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Theorem: suppose $\Omega$ is discrete, then $\omega$ is confined to a set $ \{ \omega_1 , \ldots , \omega_k \} $ iff $E[X] = \sum_{i=1}^k p_i X(\omega_i)$ such that: $p_i \geqslant 0$ and $\sum_{i=1}^k p_i = 1$

I do not quite understand what this theorem is saying, nor what it would be used for. From my understanding: $p_i \geqslant 0$ and $\sum_{i=1}^k p_i = 1$ must always be true, a probability is never negative or the probability of the universe more than $1$. So does this imply that a discrete random variable must be constrained to a finitely countable set?

(this is a theorem we proved at the beginning of the semester and have yet to use)

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    $\begingroup$ I do not know what that theorem statement means either. Likely there was some miscommunication. I suspect that was just supposed to be the definition of expectation of a random variable $X$ defined over a finite sample space $S=\{\omega_1, \ldots, \omega_k\}$ with masses $P[\{\omega_i\}]=p_i$ for $i \in \{1, \ldots, k\}$. Indeed, for a finite (or countably infinite) sample space, those probability masses would always be nonnegative and sum to 1. $\endgroup$ – Michael Oct 15 '16 at 19:44
  • $\begingroup$ You wrote $<\omega_1,\ldots,\omega_k>$ and I changed it to $\langle\omega_1,\ldots,\omega_k\rangle$, but then I noticed you said "a set". Might you have intended $\{\omega_1,\ldots,\omega_k\} \text{?} \qquad$ $\endgroup$ – Michael Hardy Oct 15 '16 at 19:53
  • $\begingroup$ You also wrote $E[x]$ where it appeared that you meant $E[X]$, so I changed that. $\endgroup$ – Michael Hardy Oct 15 '16 at 19:55
  • $\begingroup$ Thank you for edits, changed to curly braces, just figured out how to imput them! But do you have any insight on the theorem? $\endgroup$ – rannoudanames Oct 15 '16 at 19:58
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    $\begingroup$ I wouldn't worry about this. A situation where the sample space itself must be discrete does not really come up; usually the sample space is some huge set and you define discrete random variables on that set. $\endgroup$ – Ian Oct 15 '16 at 20:01

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