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I recently read that:

a normal distribution is completely specified by its mean and standard deviation.

That makes a lot of sense. But I was wondering isn't it also true that it could be completely specified by its mean and the cubic deviation? Or quadratic one? Or even the mean deviation?

If we consider the standard deviation formula: $$\sigma = (\frac{1}{N} \sum_i \lvert x_i - avg \rvert^\color{red}{p})^{1/\color{red}{p}}$$ Then:

  • p = 1: mean deviation.
  • p = 2: standard deviation.
  • p = 3: cubic deviation. I just made this name up.
  • p = 4: quartic deviation. I just made this name up.
  • p = 2.3456789: any positive non-integer value of p.

Can any of those deviations completely specify a normal distribution, in addition to the mean value of course?

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  • $\begingroup$ When $p$ is even, these are $p^\text{th}$ roots of the usual central moments. So with the mean, any even $p$ picks out a unique normal distribution. $\endgroup$ Commented Aug 31, 2020 at 23:39
  • $\begingroup$ @Bungo Your comment is correct for actual odd $p$-th moments, but not for absolute $p$-th moments as in this problem. $\endgroup$ Commented Sep 1, 2020 at 0:00
  • $\begingroup$ @kimchilover Ah, apologies, I did not notice that the OP was using absolute moments. $\endgroup$
    – user169852
    Commented Sep 1, 2020 at 0:03
  • $\begingroup$ @EricTowers What about a non-integer p? For example with any number such as p = 2.3456789? $\endgroup$ Commented Sep 1, 2020 at 0:11
  • $\begingroup$ @RossMillikan Fixed it, thanks. $\endgroup$ Commented Sep 1, 2020 at 12:10

1 Answer 1

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From the Wikipedia entry, we know that the $p$-th central absolute moment of a $N(\mu,\sigma^2)$ random variable $X$ is $$E[|X-\mu|^p]=\sigma^p\frac{2^{p/2}\Gamma(\frac{p+1}2)}{\sqrt\pi}.$$ If we know this number, and know $p$, we can determine $\tau=\sigma^p$ and then determine $\sigma=\tau^{1/p}$. It might seem paradoxical, but even when $p=1$ this is sufficient to yield $\sigma$.

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