Let $g \in R^{d}$ have $iid$ Gaussian components. Let $a \in R^{d}$, and let $b \in R^{d}$. be arbitrary vectors.
Consider the random variable $Y_{g,g}:= \frac{1}{n}\langle g,a \rangle \langle g, b \rangle$. What can be said about the tails of the random variable $Y$?
If $g$ were replaced with $\bar{g}$ a gaussian vector with iid components independent of $g$ then $Y$ is sub-exponential; absorbing $\frac{1}{\sqrt{n}}$ into each inner product shows that in this case $Y_{g, \bar{g}}$ is a product of independent Gaussians (in fact in this case, $Y_{g, \bar{g}}$ is called a decoupled Gaussian chaos I believe).
Best Answer
Let $X:=a\cdot g$ and $Y:=b\cdot g$. We want to bound the tails of the random variable (r.v.) $XY$ (the factor $\frac1n$ is clearly inessential). The r.v.'s $X$ and $Y$ are zero-mean jointly normal, with $Var\,X=|a|^2$, $Var\,Y=|b|^2$, and $Cov\,(X,Y)=a\cdot b$, where $|\cdot|$ is the Euclidean norm. By further rescaling, without loss of generality, $|a|=|b|=1$, and then \begin{equation} r:=Cov\,(X,Y)=a\cdot b\in[-1,1] \end{equation} and \begin{equation} (X,Y)=(U,rU+\sqrt{1-r^2}V) \end{equation} for some iid standard normal r.v.'s $U$ and $V$. The case of $r=\pm1$ is simple and will be henceforth excluded.
It follows that \begin{equation} M(t):=Ee^{tXY}=\frac{1}{\sqrt{1-2r t-(1-r^2)t^2}} \end{equation} for $t\in(-\frac1{1-r},\frac1{1+r})$ and $M(t)=\infty$ for the other real $t$.
So, for all real $x\ge\max[0,r]$, \begin{equation} \begin{aligned} &p_r^+(x):=P(XY\ge x)\le\inf_{t\ge0}e^{-tx}M(t)=e^{-t_x x}M(t_x) \\ & = \frac{\sqrt{2} x}{\sqrt{\sqrt{4 x^2+(1-r^2)^2}+r^2}-1} \\ &\times \exp \left(-\frac{\sqrt{4 x^2+(1-r^2)^2}-2 r x-(1-r^2)}{2 \left(1-r^2\right)}\right), \end{aligned} \end{equation} where \begin{equation} t_x:=\frac{\sqrt{4 x^2+(1-r^2)^2}-2 r x-(1-r^2)}{2 \left(1-r^2\right) x}. \end{equation} It follows that \begin{equation} p_r^+(x)\lesssim\sqrt{e x}\,e^{-x/(1+r)} \end{equation} as $x\to\infty$.
The left-tail probability $p_r^-(-x):=P(XY\le -x)$ can be estimated similarly or just using the identity $p_r^-(-x)=p_{-r}^+(x)$.