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Body, Shoulders, and Tails

Statement

For a centered normal density parameterized by variance v>0v>0

gv(x)=12πvexp⁡(−x22v),g_v(x)= \frac{1}{\sqrt{2\pi v}} \exp\left(-\frac{x^2}{2v}\right),

the body-shoulder-tail diagnostic asks where a small, mean-preserving randomization of vv adds density and where it removes density. A local version is the sign of

∂2gv(x)∂v2.\frac{\partial^2 g_v(x)}{\partial v^2}.

For the standard normal density, this second derivative changes sign at

±3−6,±3+6.\pm\sqrt{3-\sqrt6}, \qquad \pm\sqrt{3+\sqrt6}.

These four points split the line into left tail, left shoulder, peak, right shoulder, and right tail regions. Positive variance curvature means a local variance mixture raises density there; negative variance curvature means it lowers density there.

The notation gvg_v denotes the density at fixed variance vv; a mixture averages these densities over a positive random variance YY. The Normix normal-mixtures note supplies the complementary conditional construction. See the shared notation canon.

Examples

Diagnostic intuition

If a Gaussian variance is randomized while keeping the center fixed, the resulting mixture does not simply “spread out” everywhere. It tends to add mass near the peak and in the far tails, while taking mass from the shoulders. The shoulders are the moderate-deviation region that looks ordinary under a single scale but is depleted when variance uncertainty is introduced.

This diagnoses the geometry of a specified population density, not its tail index. It explains why stochastic volatility can create both a sharper center and fatter-looking tails without producing a Pareto tail. Choosing a threshold for tail estimation remains the separate judgment handled by Tail Threshold Selection.

Derivation of the variance-curvature boundaries

We derive the standard-normal variance-curvature boundaries by differentiating the density with respect to variance and solving the resulting quadratic in x2x^2. The Taylor expansion explains the local mixture interpretation. The comparison with scale perturbation below uses a different derivative; none of these boundaries is a rule for selecting a Pareto threshold.

For the normal density with variance parameter vv, write

gv(x)=12πvexp⁡(−x22v).g_v(x)= \frac{1}{\sqrt{2\pi v}} \exp\left(-\frac{x^2}{2v}\right).

Set y=x/vy=x/\sqrt v. A direct differentiation gives

∂gv(x)∂v=gv(x)2v(y2−1),\frac{\partial g_v(x)}{\partial v} = \frac{g_v(x)}{2v}(y^2-1),

and a second differentiation gives

∂2gv(x)∂v2=gv(x)4v2(y4−6y2+3).\frac{\partial^2 g_v(x)}{\partial v^2} = \frac{g_v(x)}{4v^2}(y^4-6y^2+3).

At v=1v=1, the sign changes where

x4−6x2+3=0.x^4-6x^2+3=0.

Solving the quadratic in x2x^2 gives

x2=3±6.x^2=3\pm\sqrt6.

This is a local Taylor diagnostic. A sufficient small-perturbation model is

Yτ=v0+τH,v0>0,E[H]=0,E[H2]=1,Y_\tau=v_0+\tau H,\qquad v_0>0,\quad \mathbb E[H]=0,\quad \mathbb E[H^2]=1,

with bounded local perturbation HH and τ\tau small enough that Yτ>0Y_\tau>0. For each fixed xx, Taylor’s theorem, with bounded third derivative near v0v_0, gives

E[gYτ(x)]−gv0(x)=τ22∂2gv(x)∂v2∣v=v0+O(τ3).\mathbb E[g_{Y_\tau}(x)]-g_{v_0}(x) =\frac{\tau^2}{2} \frac{\partial^2g_v(x)}{\partial v^2}\bigg|_{v=v_0} +O(\tau^3).

Thus the curvature sign determines the leading local density change away from its zeros. Small variance of a perturbation alone would not justify this remainder without control of the perturbation family. The expansion is pointwise in xx, not uniform arbitrarily far into the tails.

