Given a Ornstein-Uhlenbeck process conditionated on $V_s=v\in\mathbb R$ I'm given a Ornstein-Uhlenbeck proces $V=(V_t)_{t\geq 0}$, i.e.,
$$
V_t = \frac{\sigma}{\sqrt{2\beta}}e^{-\beta t}B_{e^{2\beta t}}.
$$
I'm told to prove that $V_{s+t}$, conditionated on $V_s=v$, follows a
$$
N(e^{-\beta t}v,\frac{\sigma^2}{2\beta}(1-e^{-2\beta t}),
$$
but I don't see why the expectation is not zero, since the expectation of the brownian motion is zero. Any hint, please?
 A: $$V_{t+s}=\frac{\sigma}{\sqrt{2\beta}}e^{-\beta(t+s)}B_{e^{2\beta(t+s)}}$$
Let us find the moments. As the expectation of the increment $B_{e^{2\beta(t+s)}}-B_{e^{2\beta s}}$ is $0$ and if $V_s$ is known then $B_{e^{2\beta s}}$ is known:
$$E[V_{t+s}|V_s]=\frac{\sigma}{\sqrt{2\beta}}e^{-\beta(t+s)}E[B_{e^{2\beta(t+s)}}|V_s]=\frac{\sigma}{\sqrt{2\beta}}e^{-\beta(t+s)}B_{e^{2\beta s}}=e^{-\beta t}V_s$$
Also
$$E[(B_{e^{2\beta(t+s)}}-B_{e^{2\beta s}})^2|V_s]=E[B_{e^{2\beta(t+s)}}^2|V_s]+B_{e^{2\beta s}}^2-2B_{e^{2\beta s}}E[B_{e^{2\beta(t+s)}}|V_s]$$
$$e^{2\beta(t+s)}-e^{2\beta s}=E[B_{e^{2\beta(t+s)}}^2|V_s]+B_{e^{2\beta s}}^2-2B_{e^{2\beta s}}^2=E[B_{e^{2\beta(t+s)}}^2|V_s]-B_{e^{2\beta s}}^2$$
Therefore
$$E[V_{t+s}^2|V_s]=\frac{\sigma^2}{2\beta}e^{-2\beta(t+s)}E[B_{e^{2\beta(t+s)}}^2|V_s]$$
$$E[V_{t+s}|V_s]^2=\frac{\sigma^2}{2\beta}e^{-2\beta(t+s)}E[B_{e^{2\beta(t+s)}}|V_s]^2$$
Thus by using $\textrm{Var}[X]=E[X^2]-E[X]^2$ we get
$$\textrm{Var}[V_{t+s}|V_s]=\frac{\sigma^2}{2\beta}e^{-2\beta(t+s)}\bigg(e^{2\beta(t+s)}-e^{2\beta s}+B_{e^{2\beta s}}^2-B_{e^{2\beta s}}^2\bigg)=\frac{\sigma^2}{2\beta}(1-e^{-2\beta t})$$
