A polynomial-tail law with finite matrix variance

By [F1], $P$ is a measure, and its total mass is $\zeta (6{)}^{-1}{\sum }_{k=1}^{\mathrm{\infty }}{k}^{-6}=1$ by [F5]. Thus [F2] makes $P$ a probability measure. The identity map is the real random variable specified by [F3], and $P(X=\pm k)=1\mathrm{/}(2\zeta (6)k^6)$ for each $k\ge 1$.

For $q=2$ and $q=4$, the truncations $Y_N=\mathrm{\mid }X{\mathrm{\mid }}^q{\mathbf{1}}_{\{\mathrm{\mid }X\mathrm{\mid }\le N\}}$ are nonnegative simple functions increasing to $\mathrm{\mid }X{\mathrm{\mid }}^q$. By [F4] and the finite atom masses, $\mathbb{E}\mathrm{\mid }X{\mathrm{\mid }}^q={\lim }_{N\to \mathrm{\infty }}\mathbb{E}{Y}_N=\frac{1}{\zeta (6)}{\sum }_{k=1}^{\mathrm{\infty }}{k}^{q-6}=\frac{\zeta (6-q)}{\zeta (6)}<\mathrm{\infty },$ where convergence is [F5]. In particular $X$ is integrable by [F7], and its symmetric atom masses give $\mathbb{E}X=0$.

By [F6], the covariance is $\mathrm{\Sigma }=\zeta (4)\mathrm{/}\zeta (6)>0$, while the scalar matrix variance is ${\sigma }_0^2=\zeta (2)\mathrm{/}\zeta (6)<\mathrm{\infty }$. Therefore $d=\frac{{\sigma }_0^2}{{\mathrm{\Sigma }}^2}=\frac{\zeta (2)\zeta (6)}{\zeta (4{)}^2}$ is finite and positive.

Fix $t>0$. The simple truncations $W_N=e^{t\mathrm{\mid }X\mathrm{\mid }}{\mathbf{1}}_{\{\mathrm{\mid }X\mathrm{\mid }\le N\}}$ increase to $e^{t\mathrm{\mid }X\mathrm{\mid }}$, so [F4] gives $\mathbb{E}{e}^{t\mathrm{\mid }X\mathrm{\mid }}=\frac{1}{\zeta (6)}{\sum }_{k=1}^{\mathrm{\infty }}{e}^{tk}{k}^{-6}=+\mathrm{\infty }\mathrm{.}$ Indeed the ratio of successive summands is $e^t(k\mathrm{/}(k+1){)}^6\to e^t>1$, so the summands do not tend to zero and the series diverges. The law has polynomially decaying atom masses, finite matrix variance, and no finite positive exponential moment.