A centered <random variable> $Y$ is <sub-Gaussian random variable>[sub-Gaussian] with variance parameter $\sigma^2$ when
$$
\log\mathbb E e^{\lambda Y}\leq\frac{\sigma^2\lambda^2}{2}
\qquad(\lambda\in\mathbb R).
$$
The <Chernoff bound>, applied to $Y$ and $-Y$, yields
$$
\mathbb P(|Y|\geq t)\leq2e^{-t^2/(2\sigma^2)}.
$$
The <tail integral formula for moments> and the substitution $u=t^2/(2\sigma^2)$ now give
$$
\begin{aligned}
\mathbb E|Y|^q
&=q\int_0^\infty t^{q-1}\mathbb P(|Y|\geq t)\,dt\\
&\leq2q\int_0^\infty t^{q-1}e^{-t^2/(2\sigma^2)}\,dt
=2\Gamma\left(\frac q2+1\right)(2\sigma^2)^{q/2}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 210 / 1 / d / Solution

For $q>0$ and a <random variable> $X$ with finite absolute $q$th moment, <Tonelli theorem> gives
$$
\mathbb E|X|^q=q\int_0^\infty t^{q-1}\mathbb P(|X|\geq t)\,dt.
$$

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 Incoming links: Tail integral formula for moments