Having trouble finding expected value and variance of a stochastic process defined by SDE:
$dX_{t} = a X_{t} dt + b dB_{t}$
$X_0 = x$, $a$ and $b$ are constant values, $B_t$~$N(0,t)$
Thank you for any help or pointers. I will now post my progresses:
1) Considering an auxiliary process $Z_t=X_te^{-at}$
2) Using Ito's formula to derive $dZ_t$ omitting quadratic covariance term due to it being between a stochastic and a determinist process (=0):
$dZ_t = e^{-at} dX_t+X_t(-ae^{-at}) dt$
3) Plugging in $dX_t$:
$dZ_t=e^{-at}(aX_tdt+bdB_t)-aX_te^{-at}dt$
$dZ_t=aX_te^{-at}dt+be^{-at}dB_t-aX_te^{-at}dt$
$dZ_t=be^{-at}dB_t$
4) Applying the integral on both sides ( change of variable from t to s inside the integral)
$\int_{0}^{t}dZ_s = b \int_{0}^{t}e^{-as}dB_s $
$Z_t-Z_0 = b \int_{0}^{t}e^{-as}dB_s $
5) Plugging back in $Z_t=X_te^{-at}$ and isolating $X_t$
$X_te^{-at}-X_0=b\int_{0}^{t}e^{-as}dB_s$
$X_t=X_0e^{at} + be^{at}\int_{0}^{t}e^{-as}dB_s$
6) Applying expectations
$\mathbb{E}(X_t)=xe^{at} + be^{at}\,\mathbb{E}(\int_{0}^{t}e^{-as}dB_s)$
I end up with this expression, now I know that I should use the Doléans exponential to prove that the stochastic integral is 0. But how can I proceed further to recover both expected value and variance?
Hint: The stochastic integral
$$M_t := \int_0^t e^{-as} \, dB_s$$
is a martingale. Hence, $\mathbb{E}M_t = \mathbb{E}M_0=0$. In order to calculate the variance of $X_t$ use Itô's isometry.
Alternative approach: Since stochastic integrals are martingales, we have
$$\mathbb{E}X_t- \mathbb{E}X_0 = \int_0^t a \cdot \mathbb{E}X_s \, ds,$$
i.e. $m(t) := \mathbb{E}X_t$ solves the ordinary differential equation (ODE)
$$m'(t) = a \cdot m(t) \qquad m(0) = \mathbb{E}X_0.$$
The (unique) solution is $m(t) =e^{at} \mathbb{E}X_0$. Similarly, using Itô's formla, one can show that
$$\mathbb{E}(X_t^2)-\mathbb{E}(X_0^2) = \int_0^t \left( 2a \mathbb{E}(X_s^2) + b^2 \right) \, ds.$$
Consequently, $\sigma(t) := \mathbb{E}(X_t^2)$ solves
$$\sigma'(t) = 2a \sigma(t)+b, \qquad \sigma(0) = \mathbb{E}(X_0^2).$$
Solving this (linear) ODE yields $\mathbb{E}(X_t^2)$.
Remark: The process $(X_t)_{t \geq 0}$ is called Ornstein-Uhlenbeck process.