By solving an SDE I want to derive the analytical results for mean and variance of the process of extended Vasicek model.
$$ dr(t) = \left(\eta - \gamma r(t) \right)dt + c dX(t) $$
where $\gamma$ is the reversion rate and $s = \frac{\eta}{\gamma}$ is the average short rate.
How can I set $X(t) = r(t) - s$ and solve by integration over both sides of the SDE with the help of the integrating factor $e^{yt}$ and in a second step derive the mean and variance?
Take your equation,
$ dr(t) = \left(\eta - \gamma r(t) \right)dt + c \, dX(t)$
and rearrange it as you suggested:
$dr(t) = \gamma \left(\frac{\eta}{\gamma} - r(t) \right)dt + c \, dX(t)$
$dr(t) = \gamma \left(s - r(t) \right)dt + c \, dX(t)$
Now if you multiply through by the integrating factor $e^{\gamma t}$ as you mentioned, you should get this expression after a little bit of manipulation:
$d \left( e^{\gamma t} r_{t} \right) = e^{\gamma t}\gamma \, s \,dt + e^{\gamma t} c \, d X(t)$
And then integrate from 0 to t to get:
$r_{t} = r_{0}e^{-\gamma t}+ s\left( 1-e^{-\gamma t }\right) +c\int_{0}^{t} {e^{-\gamma \left( t- u \right) } d X(u)}$
So the mean is just the deterministic term, and you can determine the variance via Ito isometry.
$V \left[ r_{t} \mid r_0 \right]={c}^{2} \int_{0}^{t} {e^{-2 \gamma \left(t-u \right)} du}=\frac{{c}^2}{2{\gamma}} {\left( 1- e^{-2 \gamma t} \right)}$
With the above steps in sight, could you clarify which particular part you are after?
You also use the 'Extended Vasicek', which is different in the sense that the mean is a function of time (in its simplest form). If that's what you are after then please google the derivation of the drift of the Extended Vasicek. The info here will also be useful. How to get set the theta function in the Hull-White model to replicate the current yield curve
