# Can an expected value (mean) be higher than the values used to create it?

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

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 canâ€™t get anything else!
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 $7$.