Edgeworth Corrections for Approximate Quantiles

From a CDF Approximation to a Quantile

A distribution-function approximation answers questions of the form $P(Z_n\leq x)\approx p$. In many applications the more direct question is the reverse one: which value $x$ makes the probability approximately $p$? Examples include choosing a one-sided critical value or describing the endpoints of a central interval. A small error in a CDF approximation does not translate into the same numerical error in its quantile; the local slope of the CDF matters.

For a standardized sum, the first Edgeworth correction to the normal CDF accounts for skewness. Inverting that corrected CDF gives an adjustment to the normal quantile. The adjustment depends both on the skewness and on the location of the requested probability: it can move different quantiles in different ways.

The First-Order Quantile Formula

Let $X_1,\ldots,X_n$ be independent and identically distributed with mean $\mu$, variance $\sigma^2>0$, and finite third absolute moment. Write $\gamma=E[(X_1-\mu)^3]/\sigma^3$ for their standardized third cumulant (skewness), and define the standardized sum

Z_n=\frac{\sum_{i=1}^n X_i-n\mu}{\sigma\sqrt{n}}.

Suppose the applicable first-order Edgeworth CDF expansion holds uniformly in $x$:

F_n(x)=P(Z_n\leq x) =\Phi(x)+\frac{\gamma}{6\sqrt{n}}(1-x^2)\phi(x)+o(n^{-1/2}).

Here $\Phi$ and $\phi$ are the standard normal CDF and density. For iid summands, finite third absolute moment together with a suitable nonlattice regularity condition, such as Cramér's condition, is a standard setting for this first-order expansion. The quantile argument below uses the expansion itself and its uniform remainder.

Definition: For $0<p<1$, define the $p$-quantile of $F_n$ by $q_{n,p}=\inf\{x:F_n(x)\geq p\}$. Let $z_p=\Phi^{-1}(p)$. The first-order Edgeworth approximation to this quantile is $$ q_{n,p}\approx z_p+\frac{\gamma}{6\sqrt{n}}(z_p^2-1). $$

The correction is not simply the CDF correction evaluated at $z_p$. Locally, a change in $x$ changes $\Phi(x)$ at rate $\phi(x)$, so the CDF discrepancy must be divided by that slope. The sign also reverses: if the CDF correction at $z_p$ is positive, the quantile must move left to bring the CDF back to $p$.

Theorem: Suppose $F_n(x)=\Phi(x)+n^{-1/2}a(x)+r_n(x)$, where $a$ is continuous and $\sup_x|r_n(x)|=o(n^{-1/2})$. For a fixed $p\in(0,1)$, if $z_p=\Phi^{-1}(p)$, then $$ q_{n,p}=z_p-\frac{a(z_p)}{\phi(z_p)\sqrt{n}}+o(n^{-1/2}). $$ For the first-order Edgeworth CDF above, this gives $q_{n,p}=z_p+\gamma(z_p^2-1)/(6\sqrt{n})+o(n^{-1/2})$.

Why the Inversion Works

The proof checks the CDF just to either side of the proposed quantile. These two checks bracket the generalized inverse quantile, so no density for $F_n$ is required.

Propose the correction.
Set $z=z_p$ and $c=-a(z)/\phi(z)$. We will show that $q_{n,p}=z+c/\sqrt{n}+o(n^{-1/2})$.

Evaluate on either side.
Fix any $\delta>0$ and set $x_n^\pm=z+(c\pm\delta)/\sqrt{n}$. Taylor expansion at $z$ gives $\Phi(x_n^\pm)=p+\phi(z)(c\pm\delta)/\sqrt{n}+o(n^{-1/2})$. Continuity of $a$ gives $a(x_n^\pm)=a(z)+o(1)$.

Determine the CDF signs.
Substituting these expansions and using $\phi(z)c+a(z)=0$ yields $F_n(x_n^\pm)=p\pm\phi(z)\delta/\sqrt{n}+o(n^{-1/2})$. Since $\phi(z)>0$, eventually $F_n(x_n^-)<p<F_n(x_n^+)$.

Bracket the quantile.
Monotonicity of $F_n$ and the definition of its generalized inverse imply $x_n^-\leq q_{n,p}\leq x_n^+$. This holds for every fixed $\delta>0$, so $\sqrt{n}(q_{n,p}-z)\to c$. Therefore $q_{n,p}=z+c/\sqrt{n}+o(n^{-1/2})$. $\square$

Worked Quantile Calculations

Worked Example: A One-Sided Quantile for Exponential Summands

Let $X_i$ be iid exponential random variables with mean 1, and take $n=100$. For an exponential variable, $\mu=1$, $\sigma=1$, and $\gamma=2$. We approximate the 95th percentile of $Z_{100}=(\sum_i X_i-100)/10$.

