Probability of an Ornstein-Uhlenbeck process Assume we have a probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\mathcal{F} =(\mathcal{F}_t)_{0 \leq t \leq \tau}$ is a Filtration, with $\tau < \infty$. The following definition is from an article im reading and trying to understand. On that probability space we define a Ornstein-Uhlenbeck process $D$. 
Let $W(t)_{0 \leq t \leq \tau}$ be a brownian motion on $(\Omega,\mathcal{F},\mathbb{P})$ which creates the filtration $\mathcal{F} =(\mathcal{F}_t)_{0 \leq t \leq \tau}$. Through the constants $D_0, \theta,\sigma \in \mathbb{R}_+$ and the function $\mu(t)$ we define the OU-process $D(t)$ as solution of the SDE:
\begin{align*}
& d D(t)  = \theta (\mu(t) - D(t)) d t + \sigma d W(t),  & D(0) = D_0 > 0, \quad t > 0. 
\end{align*}
The function $\mu(t)$ is defined as:
\begin{align*}
\mu(t):= a + b \cos(2 \pi t - c)- \frac{2\pi}{\theta} sin(2\pi t - c).
\end{align*} 
What I'm curious about is the probability of the OU-Process. The article makes the following argumentation:
Define $\overline{\mu(t)} := a + b \cos(2\pi t -c)$. Then the probabilty follows:
\begin{align*}
P( D(s) \leq x_1 \mid F_t ) = \Phi \left( \frac{x_1- \overline{\mu(s)}-(D(t)-\overline{\mu(t)}) e^{-\theta(s-t)}}{ \sqrt{\frac{\sigma^2}{2 \theta} (1- e^{-2 \theta(s-t)}))}} \Bigg\vert F_t \right)
\end{align*} 
where $\Phi$ denotes the cumulative distrubtion funtion of an $N(0,1)$ random variable.
I found out that a for a general OU-process $X$ we have:
\begin{align*}
X(t) \sim N\left( D(0) \cdot e^{-\overline{\theta}t} +  \overline{\mu}(t) (1-e^{-\overline{\theta}t}), \frac{\overline{\sigma}^2}{2\overline{\theta}} (1- e^{-2\overline{\theta}t}) \right).
\end{align*} 
I dont understand the defintion of $\overline{\mu}$, how does that get into $\Phi(\cdot)$.
Why do they use $s-t$ instead of just $s$? 
 A: For $s>t$, you can express $D(s)$ in terms of $D(t)$ and a term that is independent of $\mathcal{F}_t$. Then you are able to compute the conditional probability. Specifically,
\begin{align*}
e^{\theta s} D(s) = e^{\theta t} D(t) + \int_t^s \theta e^{\theta v}\mu(v) dv +\sigma\int_t^se^{\theta v}dW_v.
\end{align*}
Here, 
\begin{align*}
\mu(t) = a+ b\big[\cos(2\pi t-c) - \frac{2\pi}{\theta} \sin(2\pi t -c) \big].
\end{align*}
Then
\begin{align*}
e^{\theta s} D(s) &= e^{\theta t} D(t) + a\left(e^{\theta s} - e^{\theta t}\right)+b\left[e^{\theta s} \cos(2\pi s-c) - e^{\theta t} \cos(2\pi t-c)\right] + \sigma\int_t^se^{\theta v}dW_v\\
&=e^{\theta t} D(t) + e^{\theta s}\overline{\mu(s)} - e^{\theta t}\overline{\mu(t)} + \sigma\int_t^se^{\theta v}dW_v,
\end{align*}
where 
\begin{align*}
\overline{\mu(t)} = a + b\cos(2\pi t-c).
\end{align*}
That is,
\begin{align*}
D(s) = e^{-\theta(s-t)} D(t)+ \overline{\mu(s)}-e^{-\theta(s-t)}\overline{\mu(t)} + \sigma\int_t^se^{-\theta (s-v)}dW_v.
\end{align*}
Therefore,
\begin{align*}
E\left(D(s) \mid \mathcal{F}_t\right) &= e^{-\theta(s-t)} D(t)+ \overline{\mu(s)}-e^{-\theta(s-t)}\overline{\mu(t)},
\end{align*}
and
\begin{align*}
\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right) &=\sigma^2\int_t^se^{-2\theta (s-v)} dv\\
&=\frac{\sigma^2}{2\theta}\left(1-e^{-2\theta (s-t)}\right).
\end{align*}
Consequently,
\begin{align*}
P\left(D(s) \le x_1 \mid \mathcal{F}_t \right) &= P\left(\frac{D(s)-E\left(D(s) \mid \mathcal{F}_t\right)}{\sqrt{\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right)}} \le \frac{x_1-E\left(D(s) \mid \mathcal{F}_t\right)}{\sqrt{\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right)}} \mid \mathcal{F}_t \right)\\
&= \Phi\left(\frac{x_1-E\left(D(s) \mid \mathcal{F}_t\right)}{\sqrt{\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right)}}\right),
\end{align*}
since
\begin{align*}
\frac{D(s)-E\left(D(s) \mid \mathcal{F}_t\right)}{\sqrt{\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right)}} = \frac{\sigma\int_t^se^{-\theta (s-v)}dW_v}{\sqrt{\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right)}}
\end{align*}
is independent of $\mathcal{F}_t$, and 
\begin{align*}
\frac{x_1-E\left(D(s) \mid \mathcal{F}_t\right)}{\sqrt{\textrm{Var}\left( D(s) \mid \mathcal{F}_t\right)}}
\end{align*}
is $\mathcal{F}_t$ measurable.
