Symmetric Shifted Double Pareto Distribution

Statement¶

We use the symmetric shifted convention defined below; the name “double Pareto” is overloaded in the literature and can also refer to positive-valued size distributions with different left and right exponents. Let $Y$ have a Pareto Type I distribution with lower cutoff 1 and tail exponent $\alpha>0$ . Let $B$ be independent of $Y$ with $\mathbb P(B=1)=\mathbb P(B=-1)=1/2$ . The shifted symmetric double Pareto variable is

X = B(Y-1).

(1)

Its density on the real line is

f(x;\alpha)=\frac{\alpha}{2}(1+|x|)^{-(\alpha+1)},\qquad x\in\mathbb R.

(2)

The CDF is

F(x)= \begin{cases} \frac12(1-x)^{-\alpha}, & x<0,\\ 1-\frac12(1+x)^{-\alpha}, & x\ge 0. \end{cases}

(3)

For $x\ge0$ , each one-sided tail has probability

\mathbb P(X>x)=\frac12(1+x)^{-\alpha}, \qquad \mathbb P(X<-x)=\frac12(1+x)^{-\alpha}.

(4)

Consequently,

\mathbb P(|X|>x)=(1+x)^{-\alpha}.

(5)

More generally, a location-scale version $Z=\mu+sX$ with $s>0$ has density

f_Z(z)=\frac{\alpha}{2s} \left(1+\frac{|z-\mu|}{s}\right)^{-(\alpha+1)}.

(6)

The ordinary mean exists and equals 0 exactly when $\alpha>1$ . For $\alpha\le1$ , the positive and negative parts are both infinite, so symmetry gives a center but not an ordinary expectation. The variance is finite exactly when $\alpha>2$ , in which case

\operatorname{Var}(X)=\frac{2}{(\alpha-1)(\alpha-2)}.

(7)

The absolute moments unify these thresholds:

\mathbb E|X|^p =\frac{\Gamma(p+1)\Gamma(\alpha-p)}{\Gamma(\alpha)}, \qquad 0<p<\alpha.

(8)

Here $F$ and $f$ denote the CDF and density; $B$ is an independent sign. $\Gamma$ denotes the gamma function. Absolute moments diverge for $p\ge\alpha$ ; the displayed gamma expression only applies for $0<p<\alpha$ .

Shape and two-sided tail intuition¶

The ordinary Pareto distribution is one-sided: the rare extreme is always on the right. The double Pareto keeps the same power-law magnitude but gives the shock a sign. This makes it useful for toy return models where both large gains and large losses are possible, while preserving a transparent tail exponent.

The density is symmetric, with peak $f(0)=\alpha/2$ and

\frac{f(x)}{f(0)}=(1+|x|)^{-(\alpha+1)}.

(9)

Smaller $\alpha$ lowers the central density and puts more probability far from zero on both sides. The chance of a large positive value and the chance of an equally large negative value decay at the same power rate.

Derivation from signed Pareto magnitude¶

We derive the symmetric shifted construction used here from the signed Pareto magnitude $X=B(Y-1)$ . The tail, density, and CDF formulas follow directly from symmetry. The moment thresholds use the same Pareto moment logic as Pareto Moment Existence.

For $x\ge0$ ,

\mathbb P(X>x) =\mathbb P(B=1, Y-1>x) =\frac12\mathbb P(Y>1+x) =\frac12(1+x)^{-\alpha}.

(10)

The negative tail is identical by symmetry. Differentiating the CDF on either side gives the density. The expectation is zero only when the first absolute moment exists. When $\alpha\le1$ , $\mathbb E[X^+]=\mathbb E[X^-]=\infty$ , so the ordinary mean is undefined even though the distribution is symmetric. Since $X^2=(Y-1)^2$ ,

\mathbb E[X^2] =\mathbb E[Y^2]-2\mathbb E[Y]+1 =\frac{\alpha}{\alpha-2}-2\frac{\alpha}{\alpha-1}+1 =\frac{2}{(\alpha-1)(\alpha-2)}

(11)

for $\alpha>2$ . For $1<\alpha\le2$ , the mean exists but the variance is infinite. For $\alpha\le1$ , the ordinary mean, and hence variance about that mean, is undefined; the second raw moment is still infinite.

More generally, integration of the absolute-value density gives the beta integral

\mathbb E|X|^p=\alpha\int_0^\infty t^p(1+t)^{-\alpha-1}\,dt =\frac{\Gamma(p+1)\Gamma(\alpha-p)}{\Gamma(\alpha)}, \qquad 0<p<\alpha.

(12)

Two-sided tail calculations¶

The exact two-sided survival $(1+x)^{-\alpha}$ is regularly varying with index $-\alpha$ . At $\alpha=1.5$ , its values at $x=0,3,8$ are respectively 1, $1/8$ , and $1/27$ . Each one-sided tail has half of that probability.

For a fixed multiplier $t>0$ ,

\frac{\mathbb P(|X|>tx)}{\mathbb P(|X|>x)} =\left(\frac{1+tx}{1+x}\right)^{-\alpha}\to t^{-\alpha}.

(13)

The shift means this is a limiting power ratio, rather than the exact threshold scaling of an unshifted Pareto law.

At $\alpha=3$ , direct evaluation gives the following exact values.

$x$	Density $f(x)$	CDF $F(x)$
-4	$3/1250$	$1/250$
0	$3/2$	$1/2$
4	$3/1250$	$249/250$

The mean is zero and the variance is $2/[(3-1)(3-2)]=1$ . These values are formula evaluations, without simulation or an imported distribution package. For max-to-sum diagnostics, use $|X_i|$ or another explicitly defined nonnegative quantity; signed sums can cancel.

Caveats¶

We describe a symmetric signed construction whose magnitude $|X|$ has a Lomax law, not every distribution called “double Pareto” in the literature.
For $\alpha\le1$ , even the absolute first moment is infinite. Symmetry can make a formal location or principal-value calculation look harmless while the ordinary expectation is undefined and absolute exposure remains uncontrolled.
The distribution is a pedagogical model. Empirical returns usually need skew, truncation, volatility clustering, dependence, or threshold modeling before a tail fit is credible.

References¶

Resnick, Heavy-Tail Phenomena 2007.

Backlinks¶

Depends on: Pareto Distribution and Regular Variation.
Used by: two-sided tail examples (planned).

Source and adaptation¶

Adapted from incerto-wiki, content/concepts/distributions/double-pareto.md, revision 9717c9c (2026-09-13 Batch 2 import). Copyright (c) 2023 xshi19. Licensed under MIT. Links, notation, and qualifications were adapted for this site; executable figures and simulations were replaced with static calculations. No upstream execution or formal-proof verification is claimed for this adaptation.

References¶

Resnick, S. I. (2007). Heavy-Tail Phenomena: Probabilistic and Statistical Modeling. In Springer Series in Operations Research and Financial Engineering. Springer New York. 10.1007/978-0-387-45024-7