Infinite variance can defeat square-root-n CLT scaling

First construct the symmetric Laplace density $l(x)=e^{-\mathrm{\mid }x\mathrm{\mid }}\mathrm{/}2$. On [0,R] the primitive of $e^{-x}$ is $-e^{-x}$, so [F1]–[F3], [F5] and [F20] give integral $1-e^{-R}$; the other half interval gives the same value by the primitive $e^x$. The bridge [F2] and MCT [F6] yield $\int l=1$. The elementary exponential series bound $e^R\ge 1+R$ makes $e^{-R}\to 0$. Thus [F8] defines a probability with density l. For every bounded measurable complex $h=u+iv$, apply [F9] to $u^{+},u^{-},v^{+},v^{-}$; these four functions are bounded by $\mathrm{\mid }h\mathrm{\mid }$, so their products with $l$ have finite integral. The real/imaginary definition [F21] then gives $\int hd(ld\lambda )=\int hld\lambda$. In particular the characteristic function of the constructed law is its density integral. For real t, componentwise differentiation using [F3]–[F5] gives the primitive $e^{(-1+it)x}\mathrm{/}(-1+it)$ on the positive half-line and $e^{(1+it)x}\mathrm{/}(1+it)$ on the negative half-line. Their nonzero denominators have modulus at least one. Taking integer R limits with DCT majorant $e^{-\mathrm{\mid }x\mathrm{\mid }}$ and [F19] gives ${\phi }_l(t)=\frac{1}{2}((1-it{)}^{-1}+(1+it{)}^{-1})=1\mathrm{/}(1+t^2)$.

By [F10]–[F11], ${\int }_{-R}^R(1+t^2{)}^{-1}dt=\arctan R-\arctan (-R)<\pi$. The integrands are continuous, so the bridge and MCT make ${\phi }_l$ Lebesgue integrable. [F12] supplies the continuous density $f(y)=(2\pi {)}^{-1}\int e^{-ity}\mathrm{/}(1+t^2)dt$ for the Laplace law. It equals l everywhere: both are continuous and have the same integral on every interval, whereas any nonzero value of f-l would, by continuity, have one strict sign bounded away from zero on a nonempty interval, contradicting that integral equality and [F13]. At y=0 this identity gives $\int (1+t^2{)}^{-1}dt=\pi$. Therefore [F8] makes c a probability density. At y=-u the same identity, multiplied by two, gives $\int e^{iux}c(x)dx=e^{-\mathrm{\mid }u\mathrm{\mid }}$. No contour integral is used.

For R>=1, ${\int }_1^R{x}^{2}c(x)dx\ge (R-1)\mathrm{/}(2\pi )$ because $x^2\mathrm{/}(1+x^2)\ge 1\mathrm{/}2$ for x>=1. Nonnegative integral monotonicity and [F13] give this lower bound, tending to infinity; [F9] identifies the second moment as infinite. By [F14]–[F15], AC realizes iid copies of the c law. By [F16], their normalized-sum characteristic function is $(e^{-\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\sqrt{n}}{)}^n=e^{-\sqrt{n}\mathrm{\mid }t\mathrm{\mid }}$. It tends to zero for t different from zero and stays one at zero.

If those row laws had any weak limit probability, [F17] would make its characteristic function equal to the pointwise limit just calculated. That function is discontinuous at zero, contradicting [F18]. Thus there is no weak probability limit, and in particular no normal limit. AC was used through the Lebesgue bridge, density inversion and independent-copy construction; the direct moment lower bound and discontinuity argument make no additional selections.