Why Remainder Conditions Matter
An Edgeworth expansion adds cumulant-based corrections to a normal approximation. Its algebraic terms may be easy to write down, but the terms alone do not say how accurate the approximation is. A remainder bound depends on regularity: in particular, on how many moments exist and on how the characteristic function behaves away from the origin.
It also matters where the error is controlled. An error that is small uniformly in the argument of a cumulative distribution function (CDF) is an absolute error. It need not be small relative to a tail probability that is itself tiny. This tutorial makes those distinctions explicit for standardized sums.
Moment Conditions and the Order of the Expansion
Let \(X_1,\ldots,X_n\) be independent and identically distributed random variables with \(E[X_i]=0\) and \(\operatorname{Var}(X_i)=1\). Write \(S_n=n^{-1/2}\sum_{i=1}^n X_i\), and let \(\kappa_j\) denote the \(j\)th cumulant of \(X_1\); in particular, \(\kappa_1=0\), \(\kappa_2=1\), and \(\kappa_3=E[X_1^3]\). The terms of the Edgeworth expansion are polynomials in these cumulants, multiplied by the standard normal density \(\phi\) or distribution function \(\Phi\).
A finite \(r\)th absolute moment, \(E|X_1|^r<\infty\), supports cumulants through order \(r\). In the usual CDF expansion, that moment condition supports terms through order \(n^{-(r-2)/2}\). This is a statement about the order one can justify under the stated assumptions, not a claim that every coefficient at that order is nonzero.
The hypotheses do different jobs. The moments allow the logarithm of the characteristic function to be expanded to the required order near zero. The Cramér condition controls the contribution from frequencies away from zero when that local expansion is converted into a distributional approximation. A Taylor expansion near zero, by itself, gives no uniform CDF remainder.
What the Local Characteristic-Function Expansion Shows
The first corrections can be derived directly. With finite fourth absolute moment, the logarithm of the characteristic function has the expansion \( \log\varphi(u)=-\frac{u^2}{2}+\frac{\kappa_3(iu)^3}{6}+\frac{\kappa_4(iu)^4}{24}+o(u^4) \) as \(u\to0\). Substituting \(u=t/\sqrt n\) and multiplying by \(n\) gives, for each fixed \(t\), \( n\log\varphi(t/\sqrt n)=-\frac{t^2}{2}+\frac{\kappa_3(it)^3}{6\sqrt n}+\frac{\kappa_4(it)^4}{24n}+o(n^{-1}). \) Exponentiating and retaining terms through \(n^{-1}\) yields \( \varphi(t/\sqrt n)^n=e^{-t^2/2}\left[1+\frac{\kappa_3(it)^3}{6\sqrt n}+\frac{1}{n}\left\{\frac{\kappa_4(it)^4}{24}+\frac{\kappa_3^2(it)^6}{72}\right\}\right]+o(n^{-1}) \) for fixed \(t\). The squared-skewness term arises from the quadratic term in the exponential series. These pointwise calculations reveal the coefficients; they do not alone establish the uniform CDF theorem.
For reference, the corresponding first two CDF corrections, using probabilists' Hermite polynomials \(H_2(x)=x^2-1\), \(H_3(x)=x^3-3x\), and \(H_5(x)=x^5-10x^3+15x\), are
This is the expansion through order \(n^{-1}\), when the fourth moment exists and the conditions needed for the CDF remainder theorem hold. A formula for a coefficient and a theorem controlling the error are distinct pieces of information.
Worked Example: Exponential Summands and a Second-Order Remainder
Let \(Y_i\) be independent exponential random variables with rate 1, and set \(X_i=Y_i-1\). Then \(E[X_i]=0\), \(\operatorname{Var}(X_i)=1\), and the cumulants of \(X_i\) of orders \(j\geq2\) are \(\kappa_j=(j-1)!\). In particular, \(\kappa_3=2\) and \(\kappa_4=6\).
All absolute moments are finite. The characteristic function is \(\varphi(t)=e^{-it}/(1-it)\), so \(|\varphi(t)|=(1+t^2)^{-1/2}\to0\) as \(|t|\to\infty\). Thus the Cramér condition holds. Taking \(r=4\), the uniform CDF theorem gives an error \(o(n^{-1})\) for the expansion through order \(n^{-1}\).
Substituting the cumulants into the displayed formula gives the explicit approximation \( \Phi(x)+\phi(x)\left[\frac{1}{3\sqrt n}(1-x^2)-\frac{1}{n}\left\{\frac14H_3(x)+\frac{1}{18}H_5(x)\right\}\right]. \) The remainder statement is uniform over \(x\in\mathbb R\); it does not say that the relative error in an extreme upper tail is \(o(n^{-1})\).
Pointwise, Uniform, and Tail-Region Accuracy
A pointwise statement fixes \(x\) and lets \(n\) grow. A uniform statement takes the largest absolute error over every \(x\) before taking the limit. Uniform control is stronger and is useful when the argument is chosen after seeing the sample size. Neither statement, without further bounds, ensures small relative error where the target probability is very small.
