Subgaussian concentration of the absolute Gaussian average
= Subgaussian concentration of the absolute Gaussian average
For a standard normal variable $Y$ and $\beta=\mathbb E|Y|=\sqrt{2/\pi}$,
$$
\mathbb E e^{\pm u(|Y|-\beta)}\leq e^{Cu^2},
\qquad u\geq0.
$$
The exponential Markov inequality consequently gives Gaussian concentration for averages of independent copies of $|Y|$.