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I wonder if multivariate normal distribution is the only class of distributions whose random vectors have distributions of the same class after linear transformations? How can one justify it?

Is there a name for such property of distributions? Bodie's Investment calls this "stable".

the normal distribution belongs to a special family of distributions characterized as "stable," because of the following property: When assets with normally distributed returns are mixed to construct a portfolio, the portfolio return also is normally distributed.


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You are asking for the class of distributions which is closed under the linear transformations, aren't you? – Ilya Sep 22 '11 at 15:36
@Gor No, I think the OP wants a single random variable $X$ such that $TX$ has the same distribution as $X$ for an arbitrary linear transformation $T$. [ADDED: But the OP seems to agree with you now; perhaps I am wrong and you're right. ;)] – Srivatsan Sep 22 '11 at 15:38
No, s/he doesn't (or so I understand). I think @Gortaur means a family of distributions $\{ \mathcal D_\alpha \}$ such that if you pick an $X \in \mathcal D_\alpha$ and apply a linear transformation to it, then you get a distribution $\mathcal D_\beta$ in the same family. We may or may not have $\beta = \alpha$. Can you see the difference? – Srivatsan Sep 22 '11 at 15:42
I don't think this is going to have a clear-cut answer. For any particular distribution, can't I define the class to be all linear transformations of said distribution? For example, the class of all uniform distributions in simplices. – Rahul Sep 22 '11 at 15:47
Wikipedia article on stable distributions: – Srivatsan Sep 22 '11 at 15:53

Discrete (multivariate) distributions form one such class. Discrete distributions with at most $n$ atoms form another.

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You seem to be asking for more than the "Stable distribution" scenario, you are thinking about transformations of a multivariate variable:

$${\bf y} = {\bf A x} + {\bf b}$$

with ${\bf A}, {\bf b}$ arbitrary (${\bf A}$ square), so that the "family" (multivariate density) is preserved. I see several problems to give this a clear-cut answer, even a meaning. No only the "family" concept is rather vague, but also the "multivariate" random variable density familiy: for example, we have a definition for a multivariate gaussian, but we don't have (in general) a definition of (say) a multivariate Cauchy. Hence, its difficult to give a useful characterizaton of families of multivariate distributions.

One rather formal way of attacking it would with characteristic functions. Let $\Phi_X({\bf \omega}) = E[\exp(i {\bf \omega^t X })]$ be the (multimensional) c.f. of $X$ and $H_X({\bf \omega}) = \log(\Phi_X{\bf \omega})$. Then, we have

$$H_Y({\bf \omega}) = H_X({\bf A^t} {\bf \omega}) + i {\bf \omega}^t {\bf b}$$

Thus, we are seeking families of complex functions $H(\omega)$ (with $H(0)=0$) that are closed under the above transformation. One can see immediately (what one already knew) that the gaussian familiy fits, because in that case $H(\omega)$ is a homegeneus cuadratic, and the transformation preserves the property. But I doubt one can say something more.

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Thanks! Would you agree with Robert's reply? How will "Discrete (multivariate) distributions" be explained by characteristic functions? – Tim Oct 9 '11 at 22:29
Robert's reply is of course correct. But it must be understood: "discrete distributions" must be regarded as the full family of all discrete distributions (or, in the alternative, the family of all discrete distributions with $n$ values -atoms-). In this case the logarithm in my transformation is not very useful, it's more easy to work with the CF $\Phi_X({\bf \omega})$ itself: if the distribution is discrete, then the CF is a finite sum of complex exponentials (like a Fourier sum). And it's easy to see that the transformed CF retains this property. – leonbloy Oct 9 '11 at 23:23

I dont know if multivariate normal distribution is the only such class.....there are results in statistics which show that if $X+Y$ and $Y$ are normally distributed and $X$,$y$ are independant,then $X$ is normal...this is bernsteins theorem........there are many such interesting characterisations of normal distribution..

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