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I am trying to understand better the definition of conditional expectation, for that I want to prove the following:

Let $X$ be a random variable in the probability space $(\Omega, \Sigma, \mathbb{P})$, let $\mathcal{F}$ be a sub-$\sigma$-algebra of $\Sigma$ and let $\mathcal{P}$ be a partition of $\Omega$ such that $\mathcal{P}$ are the minimal generators of $\mathcal{F}$, i.e. $\mathcal{F} = \sigma(\mathcal{P})$.

I want to show that: $$\mathbb{E}_\mathbb{P}(X \mid \mathcal{F}) = \mathbb{E}_\mathbb{P}(X \mid \mathcal{P})$$

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  • $\begingroup$ By definition: $\mathbb{E}_P(X\mid \mathcal{P}) = \mathbb{E}_P(X\mid \sigma(\mathcal{P})) = \mathbb{E}_P(X\mid \mathcal{F})$. Actually the way conditional expectation is defined, the only way of interpreting the second argument in $\mathbb{E}_P(X\mid *)$ is as the $\sigma$-algebra generated by it. (i.e. $\sigma(*)$). $\endgroup$ – Sayantan Apr 11 at 14:51
  • $\begingroup$ Why do you interpret the second argument in $\mathbb{E}_P(X\mid *)$ as the $\sigma$-algebra generated by it? This should be my question $\endgroup$ – Robert T Apr 11 at 14:54
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    $\begingroup$ If you start from the most general definition of conditional expectation, you define $\mathbb{E}_P(X\mid \mathcal{F})$ only for sub $\sigma$-algebras $\mathcal{F}$ of $\Sigma$. Hence the only interpretation of $*$ in $\mathbb{E}_P(X\mid *)$ has to be $\sigma(*)$. $\endgroup$ – Sayantan Apr 11 at 15:00
  • $\begingroup$ What is your definition of $\mathbb E_{\mathbb P}(X|\mathcal P)$? $\endgroup$ – Mike Earnest Apr 11 at 17:27
  • $\begingroup$ I'm using the definition given by Williams - "Probability with Martingales", i.e., $\mathbb E_{\mathbb P}(X|\mathcal F)$ is the random variable $Y$ that is $\mathcal F$-measurable and satisfies that $\int_A Y d\mathbb P = \int_A X d\mathbb P, \forall A \in \mathcal F$. But for $\mathbb E_{\mathbb P}(X|\mathcal P)$ I don't have any definition since$\mathcal P$ is not a $\sigma$-algebra. $\endgroup$ – Robert T Apr 12 at 7:28

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