Static region and mixture calculations

At v=1v=1, put a=3−6≈0.742a=\sqrt{3-\sqrt6}\approx0.742 and b=3+6≈2.334b=\sqrt{3+\sqrt6}\approx2.334.

RegionVariance curvature of densityLeading change under small mean-preserving variance mixing
∣x∣<a\lvert x\rvert<a (peak)PositiveDensity increases
a<∣x∣<ba<\lvert x\rvert<b (shoulders)NegativeDensity decreases
∣x∣>b\lvert x\rvert>b (tails)PositiveDensity increases

For a concrete finite mixture, define

fmix(x)=12g0.2(x)+12g1.8(x).f_{\mathrm{mix}}(x)=\tfrac12 g_{0.2}(x)+\tfrac12 g_{1.8}(x).

Its mean is zero and its variance is one. The ratio to the unit normal is

fmix(x)g1(x)=52e−2x2+56e2x2/9.\frac{f_{\mathrm{mix}}(x)}{g_1(x)} =\frac{\sqrt5}{2}e^{-2x^2} +\frac{\sqrt5}{6}e^{2x^2/9}.

At x=0,1,3x=0,1,3, the ratio is approximately 1.491, 0.617, and 2.754. This exhibits increased center and far-tail density with depleted shoulders. The exact crossings solve the ratio-equals-one equation; they need not be the local curvature boundaries a,ba,b. This finite Gaussian mixture has a finite moment generating function at every real argument and is not regularly varying, despite having more far-tail mass than the unit normal.

Density, survival, and scale are different diagnostics

For x>0x>0, the two-sided normal survival at variance vv is Tv(x)=2Φˉ(x/v)T_v(x)=2\bar\Phi(x/\sqrt v), where Φˉ\bar\Phi is standard normal survival and ϕ\phi is its density. Differentiation gives

∂2Tv(x)∂v2=xϕ(x/v)2v5/2(x2v−3).\frac{\partial^2 T_v(x)}{\partial v^2} =\frac{x\phi(x/\sqrt v)}{2v^{5/2}}\left(\frac{x^2}{v}-3\right).

Its positive-threshold curvature boundary is x=3vx=\sqrt{3v}, about 1.732 when v=1v=1. It differs from both density boundaries.

For scale perturbation, write hσ(x)=gσ2(x)h_\sigma(x)=g_{\sigma^2}(x) and z=x/σz=x/\sigma. Then

∂2hσ(x)∂σ2=hσ(x)σ2(z4−5z2+2).\frac{\partial^2h_\sigma(x)}{\partial\sigma^2} =\frac{h_\sigma(x)}{\sigma^2}(z^4-5z^2+2).

At σ=1\sigma=1, its positive zeros are (5−17)/2≈0.662\sqrt{(5-\sqrt{17})/2}\approx0.662 and (5+17)/2≈2.136\sqrt{(5+\sqrt{17})/2}\approx2.136. Mean-preserving scale mixing is a different operation from mean-preserving variance mixing. Student scale-curvature comparisons depend on degrees of freedom, as derived below.

Student scale curvature

The scale calculation also has a closed form for a Student density. Write hs(x)=Cνs−1(1+x2/(νs2))−(ν+1)/2h_s(x)=C_\nu s^{-1}(1+x^2/(\nu s^2))^{-(\nu+1)/2}, where s>0s>0 is scale, ν>1\nu>1 is degrees of freedom, and CνC_\nu normalizes the density. Set z=x/sz=x/s and t=z2/νt=z^2/\nu. Logarithmic differentiation gives

s ∂slog⁡hs=νt−11+t,s2∂s2hshs=ν(ν−1)t2−(5ν+1)t+2(1+t)2.s\,\partial_s\log h_s=\frac{\nu t-1}{1+t},\qquad \frac{s^2\partial_s^2h_s}{h_s} =\frac{\nu(\nu-1)t^2-(5\nu+1)t+2}{(1+t)^2}.

