Gaussian Johnson–Lindenstrauss concentration argument ($\mathbb{P}(|R_u-1|\ge \epsilon )\le 2e^{-kc(\epsilon )}$)

A short concentration inequality argument explains the bound. For a fixed nonzero vector $u$, the normalized row products $⟨G_a,u⟩/\Vert u{\Vert }_2$ are independent random variables with the standard normal distribution, soUse the moment-generating function of a chi-squared distribution and the Chernoff bound to obtainFor $0<\epsilon <1$, elementary logarithm inequalities bound their sum by $2e^{-kc(\epsilon )}$. Apply the union bound to the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ nonzero pair differences to get failure probability at most $n(n-1)e^{-kc(\epsilon )}\le \delta$. Independence of different pairs is unnecessary; they use the same map. This combines fixed-vector concentration with a finite-set union bound. ^[ref]