A question on conditional probability in sde

Question

Let $s \in[a, b]$ and $x \in \mathbb{R}$ be fixed and consider the following SIE: $$ X_{t}=x+\int_{s}^{t} \sigma\left(u, X_{u}\right) d B(u)+\int_{s}^{t} f\left(u, X_{u}\right) d u, \quad s \leq t \leq b $$ Use $X_{t}^{s, x}$ to denote the solution of the SIE in above equation. Let the initial condition $x$ of equation be a constant. So 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 $$ P\left(X_{t}^{s, Z} \leq y \mid \mathcal{F}_{s}\right)=\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=Z} \quad \forall y \in \mathbb{R} $$

So, why can I have the last equality? I am confused how can I substitue an $\mathcal{F}_{s}$-measurable random variable $Z$ by a constant $x$ then put $|_{x=Z}$ outside of the conditional probability? And what is the difference between $\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=Z}$ and $P\left(X_{t}^{s, Z} \leq y\right)$?

Answer 1

Essentially, the equation $P(X_t^{s,Z} \le y | Z) = P(X_t^{s,x}\le y)|_{x=Z}$ is just stating that $X_t^{s,Z}$ has the Markov property.

The value $P(X_t^{s,x} \le y)$ is fully deterministic, and can be thought of as a deterministic function of $x$ (and $s,t,y$), i.e. for a fixed $s,t,y$ we can define $f(x) := P(X_t^{s,x} \le y)$. The notation $P(X_t^{s,x}\le y)|_{x=Z}$ means we are evaluating that deterministic function at the random variable $Z$, i.e. $P(X_t^{s,x}\le y)|_{x=Z} = f(Z)$. For clarity, I will refer to $P(X_t^{s,x}\le y)|_{x=Z}$ as $f(Z)$ from here on to avoid getting it confused with the other probabilities we encounter.

The value $P(X_t^{s,Z} \le y)$ is also deterministic, but depends on the distribution of $Z$. The easiest way to see that dependence would be to note that $X_s^{s,Z} = Z,$ so $P(X_s^{s,Z} \le y) = P(Z \le y)$. I think that $P(X_t^{s,Z} \le y) = \mathbb{E}[f(Z)]$, but I don't see any other relation between $P(X_t^{s,Z} \le y)$ and $f(Z)$.

On the other hand, $P(X_t^{s,Z} \le y|\mathcal F_s)$ is a random variable. Since $X$ is an Ito diffusion, it is Markov and hence $$P(X_t^{s,Z} \le y|\mathcal F_s) = P(X_t^{s,Z} \le y|X_s^{s,Z}) = P(X_t^{s,Z} \le y|Z).$$ Finally, since we know the deterministic function $f$ was defined by $f(x) = P(X_t^{s,x} \le y)$, the Markov property further implies $$P(X_t^{s,Z} \le y | Z) = f(Z)= P(X_t^{s,x}\le y)|_{x=Z},$$ which is the statement from the beginning.

Answer 2

Since $P\left(X_{t}^{s, Z} \leq y\right)$ is a function of $x$, we have $P\left(X_{t}^{s, Z} \leq y\right)=\left.P\left(X_{t}^{s, x} \leq y\right)\right|_{x=Z}$, thus, $P(X_{t}^{s, Z} \leq y|\mathcal{F}_{s})=\left.P(X_{t}^{s, x} \leq y|\mathcal{F}_{s})\right|_{x=Z}$

Next, since $X_{t}^{s, x}$ is independent of $\mathcal{F}_{s}$, we have $\left.P(X_{t}^{s, x} \leq y|\mathcal{F}_{s})\right|_{x=Z}=\left.P(X_{t}^{s, x} \leq y)\right|_{x=Z}$, therefore, $$P(X_{t}^{s, Z} \leq y|\mathcal{F}_{s})=\left.P(X_{t}^{s, x} \leq y)\right|_{x=Z}$$