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 $s$?
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.