For the second formula, differentiate the first, using ∂st=−2t/s\partial_st=-2t/s, and add the square of the log derivative. The numerator is a quadratic with two positive roots when ν>1\nu>1. Solving in z2z^2 gives the four density-curvature boundaries

x=±s5ν+1±(ν+1)(17ν+1)2(ν−1).x=\pm s\sqrt{ \frac{5\nu+1\pm\sqrt{(\nu+1)(17\nu+1)}}{2(\nu-1)}}.

The leading coefficient and the value at zero are positive, so curvature is positive inside the inner pair and beyond the outer pair, and negative between them. At ν=3,s=1\nu=3,s=1, the positive boundaries are 4−13≈0.628\sqrt{4-\sqrt{13}}\approx0.628 and 4+13≈2.758\sqrt{4+\sqrt{13}}\approx2.758. The standard deviation is 3\sqrt3; these numbers are in scale units. For ν≤2\nu\le2, the variance is not finite, although the four-boundary formula still applies when 1<ν≤21<\nu\le2. At ν=1\nu=1 the quadratic loses its leading term, so that formula’s domain must not be extended through the zero denominator.

Dividing numerator and denominator of the root formula by ν\nu shows that its two positive roots tend to the Gaussian scale-curvature roots as ν→∞\nu\to\infty. This is an algebraic limit, not a fitted threshold rule. The gamma-precision construction explains the Student tail exponent and moment conditions separately.

Curvature and exact density differences

Two stacked plots compare the Gaussian variance curvature divided by two with exact symmetric mixture density differences divided by squared perturbation size, and show different variance, scale and Student scale boundaries.

The upper panel compares the pointwise quadratic approximation with exact mixtures at perturbations 0.1 and 0.6. Vertical lines mark the variance curvature zeros; finite-mixture crossings need not stay on those lines. The lower panel places normal variance, normal scale, and Student-three scale boundaries on the same dimensionless axis. Student scale is 1, so its standard deviation is the square root of 3.

Use normal_peak_shoulder_tail, normal_variance_peak_shoulder_tail, t_peak_shoulder_tail.

from xmath.incerto.stats import (
    normal_peak_shoulder_tail, normal_variance_peak_shoulder_tail,
    t_peak_shoulder_tail,
)

for label, values in (("Normal, scale", normal_peak_shoulder_tail()),
                      ("Normal, variance", normal_variance_peak_shoulder_tail()),
                      ("Student 3, scale", t_peak_shoulder_tail(3))):
    print(f"{label}: [" + ", ".join(f"{v:.4g}" for v in values) + "]")

Output

Normal, scale: [-2.136, -0.6622, 0.6622, 2.136]
Normal, variance: [-2.334, -0.742, 0.742, 2.334]
Student 3, scale: [-2.758, -0.6281, 0.6281, 2.758]

These are local perturbation boundaries at unit scale, not fitted tail thresholds.

Figure regeneration: uv run python demos/incerto/finite_moments.py. These are exact density calculations on 801 fixed points, with no random sampling. The script reports the finite-mixture crossings separately from the local boundaries.

Let YY be a positive mixing variance with EY<∞\mathbb EY<\infty. Jensen’s inequality for survival needs convexity throughout the mixing interval. If 0<Y≤vmax⁡<∞0<Y\le v_{\max}<\infty and x2≥3vmax⁡x^2\ge3v_{\max}, the displayed Tv′′(x)T_v''(x) is nonnegative on that interval, and ETY(x)≥TEY(x)\mathbb E T_Y(x)\ge T_{\mathbb EY}(x). If instead Y≥vmin⁡>0Y\ge v_{\min}>0 and x2≤3vmin⁡x^2\le3v_{\min}, its curvature is nonpositive and the inequality reverses. These sufficient conditions show why a local tail-curvature statement cannot justify global Jensen claims at every threshold under unrestricted mixing.

Caveats

References

References
  1. Taleb, N. N. (2025). Statistical Consequences of Fat Tails (3rd ed.). https://arxiv.org/abs/2001.10488v4