I am reading a paper here: https://pdfs.semanticscholar.org/5f91/2d46b02b03230a4ffaaa42d655b2b6147d56.pdf
The following is my confusion.
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?
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