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I understand how one gets the mean of a random discrete random variable, but not necessarily how the process makes logical sense. For instance, how does the multiplication of x by P(X = x) give the mean of that discrete random variable? I feel like x would be an independent variable in most cases, not having any say in creating the mean.

If for instance, I was rolling a dice with sides 1-6, and let's say that all the variables have an equal probability of 16.67%, then how come the value of 6 trumps the value of 1 as more probability of the discrete random variable mean?

Caleb White
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    Welcome to cross validated, and this is an interesting question. 1. Let me ask a different question, if the outcome of a certain experiment is: {1, 4, 4, 4}, how would you calculate mean? 2. 6 does not trump 1 – as you said both are equally likely to be observed. The net contribution towards mean (x.p(x)) is higher in case of 6, as p(6)=p(1)=1/6. – PAF Apr 21 '20 at 14:43

2 Answers2

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Think about expected value as 'average'. In your example with a 6-sided die, the expected value (or average) is found by simply adding up all possible values and dividing by the number of possible values. Denote the random variable that is the value rolled as $X$:

$$E(X) = \sum_{i=1}^6 i P(X=i) = \sum_{i=1}^6 i\cdot \frac{1}{6} = 3.5$$.

The arithmetic average is a weighted average where every possible value carries the same weight $\frac{1}{N}$, where $N$ is the number of possible values. Notice the weights of all possible values sum to 1. Denote the variable that is being averaged as $Y$ and its possible values as $\{Y_1,...,Y_N\}$ :

$$\text{Arithmetic Average of } Y = \sum_{i=1}^N \frac{1}{N}Y_i. $$

Expected value is a more general version of this that allows for different weights for different possible values. Again, the weights of all possible values sum to 1. Denote the random variable as $Z$ and its possible values as $\{Z_1,...,Z_N\}$

$$\text{Expected Value of } Z = \sum_{i=1}^N P(Z=Z_i) Z_i. $$

dlnB
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A random variable is described by numbered tickets in a box. For instance, the values on the sides of a fair die correspond to a box with six tickets bearing the numbers $1,2,3,4,5,$ and $6.$

The expectation of a random variable is the (usual arithmetic) average of the numbers on its tickets.

From this it follows that

The expected value for the sum of draws made at random with replacement from a box equals $$(\text{number of draws})\times(\text{average of box}).$$

[Freedman et al. section 17.1, "The Expected Value."] This explicitly connects a definite mathematical property (the average of a box of tickets) with the possible random behavior of a quantity.

Reference

Freedman, Pisani, and Purves, Statistics. Third edition, 1998, WW Norton & Co.

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