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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.

Thanks!

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    You are asking for the class of distributions which is closed under the linear transformations, aren't you?2011-09-22
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    @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. ;)]2011-09-22
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    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?2011-09-22
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    @Srivatsan: I mean "we may or may not have β=α". Have I seen the difference?2011-09-22
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    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.2011-09-22
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    @Rahul: What change would you think will make the question to have a clear-cut answer?2011-09-22
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    Wikipedia article on stable distributions: http://en.wikipedia.org/wiki/Stable_distribution.2011-09-22
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    @Srivatsan: The Wikipedia's definition requires the random variables to be independent. I wonder what if the random variables are not required to be independent as I original asked?2011-09-22
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    "When assets with normally distributed returns are mixed to construct a portfolio" - When I read it, I imagine they are talking about taking a linear combination of (independent) assets to form a portfolio. Can you explain why you are thinking of linear combinations?2011-09-22
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    @Srivatsan: Because it means linear combination.2011-09-22
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    Well, it doesn't say "linear combination". I said so. (And by the way we seem to be taking over the comments. I would like to clean up my share of the comments, but not sure how.)2011-09-22
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    I would like to point out that your question still says "linear transformations". Wikipedia's definition involves linear *combinations*, and your text's description suggests the same. If you do mean linear *transformations*, I don't know how multiple (independent or not) random variables can get involved in a single linear transformation. If you do actually mean linear combinations, you should edit the question so that we are all on the same page.2011-09-22
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    @Rahul: By linear transformation, I view all the random variables as a single random vector, then linear transformations can be applied. But if this is still not clear enough, we can fall back on linear combination, since it is a special linear transformation.2011-09-22
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    I remember reading one definition of multivariate normal distribution as $X_1, \ldots, X_n$ are said to be jointly normal if (i) the joint density depends only on the means, variances, and covariances, (ii) the random variables are independent if they are uncorrelated, and (iii) if $(Y_1,\ldots,Y_n) = (X_1, \ldots, X_n)A + B$ where $A$ is any $n\times n$ matrix and $B$ a $n$-vector, then $Y_1, \ldots, Y_n$ are also jointly normal. I assume that constants are considered to be degenerate normal random variables with zero variance.2011-09-22
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    "uniform distributions in simplices" works for nonsingular linear transformations, but not for singular ones. If you take a uniform distribution on an $n$-simplex and project it on a simplex of lower dimension, the distribution is no longer uniform.2011-09-23

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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?2011-10-09
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    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.2011-10-09
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Discrete (multivariate) distributions form one such class. Discrete distributions with at most $n$ atoms form another.

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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..