Editing Expected value (section)

===Random variables with countably infinitely many outcomes===
Informally, the expectation of a random variable with a [[countable set|countably infinite set]] of possible outcomes is defined analogously as the weighted average of all possible outcomes, where the weights are given by the probabilities of realizing each given value. This is to say that
<math display="block">\operatorname{E}[X] = \sum_{i=1}^\infty x_i\, p_i,</math>
where {{math|''x''<sub>1</sub>, ''x''<sub>2</sub>, ...}} are the possible outcomes of the random variable {{mvar|X}} and {{math|''p''<sub>1</sub>, ''p''<sub>2</sub>, ...}} are their corresponding probabilities. In many non-mathematical textbooks, this is presented as the full definition of expected values in this context.{{sfnm|1a1=Ross|1y=2019|1loc=Section 2.4.1}}

However, there are some subtleties with infinite summation, so the above formula is not suitable as a mathematical definition. In particular, the [[Riemann series theorem]] of [[mathematical analysis]] illustrates that the value of certain infinite sums involving positive and negative summands depends on the order in which the summands are given. Since the outcomes of a random variable have no naturally given order, this creates a difficulty in defining expected value precisely.

For this reason, many mathematical textbooks only consider the case that the infinite sum given above [[absolute convergence|converges absolutely]], which implies that the infinite sum is a finite number independent of the ordering of summands.{{sfnm|1a1=Feller|1y=1968|1loc=Section IX.2}} In the alternative case that the infinite sum does not converge absolutely, one says the random variable ''does not have finite expectation.''{{sfnm|1a1=Feller|1y=1968|1loc=Section IX.2}}

====Examples====
* Suppose <math>x_i = i</math> and <math>p_i = \tfrac{c}{i \cdot 2^i}</math> for <math>i = 1, 2, 3, \ldots,</math> where <math>c = \tfrac{1}{\ln 2}</math> is the scaling factor which makes the probabilities sum to 1. Then we have <math display="block">\operatorname{E}[X] \,= \sum_i x_i p_i = 1(\tfrac{c}{2})
+ 2(\tfrac{c}{8}) + 3 (\tfrac{c}{24}) + \cdots
 \,= \, \tfrac{c}{2} + \tfrac{c}{4} + \tfrac{c}{8} + \cdots \,=\, c \,=\, \tfrac{1}{\ln 2}.</math>