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Let $X_1, \ldots, X_n$ be jointly Gaussian, each of which is marginally distributed as a standard Gaussian $N(0,1)$. It is well known that $\max |X_i|$ achieves the maximum in the stochastic sense if $X_1, \ldots, X_n$ are independent. To be more precise, $\mathbb{P}(\max|X_i| \le c)$ is minimized in the case of iid $X_1, \ldots, X_n$ for any constant $c > 0$. This fact can be directly seen from the Gaussian correlation conjecture (now a theorem), which was recently proved by Thomas Royen.

Question Are there similar results concerning the second largest $|X_i|$, the third largest, etc?

For example, is it true that the second largest $|X_i|$ attains the maximum if $X_1, \ldots, X_n$ are independent?

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    $\begingroup$ You need to be precise about the statement "achieves the maximum in the stochastic sense" $\endgroup$ – Henry.L Apr 17 '17 at 0:42
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In general, your last statement is not true. For example, consider $n=2$. Let $X_1,X_2$ be independent standard Gaussian r.v. Then, the second largest is just min. For $c>0$ we have $$ \mathbb{P}[\min(|X_1|,|X_2|)\leq c]=1-\mathbb{P}[|X_1|\geq c,|X_2|\geq c]>1-\mathbb{P}[|X_1|\geq c,|X_1|\geq c]=\mathbb{P}[\min(|X_1|,|X_1|)\leq c]. $$

Similarly, for the third largest consider $n=3$ etc.

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  • $\begingroup$ Thanks for the example. Is the statement still true for large $n$? $\endgroup$ – John Wong Apr 17 '17 at 17:10
  • $\begingroup$ @John, I don't think so. The best way to show it -- to differentiate like in the proof of Slepian's inequality using the property of a Gaussian p.d.f. $\frac{\partial f}{\partial \sigma_{ij}}=\frac{\partial^2f}{\partial x_i\partial x_j}, \quad i\ne j.$ The derivative must be zero (a necessary condition), but I believe it wouldn't. $\endgroup$ – Dmitry Zaporozhets Apr 18 '17 at 6:36

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