The First-Order Skewness Correction

Why Skewness Changes a Normal Approximation

The central limit theorem gives a normal approximation for a standardized sum of independent, identically distributed observations. But convergence to normality does not say how the finite-sample distribution departs from the normal curve. The first-order Edgeworth correction describes the leading departure when the summands have nonzero skewness.

The correction is useful because it distinguishes two distributions with the same mean and variance but different asymmetry. It is not a new exact distribution: it is an asymptotic adjustment, of order $n^{-1/2}$, to the normal density and CDF. Its sign depends on both the summand skewness and the point at which a probability is evaluated.

Standardized Sums and the Skewness Term

Let $X_1,\ldots,X_n$ be iid with mean $\mu$, variance $0<\sigma^2<\infty$, and finite third absolute moment. Write $Y_i=(X_i-\mu)/\sigma$, so $E[Y_i]=0$ and $E[Y_i^2]=1$. The standardized sum is

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

The standardized third cumulant of a summand is its skewness, denoted here by $\gamma=E[Y_i^3]$. The density and CDF of a standard normal variable are $\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}$ and $\Phi(x)$.

Definition: The first-order Edgeworth approximation for $W_n$ has density and CDF forms $$ g_n(x)\approx \phi(x)\left[1+\frac{\gamma}{6\sqrt n}H_3(x)\right], \qquad F_n(x)\approx \Phi(x)+\frac{\gamma}{6\sqrt n}(1-x^2)\phi(x), $$ where $H_3(x)=x^3-3x$ is the third probabilists’ Hermite polynomial. The correction is of order $n^{-1/2}$; it vanishes at this order when $\gamma=0$.

These are asymptotic approximations, not identities. For a CDF expansion with a uniform error smaller than order $n^{-1/2}$, a standard sufficient setting is iid summands with finite fourth moment and a nonlattice distribution satisfying Cramér’s condition. The density expansion requires additional regularity that permits density-level Fourier inversion. The algebra below identifies the correction; those regularity conditions control the remainders.

Deriving the First-Order Term

The characteristic function of $Y_i$ has the local expansion $\psi_Y(u)=1-u^2/2+\gamma(iu)^3/6+o(u^3)$ as $u\to0$. Equivalently, its logarithm is $\log\psi_Y(u)=-u^2/2+\gamma(iu)^3/6+o(u^3)$. Independence gives the characteristic function of $W_n$ as $\psi_Y(t/\sqrt n)^n$.

Derivation: For fixed $t$, substitute $u=t/\sqrt n$ into the logarithm and multiply by $n$: $$ n\log\psi_Y(t/\sqrt n) =-\frac{t^2}{2}+\frac{\gamma(it)^3}{6\sqrt n}+o(n^{-1/2}). $$ Exponentiating and retaining terms through order $n^{-1/2}$ gives $$ \psi_{W_n}(t) =e^{-t^2/2}\left[1+\frac{\gamma(it)^3}{6\sqrt n}\right]+o(n^{-1/2}). $$ The term $(it)^3$ becomes $H_3(x)\phi(x)$ under Fourier inversion. This follows from the Gaussian derivative identities: $\phi'''(x)=-H_3(x)\phi(x)$, while inversion of $(it)^3e^{-t^2/2}$ gives $-\phi'''(x)$. Thus the density correction is $+\gamma H_3(x)\phi(x)/(6\sqrt n)$. The local characteristic-function calculation determines the polynomial; suitable Fourier regularity is needed to turn it into a density approximation with a controlled remainder.

To obtain the CDF correction, integrate the density term from $-\infty$ to $x$. Since $H_2(x)=x^2-1$, differentiation gives

\frac{d}{dx}\{-\phi(x)(x^2-1)\} =x\phi(x)(x^2-1)-2x\phi(x) =\phi(x)(x^3-3x) =\phi(x)H_3(x).

The expression $-\phi(x)(x^2-1)$ tends to zero as $x\to-\infty$, so integration gives $\int_{-\infty}^x H_3(u)\phi(u)\,du=(1-x^2)\phi(x)$. This proves the stated CDF correction from the density term. Both corrections integrate or accumulate to zero at the appropriate extremes: the density adjustment has integral zero, and the CDF adjustment vanishes as $x\to\pm\infty$.

Worked Examples

Worked Example: Positive Skewness and an Upper Tail

Suppose the $X_i$ are exponential with mean $1$. Their variance is $1$, and their skewness is $\gamma=2$. For $n=25$, the first-order coefficient is $\gamma/(6\sqrt n)=2/(6\cdot5)=1/15$. At $x=2$, $H_3(2)=2^3-3(2)=2$, so the density approximation is

g_{25}(2)\approx\phi(2)\left(1+\frac{2}{15}\right) =0.053991(1.133333)\approx0.06119.

