When Edgeworth Approximations Become Unreliable

When a Good Central Approximation Becomes a Bad Tail Approximation

An Edgeworth expansion improves on a normal approximation by adding polynomial corrections involving standardized cumulants. Its power comes with a qualification: it is an asymptotic approximation in specified ranges, not a guarantee that a finite truncation behaves like a probability distribution everywhere. A correction that is small near the center can become large in the tails, where the normal density is tiny and the polynomial factors grow.

For example, for a standardized sum of $n$ independent identically distributed variables with mean zero, variance one, and third cumulant $\kappa_3$, the first-order CDF expression is

G_n(x)=\Phi(x)+\frac{\kappa_3}{6\sqrt n}(1-x^2)\phi(x),

where $\phi$ and $\Phi$ are the standard normal density and CDF. Under suitable moment and regularity conditions, this formula describes a central-range asymptotic expansion. It does not say that $G_n$ is accurate uniformly over every threshold, nor that the corresponding truncated density is nonnegative. The checks in this tutorial help identify when those limitations matter.

Four Warning Signs

A useful diagnosis begins by asking what feature of the data-generating distribution is likely to challenge the approximation. These warning signs can occur together.

The threshold is far into a tail. The correction includes a polynomial in $x$, while $\phi(x)$ and normal tail probabilities shrink rapidly. The correction can become substantial relative to the small tail probability.
The distribution is discrete or supported on a lattice. The true CDF has jumps, whereas the ordinary Edgeworth CDF expression is smooth. Lattice corrections may be needed, and a smooth CDF cannot reproduce a large jump arbitrarily well.
Standardized cumulant corrections are not small. A large skewness, especially when combined with a small effective expected count, can make the first correction comparable to the leading normal term.
A truncated density becomes negative. A genuine probability density cannot be negative. This is a direct warning that the finite polynomial correction should not be interpreted as a valid density in that region.

These are diagnostics, not a replacement for the remainder conditions governing a particular expansion. Passing a quick check does not prove a small error; failing one identifies a reason to be cautious.

Why Smooth CDFs Cannot Resolve Large Jumps

Theorem: Let $F$ be a CDF with a jump of size $\Delta=F(x_0)-F(x_0-)>0$ at $x_0$, and let $G$ be any continuous function. Then $$ \sup_x |F(x)-G(x)|\geq \frac{\Delta}{2}. $$

Proof. Suppose the supremum error is $\varepsilon$. At $x_0$, $|F(x_0)-G(x_0)|\leq\varepsilon$. As $x$ increases to $x_0$ from below, continuity of $G$ and the left limit of $F$ give $|F(x_0-)-G(x_0)|\leq\varepsilon$. The triangle inequality therefore gives

\Delta =|F(x_0)-F(x_0-)| \leq |F(x_0)-G(x_0)|+|G(x_0)-F(x_0-)| \leq 2\varepsilon.

Rearranging proves the claim. $\square$

This bound explains why a smooth CDF approximation can have an unavoidable uniform error when the distribution has appreciable point masses. It does not rule out a useful approximation at selected thresholds, nor does it rule out an expansion with an appropriate lattice correction.

Worked Example: A Sparse Binomial Sum and a Large CDF Jump

Let $B\sim\operatorname{Binomial}(100,0.01)$, and standardize it as $Z=(B-1)/\sqrt{0.99}$. Since $E[B]=1$ and $\operatorname{Var}(B)=0.99$, the event $Z\leq 0$ is exactly $B\leq 1$. Compute the true probability and compare it with the ordinary first-order Edgeworth expression at zero.

The exact probability is

P(Z\leq 0)=P(B\leq 1) =(0.99)^{100}+100(0.01)(0.99)^{99} \approx 0.7357620.

The skewness of one Bernoulli variable with success probability $p$ is $(1-2p)/\sqrt{p(1-p)}$. Here it is

\kappa_3=\frac{0.98}{\sqrt{0.0099}}\approx 9.8494.

Putting $x=0$ and $n=100$ into the first-order expression gives

G_{100}(0) =\Phi(0)+\frac{9.8494}{6\sqrt{100}}\phi(0) \approx 0.5+0.16416(0.39894) \approx 0.5654888.

The discrepancy is approximately $0.7357620-0.5654888\approx 0.1703$. Two warnings are visible: $np=1$, so the binomial is sparse, and its skewness correction is large rather than a small perturbation. Also, $Z$ is discrete. In particular, the CDF jumps at zero by $P(B=1)=100(0.01)(0.99)^{99}\approx 0.3697$; the theorem above means any continuous CDF approximation has uniform error at least about $0.1849$. This is not a setting in which the uncorrected smooth expansion should be trusted as a general CDF approximation.

Extreme Tails Magnify Polynomial Corrections

For a right-tail probability, the first-order formula implies

1-G_n(x) =1-\Phi(x)+\frac{\kappa_3}{6\sqrt n}(x^2-1)\phi(x).

For large positive $x$, the normal tail $1-\Phi(x)$ is of order $\phi(x)/x$. Thus the correction relative to that tail is roughly of order $x^3/\sqrt n$, not merely $1/\sqrt n$. This follows from the normal tail bounds $x\phi(x)/(x^2+1)\leq 1-\Phi(x)\leq\phi(x)/x$ for $x>0$. Consequently, even if an expansion works well at each fixed $x$, its correction may cease to be small when $x$ increases with $n$. A useful informal warning range is that $x^3/\sqrt n$ should not be treated as small without checking; this is a diagnostic, not a universal error bound.

