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.
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:
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
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
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
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.
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
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 \(x=1\),
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
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
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.
Check Your Understanding
Use the cumulants and Hermite polynomials in the expansion to answer these questions.
- 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?
- For a symmetric standardized summand with \(\kappa_4=-1\), which correction terms remain at order \(n^{-1}\)?
- At \(x=0\), evaluate \(H_3(0)\), \(H_4(0)\), and \(H_6(0)\). Which terms vanish for a symmetric summand?
- Explain why the fourth cumulant of a standard normal variable is zero, although its fourth moment is \(3\).
- Why does a negative value from a truncated Edgeworth density not define a valid probability density?