I have this distribution, where '$x_i$' is shorthand for $X=i$, where $i=0,1,2,3,4$ and where $X$ is a random variable with possible values $0,1,2,3,4$
$$
P(x_0) = 0.2 \\
P(x_1) = 0.25 \\
P(x_2) = 0.3 \\
P(x_3) = 0.15 \\
P(x_4) = 0.1
$$
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?
Best Answer
The expected value cannot be larger than all 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.