Let $X$ be a random variable with mean $\mu$ and probability density function $f(x)$.
The standard deviation $\sigma$ is usually defined as the root of the variance:
$$\sigma = \sqrt{\int_{-\infty}^\infty (x – \mu)^2 f(x) \,dx}$$
But how would one interpret the following term:
$$\int_{-\infty}^\infty |x – \mu| f(x) \,dx \,\,?$$
Surely this expression is not equal to $\sigma$?
Best Answer
Your expression is also a measure of dispersion. In fact it is a distance between the data points and $\mu$. It is not optimal becasuse (it is easy to prove) that the minimum distance
$$\int_{-\infty}^{\infty}(x-A)^2f_X(x)dx$$
is achieved when $A=\mu$
On the other hand, an optimal indicator is achieved when
$$\int_{-\infty}^{\infty}|x-Me|f_X(x)dx$$
Where $Me$ indicates the median. This indicator is called Median absolute deviation