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I have this distribution:

$$ P(x_0) = 0.2 \\ P(x_1) = 0.25 \\ P(x_2) = 0.3 \\ P(x_3) = 0.15 \\ P(x_4) = 0.1 $$

Using the Expected Value formula:

$$ \mu = (0)0.2 + (1)0.25 + (2)0.3 + (3)0.15 + (4)0.1 \\ \mu = 1.7 $$

How does this make sense? How can the expected value be LARGER than any probability in my distribution? Am I using the wrong formula?

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You may be interested in the related Probabilistic Method. – Raphael Feb 4 '12 at 17:58
    
In your table, you write $P(x_n)$, but it appears you mean $P(x=n)$. Is that correct? – robjohn Feb 4 '12 at 18:04
up vote 8 down vote accepted

The expected value cannot be larger than any of the possible values, but it can certainly be larger than any of the probabilities. If you roll a fair die, each of the possible values ($1,2,3,4,5$, and $6$) occurs with probability $\frac16$, so the expected value is $$(1)\frac16+(2)\frac16+(3)\frac16+(4)\frac16+(5)\frac16+(6)\frac16=3.5\;,$$ far bigger than $\frac16$.

Now imagine that the die has a $6$ on every face. The probability that it comes up $6$ is $1$, and the expected value is clearly $6$: you can’t get anything else!

The expected value can be thought of as the long run average value of the possible values, each of them weighted by its probability of occurrence; in principle it can be anywhere between the smallest possible value and the largest possible value. It is affected by the probabilities only to the extent that they affect the weighting of the possible values.

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So the 1.7 is right? Awesome :) Thanks – n0pe Feb 4 '12 at 17:51
    
I have a question regarding the example you have given. Does expected value make any sense when all the probabilities are the same? I understand that if the probabilities of getting each number was different, then we would be taking weighted averages. But, for example, in the answer: what do I infer from 3.5? – user93868 Jan 19 at 15:41
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@user93868: That if you roll a fair die many times, the average roll will be very close to $3.5$. – Brian M. Scott Jan 19 at 17:46
    
@Brian, thanks! – user93868 Jan 19 at 18:02
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@user93868: Not quite. In general the expected value need not be a possible value; however, if, for instance, you had a fair die with a $1$ on one face and $6$ on each of the other five faces, the expected value would be $\frac16\cdot1+\frac56\cdot6=\frac{31}6$, much closer to $6$ than to $1$. The probabilities are in effect weights: the more likely an outcome is, the more heavily it influences the expected value. – Brian M. Scott Jan 19 at 18:14

The expectation of your distribution is exactly $$ 0.2 x_0 + 0.25 x_1 + 0.3 x_2 + 0.15 x_3 + 0.1 x_4. $$ According to the frequentist interpretation, the expected value is the value that you get when you perform a large number of experiments and compute the average. So if for example your random variable is constant, say always equal to $10$, then the expectation is $10$. There's no problem with the expectation being bigger than $1$. However, since the expectation is a weighted average of the values of the random variable, it always lies between the minimal value and the maximal value. So if your random variable always gets a value betewen $3$ and $7$, then the expectation is also going to be between $3$ and $7$.

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