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Let $Y \sim f_Y (y)$ be a strictly continuous r.v.

Let $S \sim p(s)$ be a strictly discrete r.v.

Can you write the density $f_Y (y)$ as

$$f_Y (y) = \sum_S f(y|S=s)p(s)$$

I know the law of total probability holds when both $Y$ and $S$ are continuous random variables. But what if the conditioning r.v. is discrete, while $Y$ is continuous?

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Rather than close this question I'll leave an answer in the off chance it helps someone in future.

According to Karlin & Taylor's Introduction to Stochastic modelling, page 71, for continuous $x$ and discrete $n$, one can indeed write

$$f(x) =\sum_n f(x|n)p(n)$$

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  • $\begingroup$ This is the idea behind a mixture distribution. $p(n)$ are the $\pi_n$, and it is a sum of $N$ continuous densities. I think you asked pretty much the same question a while back. $\endgroup$
    – ken
    Commented Oct 13, 2018 at 15:02

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