Here is the basic arithmetic behind that distinction. If an approximation \(A_n(x)\) satisfies \(|F_n(x)-A_n(x)|\leq \epsilon_n\), then for an upper-tail probability, \( \frac{|(1-F_n(x))-(1-A_n(x))|}{1-F_n(x)} \leq \frac{\epsilon_n}{1-F_n(x)}. \) The right side can be large when \(1-F_n(x)\) is much smaller than \(\epsilon_n\). Thus a uniform absolute CDF bound is not automatically a relative tail bound.
A moving threshold \(x=x_n\) introduces another issue: the polynomials in an Edgeworth correction can grow with \(|x_n|\). For example, the first CDF correction contains \(\phi(x_n)(1-x_n^2)/\sqrt n\). For fixed \(x\) its polynomial factor is fixed; as \(x_n\) moves into the tails, it changes with \(n\). Claims about a growing tail region therefore require conditions and error bounds designed for that region, not just the fixed-\(x\) coefficient calculation.
Worked Example: Why an Absolute Bound Does Not Control a Tail Relatively
Suppose an approximation has a uniform absolute CDF error at most \(\epsilon_n=n^{-1}\). Consider a threshold \(x_n=\sqrt{2\log n}\) and suppose the target upper tail is of normal size, \(1-\Phi(x_n)\). The normal tail bounds \( \frac{x}{1+x^2}\phi(x)\leq 1-\Phi(x)\leq \frac{\phi(x)}{x},\quad x>0, \) show its scale. Indeed, \(\phi(x_n)=(2\pi)^{-1/2}n^{-1}\), and \(x_n/(1+x_n^2)\) is of order \(1/\sqrt{\log n}\). Hence \(1-\Phi(x_n)\) is of order \(n^{-1}/\sqrt{\log n}\).
Dividing the permitted absolute error by that tail size gives a bound of order \(\sqrt{\log n}\), which does not tend to zero. The calculation does not say the actual approximation must be poor there. It shows that the stated uniform absolute bound alone cannot certify relative accuracy at this threshold.
Smoothness and Density Remainders
A CDF expansion and a density expansion require different control. For a density expansion obtained by Fourier inversion, a useful sufficient smoothness condition is integrability of the characteristic function: if \(\int_{\mathbb R}|\varphi(t)|\,dt<\infty\), inversion gives a bounded continuous density. More generally, if \(\int_{\mathbb R}|t|^q|\varphi(t)|\,dt<\infty\) for an integer \(q\geq0\), differentiating the inversion integral under the integral sign gives a density with \(q\) continuous bounded derivatives. The justification is domination by the integrable function \(|t|^q|\varphi(t)|\).
These conditions are sufficient, not necessary, and they are not interchangeable with finite moments. Moments control derivatives of \(\varphi\) at zero; integrability controls Fourier inversion over the whole line. A distribution can have several moments without having a density, while a smooth density condition does not by itself supply high-order moments. For density remainders, one must check both the moment expansion and the Fourier control needed for the inversion argument.
Worked Example: A Student Distribution Limits the Justified Order
Let \(T\) have a Student \(t\) distribution with \(4.5\) degrees of freedom, and define \(X=T/\sqrt{4.5/(4.5-2)}\). Then \(E[X]=0\) and \(\operatorname{Var}(X)=1\). A \(t_\nu\) variable has a finite absolute moment of order \(a\) exactly when \(a<\nu\). Consequently \(E|X|^4<\infty\), but \(E|X|^5=\infty\).
This distribution has a continuous density, and its characteristic function decays to zero at infinity, so it satisfies the Cramér condition. The uniform CDF theorem with \(r=4\) justifies expansion terms through order \(n^{-1}\) and a remainder \(o(n^{-1})\). The same theorem cannot be invoked with \(r=5\): the required fifth absolute moment fails. Writing a formal term of order \(n^{-3/2}\) would not, on these assumptions, establish a remainder of that order.
Common Mistake: Reading More into the Remainder
- Wrong reasoning: “The CDF error is uniformly \(o(n^{-1})\), so the upper-tail probability is relatively accurate to \(o(n^{-1})\) at any threshold.”
- Correction: The theorem bounds an absolute error. Relative error divides that bound by the tail probability, which may be much smaller. A tail claim needs its own error analysis and assumptions appropriate to the moving threshold.
- Wrong reasoning: “The cumulant formula has a third-order correction, so a third-order remainder follows.”
- Correction: A coefficient calculation is not a remainder theorem. Check that the required moments exist and that the characteristic-function condition needed to control the inversion step holds.
Check Your Understanding
Use the assumptions and distinctions developed above to answer these questions.
- For a standardized summand with a finite fifth absolute moment and satisfying the Cramér condition, through what order does the stated uniform CDF theorem expand, and what is the remainder order?
- Why does a Taylor expansion of the characteristic function near zero not by itself establish a uniform CDF error bound?
- For the centered rate-1 exponential summand, verify the third and fourth cumulants and state the CDF remainder order justified by a finite fourth moment.
- What role does integrability of \(|t|^q|\varphi(t)|\) play in a density expansion?
- Why can a uniform absolute CDF bound fail to guarantee small relative error at a moving tail threshold?