We note that given a probability distribution function $P$ over a space $U$ the expected value of a function of the elements in U:

$$ E(f(x)) = \int_{U} f(x)P(x) $$

We thus consider the mean as the expected value of the numbers that is:

$$ E(x) = \int_{U} x P(x) $$

Now we consider "standard deviation" to be the expected difference between a variable from the mean that is

$$ Std(x) = E(|x - E(x)|) = E\left(\sqrt{(x - E(x))^2}\right) $$

Yet Standard deviation is always measured as:

$$ \sqrt{E((x - E(x)^2)} $$

The latter formula doesn't make sense to me. Can someone explain why mine is wrong and hte latter is corret?

  • 2
    $\begingroup$ Your $E(|X-E(X)|)$ is a measure of variability, so it is not wrong, and is in fact sometimes used, It is not the same as the standard deviation, which is much more useful. The main reason is that the variance of a sum of independent random variables is the sum of the variances. $\endgroup$ – André Nicolas Feb 10 '15 at 4:30

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