Tail inequality on sum of product of normal variables For independent random variables $ x_1,..,x_n$  and $y_1,...,y_n$ following normal distribution $N(0,1)$, I need a simple estimate formula for
$P(|\sum_1^n x_iy_i|\ge nt) \leq e^{(?)}$ for $t>1$. Thanks.
 A: The answer to this question is given in Berstein's inequality,
as presented in Chapter 2 in Vershynin's book.
For $z_1,...,z_n$ independent, zero-mean, sub-exponential RVs, $$P\left(\left|\sum_i{z_i}\right|\ge nt\right)\le2\exp\left( -cn\cdot\min\left\{ \frac{t^2}{K^2},\frac{t}{K}\right\} \right)$$ where $c$ is a universal constant, $K=\max_i\left\|z_i\right\|_{\psi_1}$ and $\left\|z_i\right\|_{\psi_1}$ is the sub-exponential norm of $z_i$.
Denote $z_i=x_iy_i$. As $x_i,y_i$ are standardized normal RVs, each is a sub-gaussian RV with the sub-gaussian norm $\left\|x_i\right\|_{\psi_2}=\left\|y_i\right\|_{\psi_2}=1$ so $\left\|z_i\right\|_{\psi_1}\le\left\|x_i\right\|_{\psi_2}\cdot\left\|y_i\right\|_{\psi_2}=1$ (see Lemma 2.7.7 in Vershynin's). For convenience, we'll take $\left\|z_i\right\|_{\psi_1}=1$ (lower values would yield tighter bounds). Taking $c=\frac{1}{2}$ (as done in Section 1.3 here), we eventually get:
$$P\left(\left|\sum_i{z_i}\right|\ge nt\right)\le2\exp\left( -\frac{n}{2}\cdot\min\left\{ t^2,t\right\} \right)$$
