Finite variance does not imply a sub-Gaussian tail

Let $S={\sum }_{k=1}^{\mathrm{\infty }}{k}^{-4}$. Its partial sums are increasing, and for $k\ge 2$, $k^{-4}\le k^{-2}\le 1\mathrm{/}[k(k-1)]=1\mathrm{/}(k-1)-1\mathrm{/}k$; hence $1\le S\le 2$ and $S$ is finite by [F4]. Set $c=(2S{)}^{-1}>0$.

Enumerate $\mathrm{\Omega }=\{(k,s):k\ge 1,s\in \{-1,1\}\}$ by ${\omega }_{2k-1}=(k,1)$, ${\omega }_{2k}=(k,-1)$, and assign each atom weight $w_{2k-1}=w_{2k}=c{k}^{-4}$. Define $\mathbb{P}(A)={\sum }_{j\ge 1}{w}_j{\mathbf{1}}_A({\omega }_j)$ for $A\subseteq \mathrm{\Omega }$. The weights are nonnegative and sum to $2cS=1$, so all subseries converge by [F4]. For pairwise disjoint $A_r$, put $a_{rj}=w_j{\mathbf{1}}_{A_r}({\omega }_j)$ and $q_j={\sum }_r{a}_{rj}$; at most one term in each $q_j$ is nonzero. For every finite atom prefix $J$, only finitely many $A_r$ meet that prefix, so finite regrouping gives ${\sum }_{j=1}^J{q}_j={\sum }_r{\sum }_{j=1}^J{a}_{rj}\le {\sum }_r\mathbb{P}(A_r)$. Taking the supremum over $J$ yields $\mathbb{P}({\bigcup }_r{A}_r)\le {\sum }_r\mathbb{P}(A_r)$. Conversely, for each finite event prefix $R$ and each $\delta >0$, convergence of the finitely many subseries lets us choose one atom prefix $J$ for which ${\sum }_{r=1}^R\mathbb{P}(A_r)-\delta \le {\sum }_{r=1}^R{\sum }_{j=1}^J{a}_{rj}\le {\sum }_{j=1}^J{q}_j\le \mathbb{P}({\bigcup }_r{A}_r)$. Letting $\delta \downarrow 0$ and then taking the supremum over $R$ proves the reverse inequality. Thus [F1] makes $\mathbb{P}$ a probability measure.

Define $X(k,s)=sk$ on $(\mathrm{\Omega },\mathcal{P}(\mathrm{\Omega }),\mathbb{P})$. It is a real random variable by [F2], since every preimage is a subset of the full-power-set sample space.

By [F5] applied to finite atom prefixes, $\mathbb{E}\mathrm{\mid }X\mathrm{\mid }=2c{\sum }_{k\ge 1}{k}^{-3}$ and $\mathbb{E}{X}^2=2c{\sum }_{k\ge 1}{k}^{-2}$. For $k\ge 2$, $k^{-3}\le k^{-2}\le 1\mathrm{/}[k(k-1)]$, whose partial sums telescope to at most $1$; hence both moments are finite, so $X$ is integrable and square-integrable. The two atoms at each $k$ have equal mass and opposite values; since $\mathbb{E}\mathrm{\mid }X\mathrm{\mid }<\mathrm{\infty }$, their absolutely convergent contributions cancel and $\mathbb{E}X=0$. By [F3], $X$ is centered with finite variance $\mathrm{Var}(X)=\mathbb{E}{X}^2$.

For every integer $n\ge 1$, the atoms $(n,1)$ and $(n,-1)$ give $\mathbb{P}(\mathrm{\mid }X\mathrm{\mid }\ge n)\ge 2c{n}^{-4}>0$.

Suppose for contradiction that $X$ has a finite sub-Gaussian proxy ${\sigma }^2$. If ${\sigma }^2=0$, [F6] contradicts the positive tail in step 2.1.

If ${\sigma }^2>0$, [F6] gives $2c{n}^{-4}\le \mathbb{P}(\mathrm{\mid }X\mathrm{\mid }\ge n)\le 2e^{-n^2\mathrm{/}(2{\sigma }^2)}$. By [F8], $n^4{e}^{-n^2\mathrm{/}(2{\sigma }^2)}=(n^2{)}^2\mathrm{/}{e}^{n^2\mathrm{/}(2{\sigma }^2)}$, which tends to $0$ by [F7] with $x=n^2$, $m=2$, and $a=1\mathrm{/}(2{\sigma }^2)$. It is eventually smaller than $c$, so the upper tail is eventually smaller than $2c{n}^{-4}$, contradicting step 2.1. Thus this centered finite-variance variable is not sub-Gaussian for any finite proxy.