# Solved – Example of heavy-tailed distribution that is not long-tailed

From readings about heavy-, and long-tailed distributions, I understood that all long-tailed distributions are heavy-tailed, but not all heavy-tailed distributions are long-tailed.

Could somebody please provide an example of:

• a continuous, symmetric, zero-mean density function that is long-tailed
• a continuous, symmetric, zero-mean density function that is heavy-tailed but not long-tailed

so I can better understand the meaning of their definitions?

It would be even better if both could have a unit variance.

Contents

The two definitions are close, but not exactly the same. One difference lies in the need for the survival ratio to have a limit.

For most of this answer I will ignore the criteria for the distribution to be continuous, symmetric, and of finite variance, because these are easy to accomplish once we have found any finite-variance heavy-tailed distribution that is not long-tailed.

A distribution \$F\$ is heavy-tailed when for any \$tgt 0\$,

\$\$int_mathbb{R} e^{t x} dF(x) = infty.tag{1}\$\$

A distribution with survival function \$G_F = 1-F\$ is long-tailed when

\$\$lim_{xto infty} frac{G_F(x+1)}{G_F(x)} = 1.tag{2}\$\$

Long-tailed distributions are heavy. Furthermore, because \$G\$ is nonincreasing, the limit of the ratio \$(2)\$ cannot exceed \$1\$. If it exists and is less than \$1\$, then \$G\$ is decreasing exponentially–and that will allow the integral \$(1)\$ to converge.

The only way to exhibit a heavy-tailed distribution that is not long-tailed, then, is to modify a long-tailed distribution so that \$(1)\$ continues to hold while \$(2)\$ is violated. It's easy to screw up a limit: change it in infinitely many places that diverge to infinity. That will take some doing with \$F\$, though, which must remain increasing and cadlag. One way is to introduce some upward jumps in \$F\$, which will make \$G\$ jump downwards, lowering the ratio \$G_F(x+1)/G_F(x)\$. To this end, let's define a transformation \$T_u\$ that turns \$F\$ into another valid distribution function while creating a sudden jump at the value \$u\$, say a jump halfway from \$F(u)\$ to \$1\$:

\$\$T_u[F](x) = begin{cases} F(x) & u<x \ frac{1}{2} (1-F(x))+F(x) & ugeq x end{cases}\$\$

This alters no basic property of \$F\$: \$T_u[F]\$ is still a distribution function.

The effect on \$G_F\$ is to make it drop by a factor of \$1/2\$ at \$u\$. Therefore, since \$G\$ is non-decreasing, then whenever \$u-1 le x lt u\$,

\$\$frac{G_{T_u[F]}(x + 1)}{G_{T_u[F]}(x )} le frac{1}{2}.\$\$

If we pick an increasing and diverging sequence of \$u_i\$, \$i=1, 2, ldots\$, and apply each \$T_{u_i}\$ in succession, it determines a sequence of distributions \$F_i\$ with \$F_0=F\$ and

\$\$F_{i+1} = T_{u_i}[F_i]\$\$

for \$i ge 1\$. After the \$i^text{th}\$ step, \$F_i(x), F_{i+1}(x), ldots\$ all remain the same for \$xlt u_i\$. Consequently the sequence of \$F_i(x)\$ is a nondecreasing, bounded, pointwise sequence of distribution functions, implying its limit

\$\$F_infty = lim_{itoinfty} F_i\$\$

is a distribution function. By construction, it is not long-tailed because there are infinitely many points at which its survival ratio \$G_{F_infty}(x+1)/G_{F_infty}(x))\$ drops to \$1/2\$ or below, showing it cannot have \$1\$ as a limit. This plot shows a survival function \$G(x) = x^{-1/5}\$ that has been cut down in this manner at points \$u_1 approx 12.9, u_2 approx 40.5, u_3 approx 101.6, ldots.\$ Note the logarithmic vertical axis.

The hope is to be able to choose \$(u_i)\$ so that \$F_infty\$ remains heavy-tailed. We know, because \$F\$ is heavy-tailed, that there are numbers \$0 = u_0 lt u_1 lt u_2 lt cdots lt u_n cdots\$ for which

\$\$int_{u_{i-1}}^{u_i} e^{x/i} dF(x) ge 2^{i-1}\$\$

for every \$i ge 1\$. The reason for the \$2^{i-1}\$ on the right is that the probabilities assigned by \$F\$ to values up to \$u_i\$ have been successively cut in half \$i-1\$ times. That procedure, when \$dF(x)\$ is replaced by \$dF_{j}(x)\$ for any \$jge i\$, will reduce \$2^{i-1}\$ to \$1\$, but no lower. This is a plot of \$x f(x)\$ for densities \$f\$ corresponding to the previous survival function and its "cut down" version. The areas under this curve contribute to the expectation. The area from \$1\$ to \$u_1\$ is \$1\$; the area from \$u_1\$ to \$u_2\$ is \$2\$, which when cut down (to the lower blue portion) becomes an area of \$1\$; the area from \$u_2\$ to \$u_3\$ is \$4\$, which when cut down becomes an area of \$1\$, and so on. Thus, the area under each successive "stair step" to the right is \$1\$.

Let us pick such a sequence \$(u_i)\$ to define \$F_infty\$. We can check that it remains heavy-tailed by picking \$t=1/n\$ for some whole number \$n\$ and applying the construction:

\$\$eqalign{ int_mathbb{R} e^{t x} dF_infty(x) &=int_mathbb{R} e^{x/n} dF_infty(x) \ &= sum_{i=1}^infty int_{u_{i-1}}^{u_i} e^{x/n} dF_infty(x) \ &ge sum_{i=n+1}^infty int_{u_{i-1}}^{u_i} e^{x/n} dF_infty(x) \ &ge sum_{i=n+1}^infty int_{u_{i-1}}^{u_i} e^{x/i} dF_infty(x) \ &= sum_{i=n+1}^infty int_{u_{i-1}}^{u_i} e^{x/i} dF_i(x) \ &ge sum_{i=n+1}^infty 1, }\$\$

which still diverges. Since \$t\$ is arbitrarily small, this demonstrates that \$F_infty\$ remains heavy-tailed, even though its long-tailed property has been destroyed. This is a plot of the survival ratio \$G(x+1)/G(x)\$ for the cut down distribution. Like the ratio of the original \$G\$, it tends toward an upper accumulation value of \$1\$–but for unit-width intervals terminating at the \$u_i\$, the ratio suddenly drops to only half of what it originally was. These drops, although becoming less and less frequent as \$x\$ increases, occur infinitely often and therefore prevent the ratio from approaching \$1\$ in the limit.

If you would like a continuous, symmetric, zero-mean, unit-variance example, begin with a finite-variance long-tailed distribution. \$F(x) = 1 – x^{-p}\$ (for \$x gt 0\$) will do, provided \$p gt 1\$; so would a Student t distribution for any degrees of freedom exceeding \$2\$. The moments of \$F_infty\$ cannot exceed those of \$F\$, whence it too has finite variance. "Mollify" it via convolution with a nice smooth distribution, such as a Gaussian: this will make it continuous but will not destroy its heavy tail (obviously) nor the absence of a long tail (not quite as obvious, but it becomes obvious if you change the Gaussian to, say, a Beta distribution whose support is compact).

Symmetrize the result–which I will still call \$F_infty\$–by defining

\$\$F_s(x) = frac{1}{2}left(1 + text{sgn}(x) F_infty(|x|)right)\$\$

for all \$xinmathbb{R}\$. Its variance will remain finite, so it can be standardized to the desired distribution.

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