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The bispectrum is the Fourier transform of a third-order cumulant or correlation function, with the convention specified for the process or field. It measures third-order dependence among frequency triples; the primordial bispectrum is a cosmological random-field application.

For an independent binomial distribution count $N\sim \mathrm{Binomial}(n,p)$, the aggregate can also be represented as ${\sum }_{j=1}^n{B}_j{X}_j$, with independent parameter-$p$ Bernoulli random variables $B_j$ and independent claim sizes. Its cumulant-generating function is $n\log (1-p+p{M}_X(t))$. Its third cumulant is $n(p{m}_3-3p^2{m}_1{m}_2+2p^3{m}_1^3)$ in severity moments. Constant positive claims $d$ give $np(1-p)(1-2p)d^3$, which is negative when $p>1/2$. Thus positive claim sizes do not by themselves force a positive third central moment.

Take an independent geometric distribution count with $\mathbb{P}(N=k)=p(1-p{)}^k$, $k\ge 0$, and sum its independent positive claims. Its moment-generating function is the displayed composition wherever finite. With $q=1-p$ and severity moments $m_j$, the third cumulant is $(q/p)m_3+3(q/p{)}^2{m}_1{m}_2+2(q/p{)}^3{m}_1^3$, which is strictly positive for nonzero positive claims and $0<p<1$. The law has an atom of mass $p$ at zero. Its positive-count convention would instead add a factor $M_X(t)$ to the numerator, as in the geometric-sum moment-generating function.

Conditioning on the independent Poisson distribution count in a random sum gives $\mathbb{E}{e}^{tS}=\exp (\lambda (M_X(t)-1))$ wherever finite. Differentiation at zero gives the displayed cumulants when the relevant derivatives exist. The aggregate variance is the count rate times the raw second claim moment, not just the claim variance. Positive moment-generating function arguments need an exponential moment; finite first and second claim moments suffice for the aggregate mean and variance by conditioning or right derivatives at zero of the Laplace transform of a nonnegative random variable.

For $X_t={\sum }_{j\ge 0}{\psi }_j{\eta }_{t-j}$ driven by centered unit-variance iid noise with a finite fourth moment, put ${\kappa }_4=E{\eta }^4-3$ and $m_3=E{\eta }^3$. Independence givesThe sum of the two linear-quadratic cross covariances is $m_3{\sum }_{j\ge 0}({\psi }_j^2{\psi }_{j+|h|}+{\psi }_j{\psi }_{j+|h|}^2)$. The cumulant term disappears for Gaussian noise. Merely assuming strong white noise does not justify the Gaussian formula; a fourth moment is needed for the variance of the quadratic transform.

The $n$th cumulant is the coefficient of $t^n/n!$ in the cumulant-generating function. The first two cumulants are the expected value and variance; the third is the third central moment.

For a random sum of independent claims, conditioning on the count gives $M_S(t)=\mathbb{E}[M_X(t{)}^N]$ and therefore the displayed cumulant-generating function. Write $a_j={\kappa }_{N,j}$ and $b_j={\kappa }_{X,j}$. The chain rule gives ${\kappa }_{S,1}=a_1{b}_1$, ${\kappa }_{S,2}=a_1{b}_2+a_2{b}_1^2$, and ${\kappa }_{S,3}=a_1{b}_3+3a_2{b}_1{b}_2+a_3{b}_1^3$. In raw severity moments $m_j=\mathbb{E}{X}^j$, the third cumulant is $a_1{m}_3+3(a_2-a_1)m_1{m}_2+(a_3-3a_2+2a_1)m_1^3$. Positive exponential moments justify two-sided differentiation near zero; finite ordinary moments alone suffice for the corresponding left derivatives when claims are positive.

An Edgeworth expansion approximates a probability density function by a normal distribution multiplied by Hermite polynomials encoding higher cumulants. For a centered variable of variance ${\sigma }^2$ and third cumulant ${\kappa }_3$,The correction follows from inverse-transforming the cubic term of the cumulant-generating function. It is an asymptotic approximation near the bulk of the distribution; truncation can give a negative result in far tails.

For a trace-class operator $K$ on a Hilbert space, the Fredholm determinant is the convergent product over its eigenvalues, counted with algebraic multiplicity. For self-adjoint $K$, spectral theorem for compact self-adjoint operators reduces the product to a diagonal calculation. For $\Vert K\Vert <1$, $\log \det (I+K)={\sum }_{n\ge 1}(-1{)}^{n+1}\mathrm{tr}(K^n)/n$. This identity exposes the successive cumulants in a Gaussian quadratic exponential moment.

For centered fast Fourier modes and $V=u\int {\varphi }^4/4!$, the two-fast-field vertex is $(u/4)\int {\phi }^2{\eta }^2$. Its connected second cumulant uses $⟨{\eta }^2(x){\eta }^2(y){⟩}_c=2G_{>}(x-y{)}^2$, producing $-u^2\int {\phi }^2(x){\phi }^2(y)G_{>}(x-y{)}^2/16$. Extracting the local zero-external-momentum quartic coefficient gives $u^{\prime }=b^{4-D}(u-3u^2{I}_2/2)+\cdots$, where $I_2={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}K(p{)}^{-2}>0$. At $D=4$ its sign makes weak positive quartic coupling marginally irrelevant for increasing length scale, after tuning the critical mass. This local truncation omits generated derivative and higher-order operators; it is not the exact complete finite-shell effective action.

