I have seen that expected value of a discrete random variable is equal to the arithmetic mean of the distribution provided the values it takes. Is it true for all random variables irrespective of the distribution? Is there a case or example where expected value differs from the arithmetic mean?
Secondly I think it applies only for discrete random variables. I think for continuous random variables, the pdf is zero at particular points. So in that case can I say that expected value is not equal to the mean of random variable?
Best Answer
In the discrete case the expected value is a weighted sum, where the possible values of the variable are weighted by their probability of occurring (the probability mass function), $EX=\sum_{i=1}^nx_iP(X=x_i)$. Since all weights are non-negative, smaller than untiy, and their sum equals unity, the expected value of a discrete random variable is also a specific convex combination of its possible values.
In the continuous case the expected value is a weighted integral, where the possible values of the variable are weighted by the probability density function $EY=\int_{-\infty}^{\infty}yf_Y(y)dy$.
What happens is that the arithmetic (i.e. unweighted) mean from the realization of a collection of identically distributed random variables (i.e. the "sample mean") is shown to be an unbiased and consistent estimator of the expected value, although the latter is a weighted mean.