= Exponential moment of an absolute standard normal variable
{title2=$\mathbb E e^{a|Z|}=2e^{a^2/2}\Phi(a)$}
For $Z$ with a <standard normal distribution> and $a\geq0$, split the <Gaussian integral> at zero and complete the square to obtain $\mathbb E e^{a|Z|}=2e^{a^2/2}\Phi(a)\leq2e^{a^2/2}$, where $\Phi$ is the <standard normal distribution function>. This converts a bound on an absolute <Gaussian random variable> into a two-sided <Gaussian tail bound> using <Markov inequality>.
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