Conditional probability question in Markov process proof

Question

From textbook Introduction to Stochastic Integration written by Hui-Hsiung Kuo, one equation is very unclear to me. The content of the part is quoted below. My question is that: in the middle step, author says that $$ P\left(X_{t}^{s, Z} \leq y \mid \mathcal{F}_{s}\right)=\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=Z} \quad \forall y \in \mathbb{R} $$ So why is it true? I believe the key fact is the connection between $X_t^{s,x}$ and $X_t^{s,Z}$. I know that $X_t^{s,x}$ being independent of $\mathcal{F}_{s}$ will be needed when I try to eliminate $\mathcal{F}_{s}$, but this seems to be not enough for the equation.

In this section we will study some properties of the solution of the stochastic differential (or rather integral) equation in Equation (10.3.1), namely, $$ X_{t}=\xi+\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)+\int_{a}^{t} f\left(s, X_{s}\right) d s, \quad a \leq t \leq b, $$ where $\sigma$ and $f$ satisfy the Lipschitz and linear growth conditions. We first prove that the solution $X_{t}$ is a Markov process.
Let $\left\{\mathcal{F}_{t} ; a \leq t \leq b\right\}$ be the filtration given by the Brownian motion $B(t)$, namely, $\mathcal{F}_{t}$ is defined by $\mathcal{F}_{t}=\sigma\{B(s) ; s \leq t\}$. Obviously, the solution $X_{t}$ is adapted to this filtration. Let $s \in[a, b]$ and $x \in \mathbb{R}$ be fixed and consider the following SIE: $$ X_{t}=x+\int_{s}^{t} \sigma\left(u, X_{u}\right) d B(u)+\int_{s}^{t} f\left(u, X_{u}\right) d u, \quad s \leq t \leq b . $$ To avoid confusion with the solution $X_{t}$, we use $X_{t}^{s, x}$ to denote the solution of the SIE in Equation (10.6.1). Since the initial condition $x$ of Equation (10.6.1) is a constant, we see from the approximation procedure in the proof of Theorem 10.3.5 that the solution $X_{t}^{s, x}$ is independent of the $\sigma$-field $\mathcal{F}_{s}$ for each $t \in[s, b]$. If follows that for any $\mathcal{F}_{s}$-measurable random variable $Z$, we have the equality $$ P\left(X_{t}^{s, Z} \leq y \mid \mathcal{F}_{s}\right)=\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=Z}, \quad \forall y \in \mathbb{R} $$ In particular, put $Z=X_{s}$ in this equation to get $$ P\left(X_{t}^{s, X_{s}} \leq y \mid \mathcal{F}_{s}\right)=\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=X_{s}}, \quad \forall y \in \mathbb{R} $$ Now, since $X_{t}$ is a solution of the SIE in Equation (10.3.1), we have $$ X_{t}=X_{s}+\int_{s}^{t} \sigma\left(u, X_{u}\right) d B(u)+\int_{s}^{t} f\left(u, X_{u}\right) d u, \quad s \leq t \leq b . $$ But $X_{t}^{s, X_{s}}$ is also a solution of this SIE. Hence by the uniqueness of a solution, we must have $X_{t}=X_{t}^{s, X_{s}}$. Thus Equation (10.6.2) can be rewritten as $$ P\left(X_{t} \leq y \mid \mathcal{F}_{s}\right)=\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=X_{s}}, \quad \forall y \in \mathbb{R} $$ which is measurable with respect to the $\sigma$-field generated by $X_{s}$. Therefore, we can conclude that $$ \begin{aligned} P\left(X_{t} \leq y \mid X_{s}\right) &=E\left[P\left(X_{t} \leq y \mid \mathcal{F}_{s}\right) \mid X_{s}\right] \\ &=P\left(X_{t} \leq y \mid \mathcal{F}_{s}\right), \quad \forall s<t, y \in \mathbb{R} \end{aligned} $$ Then by Lemma 10.5.9, the stochastic process $X_{t}$ is a Markov process. Hence we have proved the next theorem.

Answer 1

From Stochastic Calculus by Paolo Baldi, page 94, we have the following lemma:

Lemma 4.1 (The freezing lemma) Let $(\Omega, \mathscr{F}, \mathrm{P})$ be a probability space and $\mathscr{G}$ and $\mathscr{D}$ independent sub- $\sigma$-algebras of $\mathscr{F}$. Let $X$ be a $\mathscr{D}$-measurable r.v. taking values in the measurable space $(E, \mathscr{E})$ and $\psi: E \times \Omega \rightarrow \mathbb{R}$ an $\mathscr{E} \otimes \mathscr{G}$-measurable function such that $\omega \mapsto \psi(X(\omega), \omega)$ is integrable. Then $$ \mathrm{E}[\psi(X, \cdot) \mid \mathscr{D}]=\Phi(X), $$ where $\Phi(x)=\mathrm{E}[\psi(x, \cdot)]$.