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**hide

#### Best Answer

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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