For the CDF, the correction is $(1-2^2)\phi(2)/15=-3\phi(2)/15\approx-0.010798$. Since $\Phi(2)\approx0.977250$, the corrected CDF is about $0.966452$, and the corresponding upper tail is about $0.033548$. The normal upper tail alone is about $0.022750$. Here positive skewness increases the estimated upper-tail probability at $x=2$.

Worked Example: Negative Skewness at the Center

Let each $X_i$ have a beta distribution with parameters $(5,2)$. Its mean is $5/7$, its variance is $10/[7^2\cdot8]=5/196$, and its skewness is \[ \gamma=\frac{2(2-5)\sqrt{8}}{9\sqrt{10}}\approx-0.5963. \] For $n=25$, the CDF correction coefficient is $\gamma/30\approx-0.01988$. At $x=0$, $\Phi(0)=0.5$ and $(1-0^2)\phi(0)=0.398942$. Therefore

F_{25}(0)\approx0.5+(-0.01988)(0.398942)\approx0.49207.

The standardized sum’s CDF at its mean is estimated to be below one half. This illustrates the central effect of negative skewness: the distribution has relatively more weight above its mean, balanced by a longer left tail. The approximation does not say that every probability is shifted in the same direction.

Worked Example: Zero Skewness and the First-Order Term

If $X_i$ is uniform on $(0,1)$, its mean is $1/2$, its variance is $1/12$, and its skewness is zero by symmetry about $1/2$. Thus $\gamma/(6\sqrt n)=0$ for every $n$. For $n=36$, both first-order formulas reduce to the normal approximation:

g_{36}(x)\approx\phi(x),\qquad F_{36}(x)\approx\Phi(x).

This only means that the skewness correction is absent. The finite-sample distribution need not be exactly normal; terms of smaller asymptotic order can still matter.

How to Read the Direction of the Correction

For a CDF probability at $x$, the correction has the sign of $\gamma(1-x^2)$, because $\phi(x)>0$. Thus, when $\gamma>0$, it raises the CDF for $|x|<1$ and lowers it for $|x|>1$; for $\gamma<0$, these directions reverse. At $x=\pm1$, the first-order CDF correction is zero.

For an upper-tail probability, subtract the CDF from one. Its correction is \[ -\frac{\gamma}{6\sqrt n}(1-x^2)\phi(x) =\frac{\gamma}{6\sqrt n}(x^2-1)\phi(x). \] Consequently, positive skewness increases the first-order upper-tail estimate when $|x|>1$, but decreases it when $|x|<1$. The tail statement concerns the standardized sum and a specified threshold; skewness is not a uniform shift of the entire distribution.

The density correction has a different sign pattern, controlled by $H_3(x)=x(x^2-3)$. For example, positive skewness lowers the approximating density for $0<x<\sqrt3$, but raises it for $x>\sqrt3$. A density adjustment redistributes mass across the line, while the CDF adjustment records the accumulated change to the left of a threshold.

Common Pitfall and Scope

Reversing the CDF correction’s sign. The antiderivative is $-\phi(x)(x^2-1)$, whose derivative is $+\phi(x)(x^3-3x)$. Therefore the CDF correction is $\gamma(1-x^2)\phi(x)/(6\sqrt n)$, not its negative.
Calling the approximation an exact density. The polynomial factor can make $\phi(x)[1+\gamma H_3(x)/(6\sqrt n)]$ negative in a far tail. A function with negative values cannot be a probability density. The expansion is an asymptotic approximation, generally useful near the central range, not a globally valid replacement distribution.
Using skewness alone to predict every tail. The sign also depends on the threshold through $1-x^2$ for a CDF, or $x^2-1$ for an upper tail. State the event and evaluate the correction before describing its direction.
Ignoring lattice effects. For lattice-valued summands, a smooth CDF expansion can require an additional lattice correction. The nonlattice formula should not be applied to a lattice distribution as though that term were absent.

The first-order term matters when a normal approximation is close but its remaining error is consequential, such as when comparing tail probabilities under positively and negatively skewed models. It also supplies a diagnostic: if skewness is substantial and the threshold lies in a region where the correction is appreciable, the unadjusted normal approximation may miss an important asymmetry. The size and reliability of that adjustment depend on $n$, the threshold, and the regularity of the summand distribution.

Check Your Understanding

Use the formulas and sign rules above to answer these questions.

For $\gamma>0$, is the first-order CDF correction positive or negative at $x=0.5$? What about at $x=2$?
Write the first-order correction to an upper-tail probability $P(W_n>x)$, starting from the CDF formula.
Why does a zero third cumulant remove this correction without implying an exactly normal finite-sample sum?
For the density correction, evaluate the sign of $H_3(x)$ at $x=-1$, $x=1$, and $x=2$.
Why should the nonlattice CDF formula not be used without qualification for lattice-valued summands?