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A well-known result is that the only density that is both spherical and independent error is normal: more precisely

Let $e_i$ be errors,

If the joint probability density satisfies $$f_n(e_1,e_2, ..., e_n) = f_1(e_1)f_1(e_2)...f_1(e_n)$$ (independence) and if it is spherical: $$f_n(e_1,e_2, ..., e_n) = g_n(\Sigma_{i=1}^{n} e_i^2)$$

then, the only densities that satisfy both conditions are the normal densities:

$$f_1(e_i)={1 \over \sqrt{2 \pi}\sigma} \, \exp(-{1\over2\sigma^2} e_i^2)$$

How could I prove it?

I guess taking the derivative of $e_i$ of $f(e_1,e_2, ..., e_n) = f(e_1)f(e_2)...f(e_n)= f(\Sigma_{i=1}^{n} e_i^2)$ would be a first move..

Similar result I just found..

"Theorem [Herschel-Maxwell]: Let Z∈Rn be a random vector for which (i) projections into orthogonal subspaces are independent and (ii) the distribution of Z depends only on the length ||Z||. Then Z is normally distributed.

Cited by George Cobb in Teaching statistics: Some important tensions (Chilean J. Statistics Vol. 2, No. 1, April 2011) at p. 54."

from What is the most surprising characterization of the Gaussian (normal) distribution?

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This is a standard calculus derivation: spherical symmetry tells you that $f_1(x)$ is a function of $x^2$, i.e. $$f_1(x)=g_1(x^2).$$ Independence plus spherical symmetry tell you that $$g_1(u)g_1(0)=g_2(u) \quad\text{and}\quad g_1(u)g_1(v)=g_2(u+v)\propto g_1(u+v)$$ Therefore, rescaling $g_1$ into $h_1$ so that the above become an equality, we derive the identity $$h_1(u)h_1(v)=h_1(u+v)$$ for which the only solution is of the form $$ h_1(u) = \exp \{\alpha u\},\qquad \alpha\in\mathbb{R} $$ Thus, $$f_1(x) = \dfrac{1}{\sqrt{2\pi\sigma^2}} \exp\{-x^2/2\sigma^2\},\qquad \sigma\in\mathbb{R}_+,$$ since only negative factors $\alpha$ lead to densities.

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  • $\begingroup$ thank you! obvious typos : $g_1(u)g_1(v)=g_2(u+v)=g_1(0)g_1(u+v)$ $\endgroup$ – KH Kim Jun 6 '12 at 10:05
  • $\begingroup$ @KHKim: thanks, but this is not a typo, only the use of a proportionality sign (and the proportionality coefficient is indeed $g_1(0)$) for the next step! $\endgroup$ – Xi'an Jun 6 '12 at 10:11
  • $\begingroup$ @KHKim: thanks for the $u+1$ instead of $u+v$ typo, indeed! $\endgroup$ – Xi'an Jun 6 '12 at 11:01

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