If $X$ is <sub-Gaussian random variable>[sub-Gaussian] with variance parameter $\nu$, then
$$
\mathbb E e^{\lambda X}\leq e^{\lambda^2\nu/2}
$$
for every $\lambda\in\mathbb R$. Restricting this inequality to $|\lambda|<1/\alpha$ proves that $X$ is <sub-exponential random variable>[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> is
$$
\mathbb E e^{\lambda X}
=\frac{e^{-\lambda}}{\sqrt{1-2\lambda}},
\qquad \lambda<\frac12,
$$
and is infinite for $\lambda\geq1/2$. 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 gives
$$
\mathbb E e^{\lambda X}
\leq e^{2\lambda^2}
=e^{\lambda^2(4)/2}.
$$
Thus $X$ is sub-exponential with parameters $(4,4)$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-208/1/a/solution