|For a distribution taking on values x1, ..., xN, xi occurring with probability pi, we know the expected value, or average, is|
|x1p1 + x2p2 + ... + xNpN|
|Then for a continuous distribution p(x) the expected value, or mean, mu is|
|In general, the nth moment of p(x) is|
|So mu = M1(p).|
|The variance is the average value of
|Expanding (x - mu)2 and distributing the integrals, we see|
Return to the standard deviation.