The standard normal 95th percentile is $z_{0.95}\approx1.644854$. Its squared value is approximately $2.70554$, so $z_{0.95}^2-1\approx1.70554$. The skewness correction is

\frac{2}{6\sqrt{100}}(1.70554) =\frac{1}{30}(1.70554) \approx 0.05685.

Thus the approximate standardized quantile is $1.644854+0.05685\approx1.70170$. On the original sum scale, the corresponding 95th-percentile approximation is $100+10(1.70170)\approx117.017$. The positive correction is consistent with the right skewness of the exponential distribution: its upper quantile is shifted to the right of the normal reference value.

Worked Example: A Central Interval for Exponential Summands

Use the same exponential model and $n=100$, but now approximate the central 95% interval for $Z_{100}$. Its endpoint probabilities are $0.025$ and $0.975$, with normal quantiles approximately $-1.959964$ and $1.959964$.

At either endpoint, $z_p^2-1\approx1.959964^2-1\approx2.84146$. Since the skewness factor is $2/(6\sqrt{100})=1/30$, both endpoints receive the same correction:

\frac{1}{30}(2.84146)\approx0.09472.

The approximate lower and upper quantiles are therefore $-1.959964+0.09472\approx-1.86525$ and $1.959964+0.09472\approx2.05468$. Both move right. Their difference remains approximately $2.05468-(-1.86525)=3.91993$, the same as the normal interval's width to this order. The first-order skewness correction shifts the center of this interval, rather than changing its width.

Worked Example: The Median Under Negative Skewness

Let $E_i$ be iid exponential variables with mean 1, and set $X_i=1-E_i$. Then $E[X_i]=0$, $\operatorname{Var}(X_i)=1$, and the skewness is $\gamma=-2$, since reflection about zero reverses the sign of the third centered moment. Take $n=36$ and consider the median, for which $p=0.5$ and $z_{0.5}=0$.

The correction is

\frac{-2}{6\sqrt{36}}(0^2-1) =\frac{-2}{36}(-1) =\frac{1}{18} \approx0.05556.

Thus the approximate median of $\sum_i X_i/6$ is $0.05556$, rather than zero. This rightward median shift is consistent with negative skewness. The mean remains zero: the correction concerns a quantile, not a change in the mean.

Interpreting the Correction and Its Limits

For positive skewness, the correction $\gamma(z_p^2-1)/(6\sqrt{n})$ is positive when $|z_p|>1$, so sufficiently far-out lower and upper quantiles both shift right. For $|z_p|<1$, the correction is negative. In particular, the median correction is $-\gamma/(6\sqrt{n})$. These signs describe the first-order adjustment; they do not guarantee that the approximation is accurate for every sample size or tail probability.

The formula is useful for approximate critical values and interval endpoints when the first-order Edgeworth CDF approximation is appropriate. Its quantile precision is local: the CDF's slope $\phi(z_p)$ is small far into the tails, which makes inversion more sensitive to CDF error there. Also, the derivation applies when the stated CDF expansion holds. For lattice-valued sums, the CDF can have jumps that require additional treatment; the smooth formula should not be applied to such a case without checking its assumptions.

Common Mistakes

Using the CDF correction as the quantile correction. The CDF term is $\gamma(1-z_p^2)\phi(z_p)/(6\sqrt{n})$. Quantile inversion divides by the local slope $\phi(z_p)$ and reverses the sign, giving $\gamma(z_p^2-1)/(6\sqrt{n})$.
Assuming every quantile moves in the direction of the skewness. The factor $z_p^2-1$ changes sign. Near the center, positive skewness moves the median left; farther into either tail, it moves the quantile right.
Claiming an $O(n^{-1})$ quantile error from a first-order expansion alone. The stated uniform CDF remainder $o(n^{-1/2})$ justifies a quantile remainder $o(n^{-1/2})$, not $O(n^{-1})$. A sharper claim needs a sharper expansion and suitable conditions.
Confusing a quantile shift with a mean shift. The adjustment describes the location of a specified probability point. It does not alter the centering or mean of the standardized sum.

Key takeaway: To obtain a first-order Edgeworth quantile, evaluate the CDF correction at the normal quantile, divide by the normal density there, and reverse the sign. For a standardized iid sum, the result is $z_p+\gamma(z_p^2-1)/(6\sqrt{n})$, provided the relevant CDF expansion applies.

Check Your Understanding

Use the inversion argument and the worked examples to answer these questions.

For a standardized sum with skewness $\gamma$, write the first-order correction to its $p$-quantile in terms of $z_p$.
Why does quantile inversion divide the CDF correction by $\phi(z_p)$?
For positive skewness, what is the sign of the median correction, and how does it compare with the correction at a quantile satisfying $z_p^2>1$?
For a central interval with normal endpoints $-z$ and $z$, explain why the first-order corrections are equal and what that implies about its width.
What order of quantile remainder follows from a uniform CDF remainder of $o(n^{-1/2})$, and why is that not enough to claim an $O(n^{-1})$ remainder?