Why Hermite Polynomials Appear in Density Corrections
An Edgeworth density expansion starts from a normal approximation and adds terms that encode departures from normality. In Fourier inversion, those departures appear as powers of the Fourier variable multiplying the normal characteristic function. In the density, each such power becomes a derivative of the normal density, and those derivatives can be written as a normal density multiplied by a polynomial. The relevant polynomials are the probabilists’ Hermite polynomials.
The connection is useful because it separates two tasks: Fourier inversion determines which Hermite polynomial accompanies a term, while cumulant expansion determines that term’s coefficient. Keeping these tasks distinct also helps prevent sign errors.
The first few polynomials are \(H_0(x)=1\), \(H_1(x)=x\), \(H_2(x)=x^2-1\), and \(H_3(x)=x^3-3x\). The adjective “probabilists’” distinguishes this convention from a differently scaled convention used in some areas of mathematics and physics.
Recurrences and Derivatives
The generating function provides a compact way to derive two identities used throughout density expansions. Differentiating it with respect to \(x\) gives \(\sum_{r\geq 0}H_r'(x)t^r/r!=t e^{tx-t^2/2}\), so comparison of coefficients yields \(H_r'=rH_{r-1}\) for \(r\geq 1\). Differentiating with respect to \(t\) gives \((x-t)e^{tx-t^2/2}=\sum_{r\geq 0}H_{r+1}(x)t^r/r!\). Comparing coefficients and using the derivative identity yields the recurrence below.
Here is a direct verification of the derivative identity from the recurrence, including the initial cases. From \(H_1(x)=x\), \(H_1'(x)=1=H_0(x)\). Since \(H_2(x)=x^2-1\), \(H_2'(x)=2x=2H_1(x)\). Now suppose \(r\geq 2\) and the derivative identities hold at orders \(r\) and \(r-1\). Differentiating \(H_{r+1}=xH_r-H_r'\) gives
The recurrence at order \(r-1\) says \(H_r=xH_{r-1}-(r-1)H_{r-2}\). Substitution therefore gives \(H_{r+1}'=(r+1)H_r\), completing the induction. The generating function and the recurrence also show that \(H_r\) has degree \(r\) and leading coefficient \(1\).
Worked Example: Building the First Hermite Polynomials
Use the recurrence \(H_{r+1}=xH_r-rH_{r-1}\) to compute the polynomials through order four.
Start with \(H_0=1\) and \(H_1=x\). At \(r=1\), \(H_2=xH_1-H_0=x^2-1\). At \(r=2\), \(H_3=xH_2-2H_1=x(x^2-1)-2x=x^3-3x\).
At \(r=3\), \(H_4=xH_3-3H_2=x(x^3-3x)-3(x^2-1)=x^4-6x^2+3\). As a check, differentiating these gives \(H_2'=2H_1\), \(H_3'=3H_2\), and \(H_4'=4H_3\).
Fourier Inversion Produces the Hermite Factor
Use the characteristic-function convention \(\varphi_X(t)=E[e^{itX}]\). For an integrable characteristic function, Fourier inversion gives a density in the form \(f(x)=(2\pi)^{-1}\int_{\mathbb R}e^{-itx}\varphi_X(t)\,dt\). The standard normal characteristic function is \(e^{-t^2/2}\). Differentiating its inverse transform \(r\) times with respect to \(x\) shows that the inverse transform of \((-it)^r e^{-t^2/2}\) is \(\phi^{(r)}(x)\). Consequently, the inverse transform of \((it)^r e^{-t^2/2}\) is \((-1)^r\phi^{(r)}(x)=H_r(x)\phi(x)\).
Proof. The standard normal inversion formula is \(\phi(x)=(2\pi)^{-1}\int_{\mathbb R}e^{-itx}e^{-t^2/2}\,dt\). The integrand, and each of its derivatives with respect to \(x\), is dominated in absolute value by an integrable function of the form \(|t|^r e^{-t^2/2}\). Differentiation under the integral is therefore valid, giving \(\phi^{(r)}(x)=(2\pi)^{-1}\int_{\mathbb R}(-it)^r e^{-itx}e^{-t^2/2}\,dt\). Multiplying by \((-1)^r\) changes \((-it)^r\) to \((it)^r\). The definition of \(H_r\) then gives the identity. \(\square\)
Worked Example: Checking the Fourier Sign at Order Four
Suppose a Fourier integrand contains \(c(it)^4e^{-t^2/2}\), where \(c\) is a constant. Since \(r=4\) is even, the inverse transform is \(cH_4(x)\phi(x)\). Using the polynomial computed above, this is \(c(x^4-6x^2+3)\phi(x)\).
The sign can also be checked by differentiating: \(\phi^{(4)}(x)=(x^4-6x^2+3)\phi(x)\), and \((-1)^4\phi^{(4)}=\phi^{(4)}\). At odd orders the factor \((-1)^r\) still leads to \(H_r\phi\); omitting it or changing the Fourier convention without adjusting the calculation can produce an incorrect sign.
