I've read that the Gaussian marginal is fully determined by the mean and variance. What does this mean in reality? If we consider a Gaussian marginal PDF is given by

$$ \pi_G(\xi|\mu,\sigma) = {1\over\sqrt{{2\pi}}} exp\left( -{ (\xi-\mu)^2 \over \sqrt{2}\sigma} \right)$$

Then it only requires the mean ($\mu$) and variance($\sigma$). But what about the uniform marginal:

$$ \pi_U(\xi|\beta,\alpha) = {1 \over{ \beta-\alpha}}$$

Where $\mu = {1\over 2}(\beta+\alpha)$ and $\sigma^2 = {1\over 12}(\beta-\alpha)^2$. Then to find $\alpha$ and $\beta$ the mean and variance are also required and so surely a uniform marginal is fully determined by the mean and variance.

The beta and Gamma marginals require shape and scale parameters, which are related to the mean and variance of a process.

So surly this statement applied to many one dimensional marginals. However the text books always seems to make a big deal of this as if it only applies to Gaussian marginals. why? Is it that for higher dimensional marginals this is only true for the Gaussian?

Update: Lets consider the gamma distribution. Its PDF is given by

$$\pi(x|k,\theta) = {\frac {1}{\Gamma (k)\theta ^{k}}}x^{k\,-\,1}e^{-{\frac {x}{\theta }}}$$

Its mean and variance are $ \mu = k\theta$ and $ \sigma^2 = k\theta^2$ respectively. Therefore if I know $ \mu$ and $\sigma^2 $ then I know k and $\theta$. Since $\theta, \sigma, k >0$ there is a unique solution. Therefore surely the gamma marginal is fully determined by the mean and variance? Maybe I'm missing something?

Presumably the answer is that this isn't the case for multidimensional distributions, other than Gaussian.

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    $\begingroup$ If you used distribution instead of marginal, it would be more straightforward, IMHO. $\endgroup$ – Richard Hardy Sep 11 '16 at 11:28
  • $\begingroup$ Try some other two-parameter families (such as the gamma family) to se if there are other examples ... $\endgroup$ – kjetil b halvorsen Sep 11 '16 at 11:34
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    $\begingroup$ What you wrote in your last edit is correct, you can now post an answer yourself! $\endgroup$ – kjetil b halvorsen Sep 11 '16 at 14:43
  • $\begingroup$ The answer is: any distribution which has two free parameters, such that the system of equations formed by these parameters via mean and variance gives you a unique solution. As you rightfully point out, the gamma distribution is another such distribution. $\endgroup$ – Alex R. Sep 11 '16 at 17:49

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