I would like to solve a SDE similar to the GBM but with an additional term. Generally, the process I'm looking for should have the following recursive form:
$$ \frac{dX_t}{X_t} = \alpha dt + \gamma dW_t – \phi X_t dt, $$
$$ d X_{t} = \alpha X_t dt + \gamma X_t dW_t – \phi X_t^2 dt. $$
I have attempted solving the SDE without success using Ito's Lemma when $f(X_t) = ln(X_t) $
$$ d ln(X_t) = \frac{1}{X_t} dX_t – \frac{1}{2}\frac{1}{X_t^2} (dX_t)^2 $$
$$ d ln(X_t) = \frac{dX_t}{X_t} – \frac{1}{2X_t^2} (\gamma X_t dW_t)^2 $$
$$ d ln(X_t) = \frac{d X_t}{X_t} – \frac{1}{2} \gamma^2 dt $$
$$ \frac{d X_t}{X_t} = d ln(X_t) + \frac{1}{2} \gamma^2 dt $$
and when $ f(X_t) = X_t^{-1} $ (EDIT: fixed some mistakes below)
$$ d (X_t^{-1}) = – \frac{1}{X_t^2} dX_t + \frac{1}{2}\frac{2}{X_t^3} (dX_t)^2 $$
$$ d (X_t^{-1}) = – \frac{1}{X_t^2} (\alpha X_t dt + \gamma X_t dW_t – \phi X_t^2 dt) + \frac{1}{X_t^3} (\gamma X_t dW_t)^2 $$
$$ d (X_t^{-1}) = – \alpha X_t^{-1} dt – \gamma X_t^{-1} dW_t + \phi dt + \gamma^2 X_t^{-1} dt $$
$$ d f_t = \phi dt + (\gamma^2 – \alpha) f_t dt – \gamma f_t dW_t $$
$$ d f_t = (\phi + \gamma^2 f_t – \alpha f_t) dt – \gamma f_t dW_t $$
By substituting the result from the first attempt with Itô's Lemma into the original SDE I obtained
$$ d ln(X_t) + \frac{1}{2} \gamma^2 dt = \alpha dt + \gamma dW_t – \phi X_t dt $$
$$ ln(X_t) – ln(X_0) = (\alpha – \frac{1}{2} \gamma^2) t + \gamma W_t – \phi \int_{0}^{t} X_t dt $$
$$ ln(X_t) + \phi \int_{0}^{t} X_t dt = ln(X_0) (\alpha – \frac{1}{2} \gamma^2)t + \gamma W_t. $$
EDIT: The result from the second attempt with Itô's Lemma might be used in the same way as here https://quant.stackexchange.com/questions/50485/solve-the-following-sde-mathrmdx-t-ab-x-t-mathrmdt-c-x-t-mat and maybe here How to solve the following SDE: $\mathrm{d}X_t = a(b-X_t) \mathrm{d}t + c X_t \mathrm{d}W_t$
EDIT: How should I proceed?
Best Answer
$\ln$ is not helping you here. Use $f(x)=x^{-1}$ to obtain with Ito's Lemma that $$df=-(\alpha f-\gamma^2 f-\phi)dt - \gamma f dW,$$ which is linear and admits a closed form solution. Once you obtain the solution, you have $x = \frac{1}{f}$.