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.


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 $t\gt 0$,

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

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

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

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 $1$:

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

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 $x\lt u_i$. Consequently the sequence of $F_i(x)$ is a nondecreasing, bounded, pointwise sequence of distribution functions, implying its limit

$$F_\infty = \lim_{i\to\infty} F_i$$

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.

Figure 1: An altered survival function

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 $j\ge i$, will reduce $2^{i-1}$ to $1$, but no lower.

Figure 2: A cut-down density function

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.

Figure 3: plot of G(1+x)/G(x)

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 $x\in\mathbb{R}$. Its variance will remain finite, so it can be standardized to the desired distribution.

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  • 2
    $\begingroup$ Brilliantly explained. You offered not just an example but also the justification for it. The clarity of the explanation allowed me to understand (almost) the whole of it. I will practice it in some numerical examples. $\endgroup$ – toliveira Sep 1 '15 at 16:56

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