Past exam of the options II course of the University of Oxford / 2013 / M5 / b / ii

The two-point distribution gives $M(t)=(e^t+e^{-t})/2=\cosh t$. In the power series, $(2j)!={\prod }_{r=1}^j(2r)(2r-1)\ge 2^{j}j!$, so each coefficient satisfiesSumming the nonnegative terms proves $\boxed{M(t)\le e^{t^2/2}}$ for every real $t$. Part (a) therefore givesThis is the sub-Gaussian moment bound for a symmetric random sign.