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 |
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In general, the nth moment of p(x) is |
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So mu = M1(p). |
The variance is the average value of
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Expanding (x - mu)2 and distributing the integrals, we see |
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Return to the standard deviation.