# Would this way of evaluating this probability be correct?

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.

## 1 Answer

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.