$\newcommand\xx{\mathbf{x}}\newcommand\yy{\mathbf{y}}\newcommand\A{\mathbf{A}}\newcommand\aalpha{\boldsymbol{\alpha}}\newcommand\bbeta{\boldsymbol{\beta}}\newcommand\E{\mathbb{E}}\newcommand\inner[1]{\langle #1 \rangle}$In the scalar case, if a random variable $X$ is sub-Gaussian then we know $X^2$ is sub-exponential. [1] Can this be said for higher dimensions? Specifically, if $\xx$ satisfies $$\E\exp(\inner{\aalpha,\xx}) \leq \exp\bigl(\|\aalpha\|^2 \frac{\sigma}{2}\bigr)$$ for some $\sigma>0$, can we find $\gamma,b>0$ so that $$ \E\exp(\aalpha^*(\xx\xx^* - \E\xx\xx^*)\bbeta) \leq \exp\bigl(\|\aalpha\|^2\|\bbeta\|^2\frac{\gamma^2}{2}\bigr) $$ for all $\|\aalpha\| < b$ and $\|\bbeta\| < b$? Or (preferably, and hopefully equivalently) some self-adjoint $\A$ and $c>0$ so that $$ \mathbb{E}[(\xx\xx^* - \E\xx\xx^*)^p] \preceq \frac{p!}{2}c^{p-2}\A^2 $$ for all $p=2,3,4,\ldots$. You might recognize the above as the 'subexponential' condition given in Tropp's Matrix Bernstein inequality. [2 Thm 6.2]
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$\begingroup$ For your first question, take $\sigma=1$. Since $\alpha^*x=X_1$ is Gaussian of variance $\|\alpha\|^2$ and similarly $X_2=x^*\beta$ is Gaussian of variance $\|\beta\|^2$, you can use that in distribution $X_2=\rho_1 X_1+\rho_2 Z$ with $Z$ independent of $X_1$, and then employ the scalar bound. I have not worked out the constants. $\endgroup$– ofer zeitouniCommented Mar 6, 2018 at 6:42
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$\begingroup$ @oferzeitouni I'm somewhat confused, as $\boldsymbol{\alpha}\mathbf{x}\mathbf{x}^*\boldsymbol{\beta} = \sum \alpha_i\beta_j x_ix_j$ is a sum of products of sub-Gaussians, not a sum of sub-Gaussians. What do you mean here exactly? A result would likely require showing the product of scalar non-independent sub-Gaussians is sub-Exponential. $\endgroup$– Conner DiPaoloCommented Mar 6, 2018 at 12:35
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1$\begingroup$ And to answer your question in comments, if $X_1,X_2$ are dependent Gaussian, centered, then $X_1X_2$ has the same law as $\rho X_1^2+ \sqrt{1-\rho^2}X_1Z$, where $X_1,Z$ are independent. Since both summands are subexponential, their sum is as well (this last fact does not requyire independence of the summands). $\endgroup$– ofer zeitouniCommented Mar 7, 2018 at 7:43
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