Formula about discrete RV with parameter drawn from continuous RV

Question

Suppose $X$ is a continuous RV with density $f_X$ and $N$ is a discrete RV with parameter $X$. (For example, $N$ we could have $X$ drawn from Poisson($X$).) I read that $$P(N = n) = \int_{-\infty}^\infty P(N = n\mid X=x)f_X(x)dx.$$

How is this true? (I don't ask this from the perspective of measure theory, but from undergrad probability.)

Thanks :)

Answer 1

Suppose $X$ is discrete and takes values $\in \mathcal{X}_s$.

Then, by the total probability theorem, \begin{align*}P(N = n) &= \sum_{x \in \mathcal{X}_s} P\left (N = n, X = x \right )\\ &= \sum_{x \in \mathcal{X}_s} P(X = x) P(N = n | X = x)\end{align*}

The formula you have is just the specialization of total probability theorem to the case when $X$ is continuous. $\sum_{x \in \mathcal{X}_s} \to \int$

and

$P(X = x) \to f_X(x)$.