Expectation vs average

Expectation vs average

  2:02 AM    mayuravaani

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Let X$�$ represent the outcome of a roll of an unbiased six-sided die. The possible values for X$�$ are 1, 2, 3, 4, 5, and 6, each having the probability of occurrence of 1/6. The expectation value (or expected value) of X$�$ is then given by

(X)expected=1(1/6)+2⋅(1/6)+3⋅(1/6)+4⋅(1/6)+5⋅(1/6)+6⋅(1/6)=21/6=3.5$(�)\text{expected}=1(1/6)+2\cdot (1/6)+3\cdot (1/6)+4\cdot (1/6)+5\cdot (1/6)+6\cdot (1/6)=21/6=3.5$

Suppose that in a sequence of ten rolls of the die, if the outcomes are 5, 2, 6, 2, 2, 1, 2, 3, 6, 1, then the average (arithmetic mean) of the results is given by

(X)average=(5+2+6+2+2+1+2+3+6+1)/10=3.0$(�)\text{average}=(5+2+6+2+2+1+2+3+6+1)/10=3.0$

We say that the average value is 3.0, with the distance of 0.5 from the expectation value of 3.5. If we roll the die N$�$ times, where N$�$ is very large, then the average will converge to the expected value, i.e.,(X)average=(X)expected$(�)\text{average}=(�)\text{expected}$. This is evidently because, when N$�$ is very large each possible value of X$�$ (i.e. 1 to 6) will occur with equal probability of 1/6, turning the average to the expectation value.

Resource: stackoverflow