I am trying to follow a derivation, but I get stuck could someone take me take me through the rest:
We start with,
$$s(t,x_t)=e^{g(t)+x_t}$$
where $$dX_t=\log (J) dq_t+\left(-\text{$\alpha $X}_t\right) dt+\sigma (t) dZ_t$$
Using ito's lemma:
$$ds_t=dt e^{g(t)+x} g'(t)+\frac{1}{2} e^{g(t)+x} \left(\log (J) dq_t-\text{$\alpha $X}_t dt+\sigma (t) dZ_t\right){}^2+e^{g(t)+x} \left(\log (J) dq_t+\left(-\text{$\alpha $X}_t\right) dt+\sigma (t) dZ_t\right)$$
Knowing: $$\left\{(dt)^2\to 0,dt dZ_t\to 0,\left(dZ_t\right){}^2\to dt\right\}$$
We get:
$$ds_t=dt e^{g(t)+x} g'(t)-\log (J) \text{$\alpha $X}_t dt e^{g(t)+x} dq_t+\log (J) \sigma (t) e^{g(t)+x} dq_t dZ_t+\frac{1}{2} \log ^2(J) e^{g(t)+x} \left(dq_t\right){}^2+\log (J) e^{g(t)+x} dq_t-\text{$\alpha $X}_t dt e^{g(t)+x}+\frac{1}{2} \sigma (t)^2 dt e^{g(t)+x}+\sigma (t) e^{g(t)+x} dZ_t$$
Simplifying we get:
$$ds_t=\frac{1}{2} s_t \left(dt \left(2 g'(t)-2 \text{$\alpha $X}_t+\sigma (t)^2\right)+2 \log (J) dq_t \left(\text{$\alpha $X}_t (-dt)+\sigma (t) dZ_t+1\right)+\log ^2(J) \left(dq_t\right){}^2+2 \sigma (t) dZ_t\right)$$
Expanding:
$$ds_t=s_t dt g'(t)-\log (J) s_t \text{$\alpha $X}_t dt dq_t+\log (J) s_t \sigma (t) dq_t dZ_t+\frac{1}{2} \log ^2(J) s_t \left(dq_t\right){}^2+\log (J) s_t dq_t-s_t \text{$\alpha $X}_t dt+\frac{1}{2} s_t \sigma (t)^2 dt+s_t \sigma (t) dZ_t$$
Now what I need to figure out is how I can simplify this to:
where:
As for the covariation of $Z$ we have formally $(dq)^2 -> 0$, $(dt)\times (dq) -> 0$ and $(dq)\times (dZ) -> 0$. Hence, the terms $\frac{1}{2}\log^2(J)s_t\,(dq_t)^2$, $−\log(J)s_t\alpha X_t\,dt\,dq_t$ and $\log(J)s_t\sigma(t)\,dq_t\,dZ_t$ vanish. For the remaining, just observe that from the definition of $s$ we have $\log s_t = g_t + X_t$. This gives you the necessary simplification.