Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 208 / 1

If $X$ is sub-Gaussian with variance parameter $\nu$, thenfor every $\lambda \in \mathbb{R}$. Restricting this inequality to $|\lambda |<1/\alpha$ proves that $X$ is sub-exponential with parameters $(\nu ,\alpha )$ for every $\alpha >0$.

Now let $X=Z^2-1$ for a standard normal distribution variable $Z$. Its moment-generating function isand is infinite for $\lambda \ge 1/2$. A sub-Gaussian moment-generating function must be finite for every real $\lambda$, so $X$ cannot be sub-Gaussian with any finite parameter. For $|\lambda |<1/4$, the stated inequality givesThus $X$ is sub-exponential with parameters $(4,4)$.

For every $0<\lambda <1/\alpha$, the Chernoff bound and the sub-exponential moment-generating bound giveIf $0<t\le \nu /\alpha$, choose $\lambda =t/\nu$; at the endpoint, take a limit from below. This yieldsIf $t>\nu /\alpha$, let $\lambda \uparrow 1/\alpha$. Since $\nu /(2{\alpha }^2)<t/(2\alpha )$,and hence $\mathbb{P}(X\ge t)\le e^{-t/(2\alpha )}$.

Because the variables are independent, the moment-generating function of the sum factors. Whenever $|\lambda |<1/{\max }_i{\alpha }_i$,Therefore ${\sum }_i{X}_i$ is sub-exponential with parameters

Because $\mathbb{E}X=0$, the moment-generating function series and the Bernstein moment condition imply, for $|\lambda |b<1$,For $|\lambda |<1/(2b)$, the denominator satisfies $1-|\lambda |b>1/2$, soThus $X$ is sub-exponential with parameters $(2\nu ,2b)$.