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Edgeworth and Saddlepoint Methods · Tutorial 972 of 994

Fourier Inversion for Edgeworth Density Expansions

Learn how cumulant terms in a standardized sum’s characteristic function become explicit corrections to the normal density under suitable Fourier regularity conditions.

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What You'll Learn

  • Expand the characteristic function of a standardized independent sum through order \(n^{-1}\).
  • Identify sufficient moment and Fourier conditions for a density expansion.
  • Use Fourier inversion to derive the density correction terms and their signs.
  • Apply the expansion to exponential and uniform summands.
  • Recognize why a lattice distribution does not satisfy the density-expansion conditions.

From a Characteristic Function to a Density Approximation

The central limit theorem describes the leading behavior of a standardized sum: its distribution approaches the standard normal law. When the summands are skewed or have nonnormal tail shape, however, that leading approximation can miss systematic features. An Edgeworth density expansion expresses those features as corrections to the normal density.

The derivation has two parts. First, expand the logarithm of the characteristic function using cumulants and exponentiate the result. Second, invert each term by Fourier inversion. Conditions matter: the characteristic function must decay sufficiently for the sums to have densities, and the expansion must be controlled well enough that inversion preserves its error order.

Setup: Let \(Y_1,Y_2,\ldots\) be independent and identically distributed random variables with \(E[Y_1]=0\), \(\operatorname{Var}(Y_1)=1\), and \(E[|Y_1|^5]<\infty\). Define \(S_n=n^{-1/2}\sum_{j=1}^nY_j\). Write \(\psi(t)=E[e^{itY_1}]\) for the characteristic function, and let \(\kappa_3=E[Y_1^3]\) and \(\kappa_4=E[Y_1^4]-3\) be the third and fourth cumulants.

The fourth cumulant is not the fourth central moment: for a standardized variable, it is the fourth central moment minus \(3\). The subtraction removes the contribution that would be present for a normal distribution. For the density on the normal side of the approximation, write \(\phi_0(x)=(2\pi)^{-1/2}e^{-x^2/2}\).

The Cumulant Expansion

Since \(S_n\) is a scaled sum, its characteristic function is \(\psi_n(t)=\psi(t/\sqrt n)^n\). Near zero, a Taylor expansion of the log characteristic function gives \(\log\psi(u)=-u^2/2+\kappa_3(iu)^3/6+\kappa_4(iu)^4/24+O(|u|^5)\). The fifth-moment assumption supplies a bounded fifth-order remainder near zero. Substituting \(u=t/\sqrt n\) and multiplying by \(n\) yields

$$ \log\psi_n(t) =-\frac{t^2}{2} +\frac{\kappa_3(it)^3}{6\sqrt n} +\frac{\kappa_4(it)^4}{24n} +O\!\left(\frac{|t|^5}{n^{3/2}}\right). $$

Exponentiating is essential: the square of the order-\(n^{-1/2}\) term contributes at order \(n^{-1}\). Thus, through that order,

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

The coefficient \(1/72\) is \(\tfrac12(1/6)^2\), from the quadratic term in the exponential series. The sixth power \((it)^6\) retains its sign; replacing it by \(t^6\) would change the correction.

Worked Example: The Cumulant Terms for Exponential Summands

Let \(X_j\) have an exponential distribution with rate \(1\), and take \(Y_j=X_j-1\). The variables \(Y_j\) have mean \(0\) and variance \(1\). Find the coefficients in the characteristic-function expansion.

The cumulant-generating function of \(Y_j\) is \(K(s)=\log E[e^{sY_j}]=-s-\log(1-s)\), for \(s<1\). Its power series is \(K(s)=\sum_{r=2}^{\infty}s^r/r\). Since the coefficient of \(s^r\) is \(\kappa_r/r!\), this gives \(\kappa_r=(r-1)!\), and in particular \(\kappa_3=2\) and \(\kappa_4=6\).

