Expected value of a random variable differing from arithmetic mean 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?
 A: 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.
A: I think that an arithmetic mean approaches expected value, as the number of samples increase in number. Say, you have a die which you have rolled 10 times and the outcomes are {5,6,4,5,3,2,1,2,4,6}
The mean of the above values is 3.8.
But the expected value when a die is rolled 10 times ( for that matter any number of times) is constant and is 1*1/6+2*1/6+...6*1/6 = 3.5
Hence, we see that the mean and expected values are different. If you take a large number of samples, the mean of the sample means (population mean) will reach the expected vale
