Second moment bound from a clipped Gaussian tail
= Second moment bound from a clipped Gaussian tail
{title2=$\mathbb E Z^2\leq(1+\log A)/b$}
If $\mathbb P(|Z|\geq t)\leq A e^{-bt^2}$ for $A\geq1$ and $b>0$, then $\mathbb E Z^2\leq(1+\log A)/b$. Indeed the <tail integral formula for moments> gives $\mathbb E Z^2\leq\int_0^\infty\min(1,Ae^{-bu})\,du$; split at $u=(\log A)/b$. Clipping at one improves the constant compared with integrating the exponential bound alone.