Integrability of Gaussian sums

Question

Let $(X_1, \ldots, X_n)$ be a Gaussian vector, and $Z = \sum_{i=1}^n |X_i|$. Since the map $x \mapsto e^{x^2}$, is convex, for any $t>0$ $$ e^{tZ^2} \, = \, e^{t \big(\sum_{i=1}^n |X_i| \big)^2} \, = \, e^{ \big(\frac 1n\sum_{i=1}^n \sqrt {t} \, n|X_i| \big)^2} \, \leq \, \frac 1n \sum_{i=1}^n e^{t n^2 X_i^2} . $$ The law of each $X_i$ has density $\frac {1}{\sqrt {2\pi \sigma_i^2}} \, e^{- (x-m_i)^2/2\sigma_i^2}$, so that there exists $t_0 >0$ such that $$ E \big(e^{t_0Z^2} \big) \, \leq \, \frac 1n \sum_{i=1}^n E \big (e^{t_0 n^2 X_i^2} \big) \, < \, \infty. $$ But is there a way to find the largest $t_0 >0$ such that $E(e^{t_0Z^2}) < \infty$?

Answer 1

As Paata suggests, we write $$ \mathbb{E} e^{tZ^{2}} = 2t \int_{0}^{\infty}\lambda e^{t\lambda^{2}}P(Z>\lambda)d\lambda. \quad\quad\quad (\heartsuit) $$ Next, for any vector $(\delta_1,\ldots,\delta_n)\in \{-1,1\}^n$ we have $$P\left(\sum \delta_i X_i>\lambda\right)\leqslant P(Z>\lambda)\leqslant \sum_{\varepsilon_i=\pm 1,i=1,\ldots,n} P\left(\sum \varepsilon_i X_i>\lambda\right),$$ thus the integral $(\heartsuit)$ converges if and only if each integral $$ \int_{0}^{\infty}\lambda e^{t\lambda^{2}}P\left(\sum \delta_i X_i>\lambda\right)d\lambda $$ converges. Since each $\sum \delta_iX_i$ is a 1-dimensional Gaussian with variance which I denote $\sigma^2(\delta_1,\ldots,\delta_n)$, you may take $t<1/(2\sigma^2(\delta_1,\ldots,\delta_n))$ and can not take $t\geqslant 1/(2\sigma^2(\delta_1,\ldots,\delta_n))$. (Possibly for specific $\delta_i$'s you can take $t=2\sigma^2(\delta_1,\ldots,\delta_n)$, but then for $-\delta_i$'s you can't.)

Answer 2

It is a standard fact from the theory of Gaussian processes (see e.g. the Ledoux-Talagrand book) that if $X_s$, $s \in S$, is a (centered) Gaussian process such that $Z = \sup_s X_s$ is finite almost surely, then $E(e^{tZ^2}) < \infty$ if and only if $ t < \frac {1}{2\sigma^2}$ where $\sigma^2 = \sup_s E(X_s^2)$. By duality $ \sum_{i=1}^n |X_i| = \sup_\alpha \langle \alpha, X \rangle$ where the supremum runs over all $\alpha = (\alpha_1, \ldots, \alpha_n) \in R^n$ such $\max_{1 \leq i \leq n} |\alpha_i| \leq 1$, or only $\alpha_i = \pm 1$ (and $ X = (X_1, \ldots, X_n)$). Hence here $\sigma^2 = \sup_\alpha \langle \Sigma \alpha, \alpha \rangle$ where $\Sigma$ is the covariance matrix of the Gaussian vector $X = (X_1, \ldots, X_n)$.