Friday, January 9, 2015

Avg(X-squared) always larger than square[avg(X)]

E[X2] is always larger than E2[X]

Confused which is larger? Quick reminder -- think of a population of X ~{-5,5} uniform so E[X] = 0. More generally,

If the population has both positive and negative members, then averaging will reduce the magnitude by cancelling out a lot of extreme values.

In the common scenario where population is all positive, it's slightly less intuitive, but we can still look at an outlier. Averaging usually reduces the outlier's impact, but if we square every member first the outlier will have more impact.

One step further,

E[X2] = E2[X]      + Var[X]


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