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Let $\mathbb{R}^N$ be the ambient space and let $D_1$ and $D_2$ be different multivariate Gaussian distributions with the same mean (assume they are independent). Suppose further that $D_1$ is restricted to a proper subspace (for simplicity we can take a line) of $\mathbb{R}^N$ while $D_2$ covers the whole space. The variances are different.

Consider the mixture of these two distributions: $pD_1+(1-p)D_2$.

First question: since each of these is an elliptical distribution, is the mixture of these elliptical?

I suspect the answer is yes, but I am having trouble showing why. If it is, then the natural second question is: how can I characterize the above mixture as an elliptical distribution in an explicit way?

Note: An elliptical distribution $f$ is such that $f(x)=C(g)g(x^T\Sigma^{-1}x)\text{det}(\Sigma)^{\frac12}$, where $\Sigma$ is a positive definite matrix in $\mathbb{R}^{N\times N}$ and $g:[0,\infty]\to[0\infty]$ is such that $\int_0^\infty g(x)x^{N-1}<\infty$ and $C(g)$ is a normalization parameter.

Any general references on elliptical distributions would be fantastic.

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  • In my deleted answer I thought you had $D_1,D_2$ as random vectors as opposed to distribution functions. Can you please confirm you mean the latter? – JPi Feb 07 '14 at 01:54
  • These are probability distribution. In particular they are http://en.wikipedia.org/wiki/Multivariate_normal_distribution –  Feb 07 '14 at 01:55
  • Are they distribution FUNCTIONS or are they random vectors with those distributions? Big difference. If the latter then the sum is a normal and hence elliptical. If the former then that's a very ugly beast because of the subspace limitation. – JPi Feb 07 '14 at 02:00
  • I don't understand the distinction, perhaps. We are drawing points/vectors from these distributions. I will shortly define elliptical distributions. –  Feb 07 '14 at 02:04
  • If you're drawing vectors from these distributors and then taking their weighted average then you again get a normal. If you draw a random vector from the distribution that's characterized by a weighted average of the two different distribution functions then you get the ugly beast. – JPi Feb 07 '14 at 02:16
  • The point is that you are drawing with probability $p$ from one distribution, and with probability $(1-p)$ from the other distribution. As I understand it this is called a mixture distribution. So my first question is whether this mixture of Gaussians is elliptical (it cannot be Gaussian). And if so how can we express it explicitly by the definition of an elliptical distribution. –  Feb 07 '14 at 02:19
  • I think maybe your question was whether I was taking a convolution or a mixture. I am taking a mixture. –  Feb 07 '14 at 02:21
  • No, they're not; see http://www.math.ethz.ch/finance/summerschool/partD.pdf . – JPi Feb 07 '14 at 03:08
  • Can you be specific? Maybe write an answer? The link you showed seemed to suggest that linear combinations of elliptical functions were elliptical. If this is true than my second question remains. –  Feb 07 '14 at 03:15

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As JPi said, you'll get an ugly beast. The set of points with probability density greater than some threshold is a filled ellipsoid for an elliptical distribution, but if you consider the case where $D_1$ is the distribution of $X+\varepsilon Y$ and $D_2$ of $Y-\varepsilon X$ for $X,Y$ iid normal variables and $\varepsilon$ small, the set of points with probability density greater than some threshold will be a non-convex "crosshairs" set.

The stability of elliptical distributions under linear combination mentioned in JPi's paper refers to addition of random variables (that is, convolution of distribution functions), not linear combination of distribution functions (that is, mixture of distributions).

On the other hand, if the covariance matrices are scalar multiples of each other, you will definitely get an elliptical mixture. It's a necessary and sufficient condition.

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