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Let $\{X_1,X_2,\ldots,X_n\}$ be jointly Gaussian random variables of zero mean and variance $1$ with covariance matrix $K$. Let $Y=\max\{X_i\,:\,i=1,\ldots,n\}$.

In the case the variables are also independent ($K=I_{n}$) is a known result that $$ \mathbb{P}\Big[\lim_{n\to\infty}(2\log(n))^{-1/2}Y=1\Big]=1. $$

My question is: Is it true that for a general covariance matrix (with diagonal entries equal to one) then $$ \mathbb{P}\Big[\lim_{n\to\infty}(2\log(n))^{-1/2}Y\leq 1\Big]=1? $$ I'm only interested in the asymptotic behavior as $n$ increases.

Thanks!

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What do you mean by "as big as possible"? –  Dilip Sarwate Apr 10 '12 at 2:04
    
Among all the covariances $K$ such that $X_i$ has unit variance. –  ght Apr 10 '12 at 2:12
2  
Fix your accept rate. You could mean a couple of things by "as big as possible". Do you mean $E(Y)$ is as big as possible? In that case, no, $I_n$ does not maximize $E(Y)$. Try $n=2$. If you mean something else, clarify it. –  deinst Apr 10 '12 at 2:30
    
You might want to say, not just that they are Gaussian, but that they are jointly Gaussian, i.e. so distributed that every linear combination of them is Gaussian. –  Michael Hardy Apr 10 '12 at 2:37
    
I might interpret the question like this: $X$ is stochastically (weakly) bigger than $Y$ if for every real $x$, $\Pr(X>x)\ge\Pr(Y>x)$. That's a partial ordering. It could mean bigger in that sense. –  Michael Hardy Apr 10 '12 at 2:41
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