Worked Example: A Tail Correction That Is Not Small

For an exponential distribution standardized to mean zero and variance one, the third cumulant is $2$. Take $n=100$ and examine $x=3$. The first-order correction to the right tail is

\frac{2}{6\sqrt{100}}(3^2-1)\phi(3) =\frac{1}{30}(8)\phi(3) \approx 0.0011818.

The standard normal right tail at 3 is approximately $0.0013499$. The displayed correction is therefore about $0.0011818/0.0013499\approx 0.88$ times the normal tail itself. The formula may still be useful as an asymptotic calculation under its conditions, but the correction is plainly not a negligible adjustment here. Moreover, a large first correction does not certify accuracy: omitted higher-order terms could matter too. The appropriate conclusion is to treat the truncated tail value cautiously, not to assume the correction has made it reliable.

Large Skewness and Negative Approximate Densities

A first-order density expansion has the form

g_n(x)=\phi(x)\left[1+\frac{\kappa_3}{6\sqrt n}H_3(x)\right], \qquad H_3(x)=x^3-3x.

The factor $H_3(x)$ grows in magnitude as $|x|$ grows. Even with fixed skewness, the bracket can eventually be negative at a sufficiently extreme $x$. A negative value is not a negative probability; it is evidence that the truncated expression is not a valid density there.

Theorem: Suppose $f$ is a probability density and $g$ is an integrable signed function with $\int g(x)\,dx=1$. Define its negative mass by $m=\int_{\{x:g(x)<0\}}-g(x)\,dx$. Then $$ \int |f(x)-g(x)|\,dx\geq 2m. $$

Proof. Let $A=\{x:g(x)<0\}$. Because $f\geq 0$,

\int_A |f-g|\,dx=\int_A(f-g)\,dx\geq-\int_A g\,dx=m.

Both functions integrate to one, so $\int(f-g)=0$. Hence the integral of $f-g$ over $A^c$ equals $-\int_A(f-g)$, whose magnitude is at least $m$. Since the integral of an absolute value is at least the absolute value of the integral,

\int_{A^c}|f-g|\,dx\geq\left|\int_{A^c}(f-g)\,dx\right|\geq m.

Adding the bounds on $A$ and $A^c$ proves the result. $\square$

Worked Example: A Second-Order Density That Turns Negative

Let $Y_1,\ldots,Y_{10}$ be independent uniform variables on $(-\sqrt{3},\sqrt{3})$, each with mean zero and variance one, and let $Z=(Y_1+\cdots+Y_{10})/\sqrt{10}$. The individual variables have zero skewness and excess kurtosis $-6/5$. The second-order density expression in this symmetric case is

g_{10}(x)=\phi(x)\left[1-\frac{1}{200}H_4(x)\right], \qquad H_4(x)=x^4-6x^2+3.

At $x=5$, $H_4(5)=625-150+3=478$, so the bracket is $1-478/200=-1.39$. Thus $g_{10}(5)<0$. But the actual standardized sum has support $(-\sqrt{30},\sqrt{30})$, and $5<\sqrt{30}$; its density is positive at this interior point. The negative value is therefore a property of the truncated approximation, not of the sum's distribution. In this example the point is also near the support boundary, another reason not to read the expansion as a globally valid density.

Practical Checks and a Common Mistake

Before reporting a truncated Edgeworth result, check the target and the scale of the correction. In particular:

State whether the target is a central CDF value, a point probability, a quantile, or a tail probability; these targets can be affected differently by discreteness and tails.
Compare the correction with the leading term, not just with one. For a small tail probability, an apparently small absolute correction can be large on a relative scale.
Check whether the distribution is lattice-valued and whether a lattice correction is appropriate for the target.
Inspect the approximate density or mass expression for impossible signs, and check whether the evaluation point is near a support boundary.
Keep the asymptotic regime in view. A formula justified as $n\to\infty$ for fixed standardized $x$ does not automatically justify a threshold moving far into a tail.

Common mistake: “The Edgeworth expression includes a skewness correction, so it must be more accurate than the normal approximation, even in the far tail.” The correction is designed to improve an asymptotic expansion under its conditions; it is not guaranteed to improve every finite-sample probability at every threshold. If it is large relative to the target tail, if the support is discrete, or if the corresponding density becomes negative, the right response is to question the truncation and examine a method suited to the target—not to treat the formula as an exact probability law.

Key takeaway: Edgeworth expansions are most persuasive when the correction is controlled, the target lies in a justified range, and discreteness and support are handled appropriately. Extreme tails, large skewness, substantial jumps, and negative approximate densities are concrete reasons to withhold trust from a truncated result.

Check Your Understanding

Use the diagnostics and results above to answer these questions.

Why can a smooth CDF not uniformly approximate a distribution with a jump of size $\Delta$ to error smaller than $\Delta/2$?
For the first-order right-tail correction, explain why its size relative to a normal tail can grow on the order of $x^3/\sqrt n$.
In the sparse binomial example, identify two distinct warning signs and explain what each one indicates.
What does a negative value of a truncated Edgeworth density imply, and what does it not imply?
Why is a large first correction not evidence by itself that the corrected tail probability is accurate?