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

Its density on the real line is

The CDF is

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

Consequently,

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

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

The absolute moments unify these thresholds:

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

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.

Regenerate the card with uv run python demos/incerto/distribution_cards.py; the double_pareto example below uses the same distribution convention.

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 $x\ge0$,

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 $X^2=(Y-1)^2$,

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.

Use double_pareto.

from xmath.incerto.distributions import double_pareto

for alpha in (0.5, 1, 1.5, 2, 3):
    mean, variance = double_pareto(alpha).stats(moments="mv")
    print(f"alpha={alpha:.4g}: mean={mean:.4g}, variance={variance:.4g}")

Output

alpha=0.5: mean=nan, variance=inf
alpha=1: mean=nan, variance=inf
alpha=1.5: mean=0, variance=inf
alpha=2: mean=0, variance=inf
alpha=3: mean=0, variance=1

For $0<\alpha\le1$, inf reflects the infinite raw second moment, not variance about an undefined mean.

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

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 $t>0$,

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.

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.

Matching a Student tail or a variance¶

SCoFT v4 §5.3, p. 101 Taleb, 2025, compares a symmetric shifted power law with a Student law. We keep the normalized density $\alpha/2$ above on each side and derive both matching scales. Equal exponents do not imply equal moments or equal leading tail constants.

Let $T_\nu$ have the unit-scale Student density, with $\nu=\alpha>0$, and let $Z=sT_\alpha$, $s>0$. Its right tail satisfies

We prove the constant from the density. Factoring $x^2/(\alpha s^2)$ out of $1+x^2/(\alpha s^2)$ in the Student density gives $f_Z(x)\sim\alpha C_\alpha s^\alpha x^{-\alpha-1}$. For each $\varepsilon>0$, this density lies between $1\pm\varepsilon$ times its leading power beyond a fixed threshold. Integrating those bounds from $x$ to infinity and letting $\varepsilon\downarrow0$ gives the displayed survival asymptotic. For positive orders $p$, the same comparison proves $\mathbb E|Z|^p<\infty$ exactly when $p<\alpha$; the density is bounded near zero.

The double-Pareto right-tail constant is $1/2$ at scale one. Matching it requires

For $\alpha>2$, the Student variance is $s^2\alpha/(\alpha-2)$ (the beta-integral moment calculation is also given in the mixture note). Equating that to $2/[(\alpha-1)(\alpha-2)]$ instead gives

At $\alpha=3$, $C_3=2\sqrt3/\pi$. Variance matching uses $s_{\rm var}=1/\sqrt3$ and gives Student right-tail constant $2/(3\pi)$, versus $1/2$ for double Pareto. Their asymptotic survival ratio is $4/(3\pi)\approx0.424413$ even though both variances are one. Tail matching uses $s_{\rm tail}=(\pi/(4\sqrt3))^{1/3}$ and gives Student variance about 1.770683, versus one for double Pareto. Matching tails thus leaves a substantial variance difference in this example.

Use double_pareto, student_double_pareto_scales.

from xmath.incerto.distributions import double_pareto
from xmath.incerto.power_laws import student_double_pareto_scales

law = double_pareto(3)
mean, variance = law.stats(moments="mv")
print(f"Mean: {mean:.4g}; variance: {variance:.4g}")
print("Survival at 1, 10: [" + ", ".join(f"{v:.3e}" for v in law.sf([1, 10])) + "]")
print("Student matching:")
for name, value in student_double_pareto_scales(3).items():
    print(f"  {name}: {value:.4g}")

Output

Mean: 0; variance: 1
Survival at 1, 10: [6.250e-02, 3.757e-04]
Student matching:
  student_tail_constant: 1.103
  tail_scale: 0.7683
  variance_scale: 0.5774
  tail_matched_variance: 1.771

Figure regeneration: uv run python demos/incerto/power_laws.py. The figure uses 501 density points on $[-4,4]$ and 241 thresholds from 1 to 10⁴ for each tail panel. All values are dimensionless, exact-model evaluations.

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¶

• Taleb (2025), §5.3, p. 101: comparison context. The normalization, tail constants and matching formulas above are derived for our explicit convention.

• Resnick, Heavy-Tail Phenomena 2007.

Backlinks¶

• Depends on: Pareto Distribution and Regular Variation.

• Used by: Chapter 5 and its Student comparison.