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Is it always true that nested expectations equal expectations on the joint? $$ \mathbb{E}_{p(x)}[\mathbb{E}_{p(y \mid x)}[f(x, y)]] = \mathbb{E}_{p(x, y)}[f(x, y)] $$ Something along these lines appears at the end of a proof I am reading. Surely though one can write the LHS using the TOWER property which would give $$ \mathbb{E}_{p(y)}[f(x, y)] $$ which seems quite different..

Physics_Student
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    Research Fubini's Theorem. Alternatively, examine whatever definition of conditional expectation you are familiar with (this is one definition). – whuber Jul 03 '23 at 14:09
  • By Fubini's I know I can exchange integrals. I'm assuming I need to decompose it as such? $$ \int f(x, y) p(x, y) = \int f(x, y) p(y \mid x) p(x) = \mathbb{E}{p(x)}[\mathbb{E}{p(y\mid x}[f(x, y)]] $$ – Physics_Student Jul 03 '23 at 14:13
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    Fubini's Theorem also asserts the sequential integrals equal the joint integral, under mild conditions on the integrand. – whuber Jul 03 '23 at 14:24
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    As I argued in answers and comments of this question and this question, it is really pointless and potentially inaccurate to attach subscripts to the expectation operator "$E$". When written in standard notations, I infer what you tried to confirm is \begin{align} E[E[f(X, Y)|X]] = E[f(X, Y)], \end{align} which is exactly the law of iterative expectations. – Zhanxiong Jul 03 '23 at 15:19

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