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I'm reading Bayesian Data Analysis: http://www.stat.columbia.edu/~gelman/book/ and this equation was given

$$E(u)=\int E(u|v)p(v)dv$$

Can someone explain this to me? I don't see how they're equivalent, because to get the expectation of $u$ you need to get the average of $u$, but the RHS is the integral across all $v$, but then there could be values of $u$ not conditioned on $v$ so that means those values will be excluded will they not? Which means it's not the expectation for $u$ because it's not the weighted sum across all $u$.

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It's the simpler form of Law of Total Expectation, $E[u]=E[E[u|v]]$. The inside expression is a function of $v$, so it's like $E[u|v]=g(v)$, and we want $E[g(v)]$, which is $$E[u]=E[g(v)]=\int g(v)p(v)dv=\int E[u|v]p(v)dv$$

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  • $\begingroup$ I don't think they made it clear in the book but I'm guessing $v$ is a set of sets that at independent and together make up the entire space? If that's the case then that totally make sense. $\endgroup$ Feb 1 at 0:24
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    $\begingroup$ $v$ is another random variable, and we integrate over its all possibilities. $\endgroup$
    – gunes
    Feb 1 at 13:39

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