From a Density Approximation to a CDF Approximation
An Edgeworth density expansion describes how the density of a standardized sum differs from the standard normal density. Probabilities such as \(P(Z_n \leq x)\) and \(P(Z_n > x)\), however, are areas under that density. To use the expansion for these probabilities, each Hermite-polynomial correction must be integrated, with its sign and boundary term handled carefully.
The key feature is that a Gaussian density multiplied by a Hermite polynomial is, up to a sign, a derivative of the same density multiplied by a lower-order Hermite polynomial. Consequently, integration reduces the polynomial's order by one and produces a boundary term at the evaluation point.
The Density Expansion and Its Integrated Form
Let \(X_1,\ldots,X_n\) be independent and identically distributed with mean \(\mu\), variance \(\sigma^2>0\), and finite moments sufficient for the density expansion below. Define the standardized sum \(Z_n=\sum_{i=1}^n(X_i-\mu)/(\sigma\sqrt n)\). Write \(\gamma_1=E[(X_1-\mu)^3]/\sigma^3\) for the standardized skewness and \(\gamma_2=E[(X_1-\mu)^4]/\sigma^4-3\) for the excess kurtosis. The \(-3\) in \(\gamma_2\) makes it the fourth cumulant on the standardized scale.
Use the probabilists' Hermite polynomials \(H_k(x)=(-1)^k\phi(x)^{-1}\phi^{(k)}(x)\), where \(\phi\) and \(\Phi\) are the standard normal density and CDF. The polynomials needed here include \(H_2(x)=x^2-1\), \(H_3(x)=x^3-3x\), \(H_4(x)=x^4-6x^2+3\), and \(H_5(x)=x^5-10x^3+15x\). In this convention, a second-order Edgeworth density expansion has the form below.
The integrated remainder condition is one way to make the transfer from density to CDF precise; it is not a claim that every local density expansion automatically has this property. The calculation below is conditional on the stated expansion. Its central identity follows directly from the definition of the Hermite polynomials.
Why Integration Lowers the Hermite Order
By definition, \(\phi(x)H_k(x)=(-1)^k\phi^{(k)}(x)\). Therefore, for \(k\geq 1\), \((\phi(x)H_{k-1}(x))'=-\phi(x)H_k(x)\). The boundary term at \(-\infty\) vanishes because every derivative of \(\phi\) is a polynomial times \(\phi\), and a polynomial times a Gaussian density tends to zero as \(x\) tends to either infinity.
Here is the full derivation of the CDF formula. Integrate the density expansion from \(-\infty\) to \(x\). The normal term integrates to \(\Phi(x)\). Applying the identity to the \(H_3\), \(H_4\), and \(H_6\) terms gives, respectively, \(-\phi(x)H_2(x)\), \(-\phi(x)H_3(x)\), and \(-\phi(x)H_5(x)\). Finally, the integrated remainder is uniformly bounded: \(\left|\int_{-\infty}^{x}r_n(t)\,dt\right|\leq\int_{\mathbb R}|r_n(t)|\,dt=o(n^{-1})\). Thus the CDF formula follows, including its uniform remainder. Since \(P(Z_n>x)=1-P(Z_n\leq x)\), taking the complement reverses the signs of all the corrections and gives the upper-tail formula. \(\square\)
Worked Example: A Second-Order Approximation to an Exponential-Sum Tail
Let \(X_i\) be independent exponential random variables with rate 1, and set \(Z_n=(\sum_{i=1}^n X_i-n)/\sqrt n\). Approximate \(P(Z_{100}>1.5)\) using the second-order formula. An exponential variable with rate 1 has mean 1, variance 1, standardized skewness \(\gamma_1=2\), and excess kurtosis \(\gamma_2=6\).
At \(x=1.5\), the needed Hermite values are \(H_2(1.5)=1.5^2-1=1.25\), \(H_3(1.5)=1.5^3-3(1.5)=-1.125\), and \(H_5(1.5)=1.5^5-10(1.5^3)+15(1.5)=-3.65625\). The correction inside the braces in the upper-tail formula is therefore \(2(1.25)/(6\sqrt{100})+[6(-1.125)/24+4(-3.65625)/72]/100\).
The first-order term is \(2(1.25)/60=0.0416667\). The two second-order terms are \(6(-1.125)/2400=-0.0028125\) and \(4(-3.65625)/7200=-0.00203125\). Their sum with the first-order term is \(0.0368229\). Since \(\phi(1.5)\approx0.1295176\), the density-correction contribution to the upper tail is approximately \(0.1295176(0.0368229)=0.00477\).
Using \(1-\Phi(1.5)\approx0.0668072\), the resulting approximation is \(P(Z_{100}>1.5)\approx0.0668072+0.00477=0.07158\). The positive correction is consistent with the positive skewness of the exponential distribution at this threshold. This calculation applies the smooth Edgeworth formula; it does not include a lattice correction, which is relevant for sums with discrete distributions.
