Theta}-1\right)(\ln P(t)-\mu t) + \sigma\int_{t}^{t+h}e^{-\Theta\,(t+h-u)}dW_u $$