Would this way of evaluating this probability be correct?

Question

Suppose I have a discrete variable $S_t$ and a continuous variable $X_t$. Further, suppose I wish to evaluate $P(S_t=s_t)$. Would the below derivations be correct?

\begin{align} P(S_t=s_t)&=\int P(S_t=s_t,X_t)dX_t\\ &=\int P(S_t=s_t\mid X_t)P(X_t)dX_t \end{align} Now suppose I have a realisation $x_t$ for $X_t$, such that I can evaluate $P(S_t=s_t\mid X_t=x_t)=c$, with $c\in[0,1]$. Can I then express the above as

\begin{align} P(S_t=s_t)&\overset{?}{=}\int \underbrace{P(S_t=s_t\mid X_t=x_t)}_{c}P(X_t)dX_t\\ &\overset{?}{=}c\int P(X_t)dX_t \end{align} ? I am pretty confident the above is incorrect. However, I am hoping someone sees what I am trying to accomplish and help out with an alternative. Thanks in advance.

Answer 1

I do not understand what you mean by $\mathbb{P}\left(S_t=s_t\vert X_t=x_t\right)=c.$

This definitely looks wrong as you end up with $\mathbb{P}\left(S_t=s_t\right)=c \int_{-\infty}^{+\infty} f_{X_t}(x_t)dx_t=c$.

Maybe you want to use your realisations in order to use an estimator of $\mathbb{P}\left(S_t=s_t\vert X_t=x_t\right)$ equal to $c(t,x_t)$ (so it would vary as a function of time and of realisation of $X_t$ ? But estimating $c(t,x_t)$ properly would require multiple observations of the processes $S_t$ and $X_t$, and the estimation could be tricky since $X_t$ is a continuous random variable.