Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

In my research I reached some very nice results for IID complex Gaussian vectors $\bf{x}$. Now I realize that my results hold for any random vectors that are unaffected by a unitary map, i.e., $\bf{z}\sim \bf{x}$, where $\bf{z}=\bf{Q}\bf{x}$, and where $\bf{Q}$ is unitary.

My question is simple, and perhaps a bit naive, what other interesting distributions are there?

share|improve this question
3  
In my research I've proved some very nice results for curves of genus 3. Now I realize that my results hold also for curves of genus 4. What other interesting genuses are there? –  Steven Landsburg Jul 5 '12 at 21:51
add comment

2 Answers

You're looking at distributions with rotational symmetry. The answer is, any distribution that depends only on the norm of ${\bf x}$. You can get such distributions by taking a symmetric distribution $p$ over $\mathbb{R}$ and letting $P({\bf x}) \sim p(||{\bf x}||)$

share|improve this answer
    
They are also known as spherical distributions. –  Federico Poloni Jul 30 '12 at 19:34
add comment

If you restrict your attention to random vectors, which have independent entries, then the only possible case is Gaussian. First, note that you can permute coordinates to deduce that they are in fact IID. Multiplying one coordinate (say $z_1$) by $i$ and $-i$ we can conclude that real and imaginary parts are identically distributed and symmetric. Now, apply transformation $(z_1, \dots z_n) \to (\frac{z_1+z_2}{\sqrt{2}}, \frac{z_1 - z_2}{\sqrt{2}}, \dots z_n)$, which gives as independence of $z_1+z_2$ and $z_1 - z_2$. Also their real and imaginary parts are independent, so we need a lemma:

Lemma. Let $X,Y$ be IID, symmetric random variables, such that $X-Y$ and $X+Y$ are independent. Then $X$ is Gaussian.

Proof. Let's compute characteristic function of X: $$ \varphi_X(2t) = \varphi_{X+Y + X-Y} (t) = \varphi_X(t)^{4}$$ $\Psi(t) := \varphi_X(t)^\frac{1}{t^2}$ satisfies $\Psi(t) = \Psi(2t)$, so we have $A^{t^2} \leqslant \varphi_X(t) \leqslant 1$ for some $A\leqslant 1$ which means that around zero we have $\varphi_X(t) \geqslant 1 - \varepsilon t^2$ for some $\epsilon >0$ (or $\varphi_X(t)\equiv 1$), hence $\mathbb{E}X^2 < \infty$. Now let $X_n$ be a sequence of IID random variables such that $X_n \sim X$. Then it can be easily seen that $\varphi_{\frac{X_1 + X_2 + X_3 + X_4}{2}}(t) = \varphi_X(t)$ and, inductively, $\varphi_{\frac{X_1 + \dots + X_{4^{n}}}{2^{n}}} = \varphi_X(t)$. But $X$ has finite second moment, hence by CLT the LHS converges to the characteristic function of a Gaussian, and the result follows.

Actually, what I've just written above, is only sufficient to write $(z_1, \dots, z_n)$ in a form $\mathbf{X} + i \mathbf{Y}$, where $\mathbf{X}$ and $\mathbf{Y}$ are identically distributed standard (possibly rescaled) Gaussian vectors, but one can exploit unitary invariance, to prove that joint distribution is also Gaussian.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.