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Edgeworth and Saddlepoint Methods · Tutorial 975 of 994

Second-Order Kurtosis and Skewness Corrections

See how fourth-cumulant and squared-skewness terms extend the first-order Edgeworth density approximation, and calculate the corrections for explicit standardized sums.

Advanced 11 min read

What You'll Learn

  • Identify the fourth cumulant and distinguish it from ordinary kurtosis.
  • Derive the order-\(n^{-1}\) Fourier expansion for a standardized iid sum.
  • Convert Fourier powers into Hermite-polynomial density corrections.
  • Calculate second-order corrections for normal, exponential, and uniform summands.
  • Recognize when an Edgeworth density is a formal approximation rather than a guaranteed nonnegative density.

Why a Second-Order Correction Has Two Parts

The central limit theorem describes the leading shape of a standardized sum: a standard normal density. When the summands are not normal, their skewness produces a correction of order \(n^{-1/2}\). At the next order, \(n^{-1}\), two effects appear: the fourth cumulant of one summand and the square of its third cumulant. The second effect arises because exponentiating the cumulant expansion creates a product of two first-order skewness terms.

This tutorial derives those density terms and calculates them for specified sums. The approximation concerns densities, not cumulative distribution functions. It is most useful as a local correction near the central part of a distribution; it should not be treated as an exact density or automatically trusted far into the tails.

Definition: Let \(X_1,\ldots,X_n\) be independent and identically distributed random variables with \(E[X_i]=0\), \(\operatorname{Var}(X_i)=1\), and finite fourth moment. Write \(\kappa_3=E[X_i^3]\) for their third cumulant (the skewness) and \(\kappa_4=E[X_i^4]-3\) for their fourth cumulant (the excess kurtosis). The standardized sum is \(S_n=n^{-1/2}\sum_{i=1}^n X_i\).

The distinction between \(\kappa_4\) and the fourth standardized moment matters: the latter is \(E[X_i^4]\), while the fourth cumulant subtracts \(3\). For a standard normal variable, \(E[X_i^4]=3\), so \(\kappa_4=0\). We use the probabilists' Hermite polynomials:

$$ H_3(x)=x^3-3x,\qquad H_4(x)=x^4-6x^2+3,\qquad H_6(x)=x^6-15x^4+45x^2-15. $$

Deriving the Density Terms

Let \(\varphi_n(t)=E[e^{itS_n}]\) be the characteristic function of \(S_n\). For a mean-zero, variance-one summand with finite fourth moment, the logarithm of its characteristic function has, near zero, the cumulant expansion

$$ \log E[e^{iuX_1}] =-\frac{u^2}{2} +\frac{\kappa_3(iu)^3}{6} +\frac{\kappa_4(iu)^4}{24} +o(u^4). $$

Independence gives \(\varphi_n(t)=\{E[e^{itX_1/\sqrt n}]\}^n\). Substituting \(u=t/\sqrt n\) and multiplying the logarithm by \(n\), for each fixed \(t\) we obtain

$$ \log\varphi_n(t) =-\frac{t^2}{2} +\frac{\kappa_3(it)^3}{6\sqrt n} +\frac{\kappa_4(it)^4}{24n} +o(n^{-1}). $$

To recover the characteristic function rather than its logarithm, set \(a=\kappa_3(it)^3/6\) and \(b=\kappa_4(it)^4/24\). Expanding the exponential through order \(n^{-1}\) gives \(1+a/\sqrt n+(b+a^2/2)/n\). In particular, \(a^2/2=\kappa_3^2(it)^6/72\). Thus the Fourier-side expansion is

$$ \varphi_n(t) =e^{-t^2/2}\left[ 1+\frac{\kappa_3(it)^3}{6\sqrt n} +\frac{1}{n}\left\{ \frac{\kappa_4(it)^4}{24} +\frac{\kappa_3^2(it)^6}{72} \right\}\right] +\text{terms of smaller formal order}. $$

The word “formal” is important here. Finite fourth moment establishes the fixed-\(t\) coefficient expansion, but by itself does not justify integrating the remainder over all frequencies to prove a density error bound. Turning Fourier coefficients into a rigorous uniform density approximation requires additional regularity and remainder control. The calculation below identifies the Edgeworth terms; it does not claim those stronger error bounds.

Second-order density expansion: For a standardized iid sum whose density admits the relevant Fourier inversion and termwise approximation, the terms through order \(n^{-1}\) are
$$ f_{S_n}(x)\ \approx\ \phi(x)\left[ 1+\frac{\kappa_3}{6\sqrt n}H_3(x) +\frac{1}{n}\left\{ \frac{\kappa_4}{24}H_4(x) +\frac{\kappa_3^2}{72}H_6(x) \right\}\right], \qquad \phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}. $$

The conversion uses the Fourier inversion convention \(f(x)=(2\pi)^{-1}\int e^{-itx}\varphi(t)\,dt\). With that convention, the inverse transform of \((it)^r e^{-t^2/2}\) is \(H_r(x)\phi(x)\). Indeed, the \(r\)th derivative of \(\phi\) is \((-1)^rH_r\phi\); integration by parts in the Fourier transform shows that multiplication by \((it)^r\) in the integrand corresponds to \((-1)^r\phi^{(r)}\), which is \(H_r\phi\). Applying this identity to powers \(3,4,6\) gives the displayed density terms. \(\square\)

The expansion includes the first-order skewness correction as well as both second-order terms. The fourth-cumulant contribution is proportional to \(H_4\); the squared-skewness contribution is proportional to \(H_6\). Omitting the latter would discard the \(a^2/2\) term created when the logarithmic characteristic function is exponentiated.

