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).



Non-independent Sub-gaussian variables and concentration
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 $p\_r^-(-x)=p\_{-r}^+(x)$.