How Standardized Cumulants Set the Coefficients
Let \(X_1,\ldots,X_n\) be independent and identically distributed, with mean \(\mu\), variance \(\sigma^2>0\), and cumulants \(\kappa_r\) through the orders being used. Define \(Y_i=(X_i-\mu)/\sigma\), and write \(\lambda_r=\kappa_r/\sigma^r\) for the standardized cumulant of order \(r\). In particular, \(\lambda_1=0\) and \(\lambda_2=1\). The standardized sum is \(Z_n=n^{-1/2}\sum_{i=1}^nY_i\).
For example, if the moment generating function exists in a neighborhood of zero, the cumulants are the coefficients in the Taylor expansion of its logarithm. Equivalently, near zero the logarithm of the characteristic function of \(Y_i\) has terms \(\lambda_r(it)^r/r!\). Independence multiplies characteristic functions, so the logarithm for \(Z_n\) is \(n\) times the one-observation logarithm evaluated at \(t/\sqrt n\). Its formal expansion is
This expression explains the bookkeeping: an order-\(r\) standardized cumulant contributes a power \((it)^r\), which Fourier inversion turns into \(H_r(x)\phi(x)\), and its sample-size factor is \(n^{-(r-2)/2}\). Exponentiating the cumulant terms creates products, and a product of powers of \(it\) becomes a Hermite polynomial whose order is the sum of those powers.
This is a coefficient and inversion rule, not by itself a guarantee that a truncated series approximates a particular density to a specified error. To interpret a finite expansion as a valid density approximation requires the regularity and remainder conditions relevant to the distribution and expansion under study.
Worked Example: The Term From a Fifth Standardized Cumulant
Consider the single fifth-cumulant term in the log characteristic function of \(Z_n\). Its coefficient is \(\lambda_5(it)^5/(5!n^{3/2})\), because \((5-2)/2=3/2\). In the exponential expansion, this term appears linearly with the same coefficient.
Since \(5!=120\), Fourier inversion gives \(\lambda_5 n^{-3/2}H_5(x)\phi(x)/120\). The recurrence gives \(H_5=xH_4-4H_3=x^5-10x^3+15x\). Thus this contribution is \(\lambda_5 n^{-3/2}(x^5-10x^3+15x)\phi(x)/120\).
This calculation identifies one contribution only; products of lower-order cumulant terms can occur at the same power of \(n^{-1/2}\) and must be included when forming a complete expansion at that order.
Worked Example: A Product of Cumulant Terms
In the exponential expansion, select one third-cumulant term and one fourth-cumulant term. Their product has sample-size factor \(n^{-1/2}n^{-1}=n^{-3/2}\), Fourier power \((it)^3(it)^4=(it)^7\), and coefficient \(\lambda_3\lambda_4/(3!4!)\). There is no additional factorial from repeated copies because each selected term occurs once.
Since \(3!4!=6\cdot24=144\), inversion gives \(\lambda_3\lambda_4 n^{-3/2}H_7(x)\phi(x)/144\). Using the recurrence, \(H_7=xH_6-6H_5=x^7-21x^5+105x^3-105x\).
The example illustrates why both indices matter: the sample-size order is determined by \(3-2\) and \(4-2\), while the Hermite order is \(3+4=7\).
Orthogonality and a Useful Check
Hermite polynomials are orthogonal with respect to the standard normal density. This fact gives a quick check on the form of density corrections: every nonconstant Hermite term has integral zero against the normal density.
Proof. Suppose first that \(m\geq k\). From the definition, \(H_m\phi=(-1)^m\phi^{(m)}\). Integrating by parts \(m\) times transfers the derivatives from \(\phi\) to \(H_k\), yielding \(\int H_mH_k\phi=\int H_k^{(m)}\phi\). All boundary terms vanish: each derivative of \(\phi\) is a polynomial times \(\phi\), and a polynomial times the Gaussian density tends to zero at both tails. If \(m>k\), the \(m\)th derivative of the degree-\(k\) polynomial \(H_k\) is zero. If \(m=k\), its leading coefficient is \(1\), so \(H_k^{(k)}=k!\), and the integral is \(k!\int\phi=k!\). When \(k>m\), interchange the roles of \(m\) and \(k\) and use the symmetry of the integral. \(\square\)
Common Pitfall
A common error is to match \((it)^r\) with \(\phi^{(r)}\) directly and overlook the sign. With the stated Fourier convention, differentiating the inverse transform produces \((-it)^r\), not \((it)^r\). The correction is \((it)^r\mapsto(-1)^r\phi^{(r)}=H_r\phi\). In particular, the sign factor is essential at odd orders. A second useful check is that \(H_r\phi\) integrates to zero for every \(r\geq1\), by the orthogonality result with \(k=0\); a proposed nonconstant correction that changes total mass has an error or is missing compensating terms.
Check Your Understanding
Use the definitions and coefficient rule to answer the following.
- Use the recurrence to calculate \(H_3(x)\) from \(H_1(x)\) and \(H_2(x)\).
- Under the stated Fourier convention, what is the inverse transform of \((it)^6e^{-t^2/2}\)?
- For a single standardized cumulant of order \(r\), what power of \(n\) and what Hermite order accompany its density term?
- One third-cumulant term and one fifth-cumulant term are multiplied in the exponential expansion. What is the resulting power of \(n\), Hermite order, and cumulant coefficient, including factorials?
- Why must a term proportional to \(H_r(x)\phi(x)\), with \(r\geq1\), integrate to zero?