Substitution gives \(\kappa_3/6=1/3\), \(\kappa_4/24=1/4\), and \(\kappa_3^2/72=4/72=1/18\). Therefore the characteristic-function approximation is

$$ \psi_n(t)\approx e^{-t^2/2}\left[ 1+\frac{(it)^3}{3\sqrt n} +\frac{1}{n}\left\{\frac{(it)^4}{4}+\frac{(it)^6}{18}\right\} \right]. $$

For this example \(|\psi(t)|=(1+t^2)^{-1/2}\), so \(\int_{\mathbb R}|\psi(t)|^2\,dt<\infty\). Also \(|\psi(t)|<1\) for every \(t\ne0\). Thus the Fourier regularity conditions used below hold.

Fourier Inversion and the Density Expansion

Use the convention that a density \(f\) with an integrable characteristic function \(\chi\) is recovered as \(f(x)=(2\pi)^{-1}\int_{\mathbb R}e^{-itx}\chi(t)\,dt\). In this convention, each factor \((it)^r\) in a characteristic function corresponds to \((-1)^r\) times the \(r\)th derivative of the density. Indeed, differentiating the inversion formula \(r\) times introduces \((-it)^r\), so

$$ \frac{1}{2\pi}\int_{\mathbb R}e^{-itx}(it)^r e^{-t^2/2}\,dt =(-1)^r\phi_0^{(r)}(x). $$

For \(r=3,4,6\), evaluating these derivatives gives

$$ \begin{aligned} -\phi_0'''(x)&=(x^3-3x)\phi_0(x),\\ \phi_0^{(4)}(x)&=(x^4-6x^2+3)\phi_0(x),\\ \phi_0^{(6)}(x)&=(x^6-15x^4+45x^2-15)\phi_0(x). \end{aligned} $$
Theorem: Under the setup above, suppose in addition that the characteristic function satisfies Cramér’s condition \(\limsup_{|t|\to\infty}|\psi(t)|<1\), and that \(\int_{\mathbb R}|\psi(t)|^m\,dt<\infty\) for some positive integer \(m\). Then, for all sufficiently large \(n\), \(S_n\) has a continuous density \(f_n\), and
$$ \begin{aligned} f_n(x)=\phi_0(x)\Bigg[1 &+\frac{\kappa_3}{6\sqrt n}(x^3-3x)\\ &+\frac{1}{n}\left\{ \frac{\kappa_4}{24}(x^4-6x^2+3) +\frac{\kappa_3^2}{72}(x^6-15x^4+45x^2-15) \right\}\Bigg]+o(n^{-1}), \end{aligned} $$

where the remainder is uniform in \(x\): its supremum in absolute value is \(o(n^{-1})\). The integrability condition on a power of \(\psi\) ensures that \(\psi(t/\sqrt n)^n\) is integrable for \(n\ge m\), so Fourier inversion yields a continuous density.

Proof. The log-characteristic-function expansion above, followed by exponentiation, gives the displayed approximation to \(\psi_n(t)\). To justify inversion with a uniform error, let \(G_n(t)\) denote the displayed Gaussian-times-polynomial approximation and consider \(\int_{\mathbb R}|\psi_n(t)-G_n(t)|\,dt\).

Choose a sufficiently small \(\delta>0\). Because \(E[Y_1]=0\) and \(\operatorname{Var}(Y_1)=1\), the expansion near zero implies \(|\psi(u)|\le e^{-cu^2}\) for some \(c>0\) and \(|u|\le\delta\). Split the integral into three regions. On \(|t|\le n^{1/20}\), substitute \(u=t/\sqrt n\) in the log expansion and expand the exponential through order \(n^{-1}\). The omitted terms are bounded by \(C n^{-3/2}\) times a fixed polynomial in \(|t|\), multiplied by \(e^{-t^2/2}\), for all sufficiently large \(n\). This bound is integrable, so its integral is \(O(n^{-3/2})=o(n^{-1})\).

