I have the following Schwartz model: $$dS_t=a(\mu-\ln S_t)S_tdt+\sigma S_tdW_t$$ $$X_t=\ln S_t$$ $$dX_t=a(\hat{\mu}-X_t)dt+\sigma dW_t$$ with $\hat{\mu}=\mu-\frac{\sigma^2}{2a}\sigma$ $$F_t(T)= \exp\left(e^{-a(T-t)}X_t+\hat{\mu}(1-e^{-a(T-t)})+\frac{\sigma^2}{4a}(1-e^{-2a(T-t)})\right)$$ and I want to derive the value of $d \ln F(t,T)$
I do the following:
Let $Y_t=\ln F(t,T)$
Then $dY=\frac{dF}{F}-\frac{1}{2}\frac{<dF,dF>}{F^2}$
We have $\frac{dF}{F}=e^{-a(T-t)} dX_t=e^{-a(T-t)}\left(a(\hat{\mu}-\ln S_t)dt+\sigma dW_t\right)$
and $\frac{<dF,dF>}{F^2}= e^{-2a(T-t)}\sigma^2dt$
So that $d\ln F(t,T)=e^{-a(T-t)}\left(a(\hat{\mu}-\ln S_t)dt+\sigma dW_t\right)-\frac{1}{2}e^{-2a(T-t)}\sigma^2dt$
However I believe that this is wrong as the answer given in correction is: $$ d\ln F(t,T) = e^{-a(T-t)}\sigma dW_t-\frac{1}{2}e^{-2a(T-t)}\sigma^2dt$$
This means that there is probably a mistake in my $\frac{dF}{F}$, can you help me?
$d \log F(t,T) = d \left[ e^{a(t-T)} X_t + \hat{\mu} \left( 1 - e^{a(t-T)} \right) + \frac{\sigma^2}{4a} \left( 1 - e^{2a(t-T)} \right) \right]$
Using the Ito rule on this gives us $d \log F(t,T) = e^{a(t-T)}dX_t + \left[ a e^{a(t-T)} X_t - \hat{\mu} a e^{a(t-T)} + \frac{\sigma^2}{4a} 2a e^{2a(t-T)} \right] dt$
Using the fact that $dX_t = a(\hat{\mu} - X_t) dt + \sigma dW_t$ gives us
$d \log F(t,T) =e^{a(t-T)}\left[a(\hat{\mu} - X_t) dt + \sigma dW_t \right] + \left[ a e^{a(t-T)} X_t - \hat{\mu} a e^{a(t-T)} + \frac{\sigma^2}{4a} 2a e^{2a(t-T)} \right] dt$
This can be simplified to give the answer as requested.