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For known location, we can find the scale parameter of a normal distribution by calculating the sum of squared differences to the location, then dividing by n-1 and taking the square root. This is the maximum-likelihood and most-efficient scheme for parameter estimation for the normal distribution.

However, for the double-exponential distribution (Laplace) using the sum of absolute values is the MLE scheme. So the power p is not 2 but 1! And for the uniform distribution the maximum difference to the center gives the most-efficient parameter estimation scheme. Also that can be described as sum of power, by let the power p go to infinite (Chebyshev norm).

So for these 3 distributions with $p=1,2,\ldots,\infty$ we know both the MLE and the PDF well.

Question: If we generalize the estimation scheme by using any positive power (like p=4, 6, or whatever), what is the according distribution PDF(p,x)?

This would lead to quite a general PDF and can form a generalized normal distribution (like the Student-t extends the normal for more long-tailed distributions).

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    $\begingroup$ Division by $n-1$ is not ML. The MLE has division by $n$. $\endgroup$ – Glen_b Jun 26 '14 at 13:14
  • $\begingroup$ Hi Glen, probably you know better on N vs N-1, but using N-1 gives less bias and is preferable for me. $\endgroup$ – user32038 Jul 17 '14 at 9:31
  • $\begingroup$ In your first sentence you said this: "For known location, we can find the scale parameter of a normal distribution by calculating the sum of squared differences to the location, then dividing by n-1 and taking the square root. This is the maximum-likelihood ...*". The problem is that claim is untrue. It's NOT ML. Then in your comment you said "using N-1 gives less bias"... and sorry, but that's ALSO false, because you specified "known location". In that case, it's the $n$ divisor that's unbiased, and the $n-1$ divisor is biased... (ctd) $\endgroup$ – Glen_b Jul 17 '14 at 9:57
  • $\begingroup$ (ctd)... You can prefer whatever you wish --- so you can prefer $n-1$ even when it's neither ML nor unbiased if you want, but you can't claim something is ML when it isn't and you can't call unbiased things biased and vice-versa (or at least not without some vigorous opposition). Those things are not matters of preference. $\endgroup$ – Glen_b Jul 17 '14 at 10:01
  • $\begingroup$ Thanks for the correction on N vs N-1. I was wrong telling that N-1 is related to MLE. $\endgroup$ – user32038 Dec 12 '18 at 11:46
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See the generalized normal distribution (see version 1, not version 2):

$$f_X(x)=\frac{\beta}{2\alpha\Gamma(1/\beta)} \; e^{-(|x-\mu|/\alpha)^\beta}\quad \alpha,\beta>0,\,\,-\infty<x,\mu<\infty$$

The location parameter is $\mu$ and the scale is $\alpha$.

Given $\beta$:

$\hat{\mu}$ is the value of $\mu$ that minimizes $\sum_{i=1}^{n} |x_i-\mu|^{\beta}$

$\hat{\alpha} = ( \frac{\beta}{n} \sum_{i=1}^{n}|x_i-\mu|^{\beta})^{\frac{1}{ \beta}}$

If you want a scale, (denoted by $\sigma$, say) such that $\hat{\sigma}= ( \frac{1}{n} \sum_{i=1}^{n}|x_i-\mu|^{\beta})^{\frac{1}{ \beta}}$, you have to scale $\alpha$ by $\beta^\frac{1}{\beta}$ (i.e. $\sigma=\alpha/\beta^\frac{1}{\beta}$).

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  • $\begingroup$ Perfect answer, this helps me a lot! $\endgroup$ – user32038 Jun 26 '14 at 14:52

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