What is the correct way of expressing this probability mass function?

Question

Suppose I have a discrete random variable $S$. Now further suppose that I have a random variable $X$ which is a continuous random variable on the real line. I wanted to express the p.m.f. $P[S=s]$ for $S$ using the expression: \begin{equation} P[S=s]= \int_{\mathbb{R}}P[S=s\mid x]P[x\mid s]dx \end{equation}

A professor I spoke to mentioned that if both $S$ and $X$ were discrete, this expression would have been correct, however, given that $X$ is a continuous random variable, the way I have expressed it implies a zero probability (I suppose because of $P[x\mid s]$). I was told to express this correctly, I have to express it in terms of probability measures.

I have been spending the better part of yesterday on Shiryaev's probability book and on probability measures and Lebesgue integrals; but while I know he is correct about my incorrect expression, I am not sure exactly how this needs to be expressed in terms of probability measures.

Answer 1

It sounds like you are trying to write the law of total probability. For a discrete random variable $S$ and a continuous conditioning random variable $X$ with marginal density $f_X$, you can write this as:

$$\mathbb{P}(S=s) = \int \limits_\mathbb{R} \mathbb{P}(S=s|X=x) f_X(x) \ dx.$$

Observe that the integrand here consists of the conditional probability mass for $S|X$ multiplied by the marginal density of $X$ (not the conditional density). I see no reason you would need to write this in terms of the probability measures or the Lebesgue integral (since you know that $X$ is continuous here), but if you really want to do so then you can take the probability measure $\mu_X$ or the distribution function $F_X$ for $X$ and write the relevant result as:

$$\mathbb{P}(S=s) = \int \limits_\mathbb{R} \mathbb{P}(S=s|X=x) \ d F_X(x) = \int \limits_\mathbb{R} \mathbb{P}(S=s|X=x) \ d \mu_X.$$