On \(n^{1/20}<|t|\le\delta\sqrt n\), the bound \(|\psi(t/\sqrt n)|^n\le e^{-ct^2}\) makes the integral of the exact characteristic function exponentially small in \(n^{1/10}\). The Gaussian polynomial \(G_n\) has a tail with the same negligible property on this region. Finally, for \(|t|>\delta\sqrt n\), change variables to \(u=t/\sqrt n\). Cramér’s condition and continuity give a \(\rho<1\) such that \(|\psi(u)|\le\rho\) for \(|u|\ge\delta\). For \(n\ge m\),

$$ \int_{|t|>\delta\sqrt n}|\psi(t/\sqrt n)|^n\,dt \le \sqrt n\,\rho^{\,n-m}\int_{\mathbb R}|\psi(u)|^m\,du, $$

which is exponentially small. The Gaussian polynomial’s integral over this last region is also negligible. Consequently \(\int_{\mathbb R}|\psi_n(t)-G_n(t)|\,dt=o(n^{-1})\). Fourier inversion bounds the supremum error between the corresponding densities by \((2\pi)^{-1}\) times this integral. Inverting the terms of \(G_n\) using the derivative identities gives exactly the stated density expansion. \(\square\)

Worked Example: A Symmetric Uniform Distribution

Take \(Y_j\) uniform on \((-\sqrt3,\sqrt3)\). Its mean is \(0\), its variance is \(1\), its third cumulant is \(0\), and \(E[Y_j^4]=9/5\). Hence \(\kappa_4=9/5-3=-6/5\). The characteristic function is \(\psi(t)=\sin(\sqrt3t)/(\sqrt3t)\); its squared absolute value is integrable, and it satisfies Cramér’s condition.

At \(x=0\), the terms involving \(\kappa_3\) vanish, and \(x^4-6x^2+3=3\). The density approximation at zero is therefore

$$ f_n(0)=\frac{1}{\sqrt{2\pi}}\left(1+\frac{1}{n}\frac{-6/5}{24}\,3\right)+o(n^{-1}) =\frac{1}{\sqrt{2\pi}}\left(1-\frac{3}{20n}\right)+o(n^{-1}). $$

For \(n=100\), the displayed approximation is \(0.3989423(0.9985)\approx0.3983439\). This is an asymptotic approximation, not an exact value for the density at that sample size.

Conditions and Common Pitfalls

The expansion is a density approximation, so applying it requires more than finite moments. The moment conditions justify the cumulant expansion near zero; the Fourier conditions control the rest of the inversion integral. They are not interchangeable. In particular, a distribution concentrated on a lattice can have moments of every order and still fail Cramér’s condition.

Worked Example: Why a Lattice Sum Has No Density Expansion Here

Let \(B_j\) be independent Bernoulli random variables with success probability \(1/2\), and set \(Y_j=2B_j-1\). Each \(Y_j\) takes values \(-1\) and \(1\), has mean \(0\), variance \(1\), and moments of every order. But its characteristic function is \(\psi(t)=\cos t\), and \(\limsup_{|t|\to\infty}|\cos t|=1\), violating Cramér’s condition.

Writing \(K=\sum_{j=1}^nB_j\), the standardized sum is \(S_n=(2K-n)/\sqrt n\). It takes only the finitely many values \((2k-n)/\sqrt n\), for \(k=0,\ldots,n\). Therefore its distribution is discrete and has no density to which the theorem’s Fourier inversion argument could apply. The moment expansion alone does not make this a valid density approximation.

A second common mistake is to retain the fourth-cumulant term but omit the square of the skewness term. That loses a contribution of the same order, \(n^{-1}\). Another is to change the signs in Fourier inversion: the factor \((it)^r\) corresponds to \((-1)^r\phi_0^{(r)}\), not always to the derivative with the same sign. Finally, the polynomial correction need not keep the approximating expression nonnegative in the far tails. An asymptotic density expansion should not be mistaken for an exact density there.

The essential calculation is systematic: scale the cumulants of the sum, expand and exponentiate its characteristic function to the desired order, then invert each polynomial-times-Gaussian term with a consistent Fourier convention. The regularity assumptions ensure that this calculation describes densities rather than only a formal local expansion of characteristic functions.

Check Your Understanding

Use the expansion and its assumptions to answer these questions.

  1. Why does the square of the third-cumulant term contribute at order \(n^{-1}\)?
  2. For a standardized summand, how is \(\kappa_4\) related to its fourth central moment?
  3. Under the Fourier convention used here, what density derivative corresponds to a factor \((it)^4\)?
  4. Which assumptions in the theorem support the cumulant expansion, and which control Fourier inversion?
  5. Why do finite moments alone not justify applying this density expansion to a standardized Bernoulli sum?
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