x is a binary variable that follows a Bernoulli distribution of parameter $\mu$ . In a demonstration I see this equality : $$ \mathsf p(x=1|D) = \int_{0}^{1} p(x=1 | \mu) p(\mu | D) d\mu $$
I don't really understand how to apply the sum rule of probability to get this result. Could you help me ?
A previous answer had the main ideas using shorthand notation. That answer also emphasized that an additional assumption about conditional independence is needed. That answer was deleted. I will give an alternative derivation that is a bit longer.
Preliminary facts: Let $A$ and $D$ be events with positive probability. Let $X$ be a continuous random variable with density $f_X(x)$. Then
\begin{align} 1. &P[A|X=x] = \frac{f_{X|A}(x)P[A]}{f_X(x)}\quad \mbox{(assuming $f_X(x)>0$)} \\ 2. &P[A] = \int_{-\infty}^{\infty} P[A|X=x]f_X(x)dx\\ 3. &P[A|D] = \int_{-\infty}^{\infty} P[A|X=x, D]f_{X|D}(x)dx \end{align}
Fact 3 is a conditioned version of Fact 2. Intuitively, Fact 3 holds because a conditional probability measure, given $D$, is still a valid probability measure.
Let $M$ be a continuous random variable that determines the probability that $X=1$: $$ P[X=1|M=\mu] = \mu \quad \forall \mu \in [0,1]$$
Method 1 (Using fact 3):
$$P[X=1|D] = \int_0^1 P[X=1|M=\mu, D]f_{M|D}(\mu)d\mu$$ This is as far as we can go. But if we know that the events $\{X=1\}$ and $D$ are conditionally independent given $M$, then $$ P[X=1|M=\mu, D] = P[X=1|M=\mu] \quad \forall \mu \in [0,1]$$ and we get the desired result $$ P[X=1|D] = \int_0^1 P[X=1|M=\mu]f_{M|D}(\mu)d\mu$$
Method 2 (Using only facts 1 and 2): We have \begin{align} P[X=1|D] &\overset{(a)}{=} \frac{P[\{X=1\}\cap D]}{P[D]}\\ &\overset{(b)}{=}\frac{1}{P[D]}\int_0^1 P[\{X=1\}\cap D|M=\mu]f_M(\mu)d\mu\\ \end{align} where (a) holds by definition of conditional probability; (b) holds by Fact 2. This is as far as we can go. But if we know that $\{X=1\}$ and $D$ are conditionally independent given $M$ then $$P[\{X=1\}\cap D| M=\mu] = P[X=1|M=\mu]P[D|M=\mu]$$ and so \begin{align} P[X=1|D] &= \frac{1}{P[D]}\int_0^1 P[X=1|M=\mu]P[D|M=\mu]f_M(\mu)d\mu\\ &= \int_0^1 P[X=1|M=\mu] \left(\frac{P[D|M=\mu]f_M(\mu)}{P[D]}\right)d\mu\\ &=\int_0^1 P[X=1|M=\mu] f_{M|D}(\mu)d\mu \end{align} where the last equality uses Fact 1.