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Below is the definition of the law of total expectation from Wiki.

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The first equation states that for any $X, Y$ on the same probability space, then \begin{equation} E(X) = E(E(X|Y)) \end{equation}

It then states that one special case is

\begin{equation} E(X) = \sum_i E(X|A_i)P(A_i) \end{equation} if $\{A_i\}_i$ is a finite or countable partition of the sample space.

The way that the second equation is prefaced by "one special case.." leads me to think that it is a special case of the first equation.

My question is: how can one relate $E(X) = E(E(X|Y))$ to $E(X) = \sum_i E(X|A_i)P(A_i)$?

That is, if I define $Y = \{A_i\}_{i=1}^{n}$ as the set of partitions on the sample space, is the following correct? $$E(X) = E(E(X|Y)) = \sum_i E(X|A_i)P(A_i)$$

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Since the law of total expectation $E(E(X | Y) ) = E(X)$ is more general, it makes sense to show how this implies $E(X) = \sum_{i = 1}^n E(X | A_i)P(A_i)$.

Consider $Y_i = I_{A_i}$ using $\displaystyle \Omega = \dot{\cup}_{i =1}^n A_i$ it follows that $ P \left( \sum_{i =1}^n Y_i = 1 \right) = 1$. Then you have

\begin{align} E(X) & = \sum_{i=1}^n E(X Y_i) \\ & = \sum_{i=1}^n E(X Y_i | Y_i = 1) P(Y_i =1) + E(X Y_i | Y_i = 0) P(Y_i = 0) \\ & = \sum_{i=1}^n E( X | Y_i = 1) P(Y_i =1) + E( 0 |Y_i = 0) P(Y_i = 0) \\ & = \sum_{i=1}^n E( X | A_i) P(A_i). \end{align}

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  • $\begingroup$ Thanks. Is it correct to add one last line to the derivation above: i.e., $\sum_{i=1}^n E(X|A_i)P(A_i) = E(E(X|Y))$ $\endgroup$ – Devon Mar 12 at 1:20
  • $\begingroup$ That is just because $E(X) = E(E(X|Y))$. $\endgroup$ – Manuel Mar 12 at 2:14

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