Signs, Endpoints, and Boundary Terms
For a CDF, the integral runs from \(-\infty\) up to \(x\). The boundary at \(-\infty\) vanishes, leaving \(-\phi(x)H_{k-1}(x)\). For an upper tail, the integral runs from \(x\) to \(+\infty\); the boundary at \(+\infty\) vanishes instead, leaving \(+\phi(x)H_{k-1}(x)\). The two correction signs are opposite because the CDF and upper tail are complements.
A useful check is that each correction integrates to zero over the entire real line: the boundary expression \(\phi(x)H_{k-1}(x)\) vanishes at both ends. Thus the density corrections do not change total mass. But this does not mean the CDF correction has the same sign everywhere: the Hermite polynomials change sign, and the sign also depends on the skewness and kurtosis coefficients.
Worked Example: The First-Order Correction at the Center
For the standardized exponential sum above, find the first-order Edgeworth approximation to \(P(Z_{25}\leq0)\). At \(x=0\), \(\Phi(0)=1/2\), \(\phi(0)=1/\sqrt{2\pi}\approx0.3989423\), and \(H_2(0)=-1\).
The first-order CDF formula gives \(P(Z_{25}\leq0)\approx\frac12-\phi(0)\frac{\gamma_1}{6\sqrt{25}}H_2(0)\). Since \(\gamma_1=2\), the product multiplying \(H_2(0)\) is \(2/30=1/15\). Hence \(P(Z_{25}\leq0)\approx\frac12-(0.3989423)(1/15)(-1)=0.526596\).
The positive skewness produces a positive first-order correction at the center. In the second-order formula, \(H_3(0)=H_5(0)=0\), so the displayed second-order terms also contribute zero at this particular threshold.
Worked Example: A Symmetric Interval and Opposite Endpoint Corrections
Suppose the summands are uniform on \([-\sqrt3,\sqrt3]\), so each has mean 0 and variance 1. The distribution is symmetric, giving \(\gamma_1=0\), and its fourth standardized moment is \(9/5\), so its excess kurtosis is \(\gamma_2=9/5-3=-6/5\). Use the second-order formula to approximate \(P(-1<Z_{25}\leq1)\).
At \(x=1\), \(H_3(1)=1-3=-2\). The CDF correction is \(-\phi(1)[\gamma_2 H_3(1)/(24\cdot25)]\). The expression in square brackets is \((-6/5)(-2)/600=0.004\), so \(P(Z_{25}\leq1)\approx\Phi(1)-0.004\phi(1)\).
At \(x=-1\), \(H_3(-1)=-1+3=2\), while \(\phi(-1)=\phi(1)\). The expression in square brackets is \((-6/5)(2)/600=-0.004\), giving \(P(Z_{25}\leq-1)\approx\Phi(-1)+0.004\phi(1)\). Subtract the latter approximation from the former: \(P(-1<Z_{25}\leq1)\approx\Phi(1)-\Phi(-1)-0.008\phi(1)\). Using \(\Phi(1)-\Phi(-1)\approx0.6826895\) and \(\phi(1)\approx0.2419707\), this is \(0.6826895-0.0019358\approx0.680754\). The interval's two endpoint corrections reinforce when the lower CDF is subtracted; dropping either endpoint would give the wrong interval correction.
Common Mistake: Reusing the CDF Sign for the Upper Tail
A common error is to write the upper-tail approximation with the same minus sign as the CDF correction. For example, integrating \(\phi(t)H_k(t)\) from \(x\) to \(+\infty\) and writing \(-\phi(x)H_{k-1}(x)\) overlooks which endpoint remains after integration. The correct calculation is \(\int_x^\infty\phi(t)H_k(t)\,dt=[-\phi(t)H_{k-1}(t)]_x^\infty=+\phi(x)H_{k-1}(x)\). Alternatively, take the complement of the CDF approximation; this also reverses every correction sign.
A second mistake is to integrate only the polynomial part and omit the Gaussian factor's boundary contribution. Hermite corrections are not ordinary polynomial corrections by themselves: their integrals are evaluated through the derivative of \(\phi H_{k-1}\). Retaining that factor is what makes the correction vanish at the tails and preserves the total mass of the expansion.
Check Your Understanding
Use the integration identity and the CDF or upper-tail formula to answer these questions.
- What is \(\int_{-\infty}^x \phi(t)H_4(t)\,dt\) in terms of \(\phi(x)\) and a Hermite polynomial?
- Why does the upper-tail correction have the opposite sign from the CDF correction?
- At \(x=0\), which second-order Hermite terms in the CDF expansion vanish, and why?
- What condition on the integrated density remainder ensures a uniform \(o(n^{-1})\) CDF remainder?
- For a CDF correction with a positive value at \(x\), what is the corresponding correction's sign in the upper-tail probability?