The paper has the following continuous time model for the price of a commodity: $$\frac{dS(t)}{S(t)}=a(lnL(t)+\mu(t)-lnS(t))dt+\sigma(t)dW(t)$$ Then the paper converts this continuous time model into a discrete time model: $$lnS(t+\Delta t)=e^{-a\Delta t}lnS(t)+(1-a^{-a\Delta t})(lnL(t)+\mu(t))-\frac{\eta_t^2}{2}+\tilde{W}_{\Delta t}$$ where $$\eta_t^2=\sigma^2(t)(\frac{1-e^{-2a(\Delta t)}}{2a})$$ $$\tilde{W}_{\Delta t}\sim N(0, \eta_t^2)$$ My confusion is how to get the second to the last term $$-\frac{\eta_t^2}{2}$$. Here is what I have tried to discretize the continuous model, and I did not get the 2 in front of $$a$$:
I let $$Y(t)=lnS(t)$$. Then $$dY(t)=(a(lnL(t)+\mu(t)-Y(t))-\frac{1}{2}\sigma^2(t))dt+\sigma(t)dW(t)$$ And $$d(Y(t)e^{at})=e^{at}((a(lnL(t)+\mu(t))-\frac{1}{2}\sigma^2(t))dt+\sigma(t)dW(t))$$ So $$Y(t+\Delta t)e^{a(t+\Delta t)}=Y(t)e^{at}+\int_{t}^{t+\Delta t}e^{as}(a(lnL(s)+\mu(s))-\frac{1}{2}\sigma^2(s))ds+\int_t^{t+\Delta t}e^{as}\sigma(s)dW(s)$$ I approximated $$lnL(s)+\mu(s)\approx lnL(t)+\mu(t)$$ on $$[t,t+\Delta t]$$, and obtained $$lnS(t+\Delta t)=Y(t+\Delta t)\approx e^{-a\Delta t}Y(t)+(1-e^{-a\Delta t})(lnL(t)+\mu(t))$$ $$-e^{-a(t+\Delta t)}\frac{1}{2}\int_t^{t+\Delta t}e^{as}\sigma^2(s)ds+e^{-a(t+\Delta t)}\int_t^{t+\Delta t}e^{as}\sigma(s)dW(s)$$ The last term has variance $$e^{-2a(t+\Delta t)}\int_t^{t+\Delta t}e^{2as}\sigma^2(s)ds\approx \frac{\sigma^2(t)}{2a}(1-e^{-2a(\Delta t)})=\eta_t^2$$ This agrees with the result - the variance of the last term is indeed $$\eta_t^2$$. However, for the second to the last term, it seems the approximate for that is instead $$-\frac{\sigma^2(t)}{2a}(1-e^{a(\Delta t)})$$ The result in the paper says this should be $$-\frac{1}{2}\eta_t^2=-\frac{\sigma^2(t)}{4a}(1-e^{-2a(\Delta t)})$$ Can someone help me reconcile the discrepency?