Worked Density Calculations

Worked Example: A Normal Sum Has No Corrections

Suppose \(X_i\sim N(0,1)\), independently, and take \(n=9\). Calculate the expansion at \(x=1.2\).

For a standard normal summand, \(\kappa_3=0\) and \(\kappa_4=0\). Therefore both the \(n^{-1/2}\) term and the two \(n^{-1}\) terms vanish, regardless of the values of the Hermite polynomials. The approximation reduces to \(\phi(1.2)\).

The value is

$$ \phi(1.2)=\frac{1}{\sqrt{2\pi}}e^{-1.2^2/2} =\frac{1}{\sqrt{2\pi}}e^{-0.72} \approx 0.1942. $$

This is exact, not merely an approximation: the standardized sum of independent standard normal variables is itself standard normal.

Worked Example: An Exponential Sum at \(x=1\)

Let \(Y_i\) be independent exponential random variables with rate \(1\), and standardize them as \(X_i=Y_i-1\). Find the second-order density approximation for \(S_{25}\) at \(x=1\).

The cumulant generating function of \(Y_i\) is \(-\log(1-t)\), so its \(r\)th cumulant is \((r-1)!\). Centering changes only the first cumulant. Since the variance is \(1\), the standardized \(X_i\) have \(\kappa_3=2\) and \(\kappa_4=6\). At \(x=1\),

$$ H_3(1)=-2,\qquad H_4(1)=-2,\qquad H_6(1)=16. $$

The first-order contribution is \((2/(6\sqrt{25}))(-2)=-2/15\). The fourth-cumulant contribution is \((6/(24\cdot25))(-2)=-1/50\). The squared-skewness contribution is \((2^2/(72\cdot25))16=16/450\). Therefore the bracket multiplying \(\phi(1)\) is

$$ 1-\frac{2}{15}-\frac{1}{50}+\frac{16}{450} =1-\frac{53}{450} =\frac{397}{450}\approx 0.88222. $$

Since \(\phi(1)\approx0.24197\), the density approximation is \((397/450)\phi(1)\approx0.21347\). For comparison, the exact density exists: \(\sum_{i=1}^{25}Y_i\) has a gamma density with shape \(25\) and rate \(1\), so the density of \(S_{25}\) at \(1\) is \(5\cdot30^{24}e^{-30}/\Gamma(25)\). The Edgeworth calculation provides a simpler approximation, not an identity with this exact value.

Worked Example: A Symmetric Uniform Sum at the Center

Let \(X_i\) be independent and uniform on \((-\sqrt3,\sqrt3)\). This distribution has mean \(0\) and variance \(1\). Calculate the second-order approximation for \(S_{16}\) at \(x=0\).

Symmetry gives \(\kappa_3=0\). The fourth moment of a uniform variable on \((-a,a)\) is \(a^4/5\), so here \(E[X_i^4]=9/5\) and \(\kappa_4=9/5-3=-6/5\). At the center, \(H_4(0)=3\), while the skewness terms are zero. The density estimate is therefore

$$ \phi(0)\left[1+\frac{-6/5}{24\cdot16}\,3\right] =\phi(0)\left(1-\frac{3}{320}\right) =\phi(0)\frac{317}{320} \approx 0.39520. $$

The negative correction reflects the negative fourth cumulant of this bounded, relatively flat-tailed summand distribution. The calculation concerns the central density of the sum; it does not say that every pointwise correction must be negative.

Interpreting the Terms and Avoiding Common Errors

The two \(n^{-1}\) contributions represent distinct features of the summand distribution. If \(\kappa_3=0\), the squared-skewness term disappears, but the fourth-cumulant term can remain, as in the uniform example. If \(\kappa_4=0\) but \(\kappa_3\ne0\), the squared-skewness term remains. If both cumulants vanish, the expansion has no corrections through this order.

A common mistake is to treat “second order” as meaning only the fourth-cumulant term. The correction is obtained by expanding the exponential of the cumulant series, not by reading off its logarithm term by term. Squaring the order-\(n^{-1/2}\) skewness term produces an order-\(n^{-1}\) term, so dropping it gives an incomplete expansion. Another frequent error is to substitute ordinary kurtosis for \(\kappa_4\) without subtracting \(3\).

Finally, an Edgeworth expression is a polynomial times a normal density. The polynomial factor can become negative for some \(x\), even though a genuine density cannot. This signals a limitation of the truncated approximation, especially away from the central region; it is not a reason to reinterpret a negative value as a probability or to alter the derived coefficients.

Key takeaway: Through order \(n^{-1}\), the density correction contains \(\kappa_4H_4/24\) and \(\kappa_3^2H_6/72\), both multiplied by \(1/n\). The squared-skewness term is required because the characteristic function is the exponential of the cumulant expansion.

Check Your Understanding

Use the cumulants and Hermite polynomials in the expansion to answer these questions.

  1. Why does the second-order expansion contain a squared-skewness term even though the logarithm of the characteristic function has no separate squared-skewness cumulant?
  2. For a symmetric standardized summand with \(\kappa_4=-1\), which correction terms remain at order \(n^{-1}\)?
  3. At \(x=0\), evaluate \(H_3(0)\), \(H_4(0)\), and \(H_6(0)\). Which terms vanish for a symmetric summand?
  4. Explain why the fourth cumulant of a standard normal variable is zero, although its fourth moment is \(3\).
  5. Why does a negative value from a truncated Edgeworth density not define a valid probability density?