Where the moment-generating function $M_Y(t)=\mathbb{E}{e}^{tY}$ is finite in a neighborhood of zero, define the cumulant-generating function ${\kappa }_Y(t)=\log M_Y(t)$ and the $r$th cumulant by ${\kappa }_r(Y)={\kappa }_Y^{(r)}(0)$. Differentiation givessince $M_Y(0)=1$. Thus the first two cumulants are the mean and variance.

Conditional on $N=n$, independence gives $\mathbb{E}(e^{tT}\mid N=n)=M_V(t{)}^n$. Summing over the Poisson distribution yieldsHence the compound Poisson cumulants giveFor the moment statements, finite second moments suffice even when a moment-generating function does not exist near zero: conditioning gives $\mathrm{Var}T=\mathbb{E}N\mathrm{Var}V+\mathrm{Var}N(\mathbb{E}V{)}^2=\nu \mathbb{E}{V}^2$. This distinguishes the raw second moment from the variance of an individual claim.

For a single flood, the law of total expectation and law of total variance similarly giveCondition on the number of floods and apply the same laws once more. All floods have these moments and are independent of $M$, so the annual total satisfiesThe additional $\nu {\lambda }^2{\mu }^2$ term is the variation in the number of entire flood clusters; it would be lost by treating individual claims as an ordinary Poisson stream.

For the annual count $N={\sum }_{i=1}^M{N}_i$, its probability generating function isThis is the nested Poisson flood count. Differentiating, or conditioning on $M$, gives $\mathbb{E}N=\nu \lambda$ and $\mathrm{Var}N=\nu \lambda (1+\lambda )$. Thus $N$ is not Poisson when $\nu ,\lambda >0$, since its variance exceeds its mean. If either parameter is zero it is identically zero, a degenerate Poisson distribution.

Let $K_i$ be the number of valid claims in flood $i$. Independent validity marking gives $K_i\mid N_i\sim \mathrm{Binomial}(N_i,1-p)$. Its generating function isso Poisson thinning gives $\boxed{K_i\sim \mathrm{Poisson}(\lambda (1-p))}$. With validity independent also of claim amount, each claim contributes expected paid amount $(1-p)\mu$. Therefore the annual expected payment is $\boxed{\nu \lambda (1-p)\mu }$. The independent-mark assumption is needed for this multiplication; a size-dependent invalidity rule would require the joint expected value of amount and validity instead.

An Edgeworth expansion refines a central normal approximation using higher cumulants. Let $Y_i$ be independent and identically distributed, with mean $\mu$, positive variance ${\sigma }^2$, and cumulants ${\kappa }_r$. DefineThe logarithm of its characteristic function has the expansionExpanding the exponential adds the squared cubic term ${\gamma }_3^2(it{)}^6/(72n)$. Fourier inversion turns $(it{)}^r{e}^{-t^2/2}$ into $\varphi (z)H_r(z)$, where the probabilists' Hermite polynomials satisfyThus the density Edgeworth expansion through order $n^{-1}$ isHere $H_3=z^3-3z$, $H_4=z^4-6z^2+3$, and $H_6=z^6-15z^4+45z^2-15$. Since $(\varphi H_{r-1}{)}^{\prime }=-\varphi H_r$, the corresponding cumulative distribution function expansion isThese remainder statements require adequate moments and smooth nonlattice conditions allowing transform inversion and error control, not merely a formal cumulant expansion. The density version needs the corresponding local smoothness. A symmetric distribution has ${\gamma }_3=0$, removing the first correction. Truncation can produce negative densities or nonmonotone distribution functions far from the center; it is not automatically a uniformly reliable relative tail approximation. A lattice distribution needs its lattice correction.

A Laplace approximation instead approximates an integral dominated by a sharp maximum. Suppose $h$ has a unique interior maximizer $\hat{\theta }\in {\mathbb{R}}^d$, with $A=-h^{\prime \prime }(\hat{\theta })$ positive definite, and the integral is localized near that point. For smooth amplitude $g$,Set $\theta =\hat{\theta }+z/\sqrt{n}$. A Taylor expansion givesExpanding $g$ and integrating the resulting Gaussian integral givesUnder sufficient smoothness and localization the relative correction is $O(n^{-1})$: odd terms of order $n^{-1/2}$ integrate to zero over the limiting symmetric Gaussian region. Higher terms use derivatives of $h,g$ and Gaussian moments.

For a likelihood function with ${\ell }_n=nh$ and a smooth positive prior density, this gives the integrated likelihood approximationwhere $j=-{\ell }_n^{\prime \prime }$ is the observed information. Ratios of such integrals give posterior expectation by Laplace approximation; the same method is useful when integrating nuisance variables. Boundary maxima, singular Hessian matrices, nonlocalized tails and several dominant modes require altered formulas, with contributions summed when appropriate.

Both methods start with a quadratic Gaussian approximation and then retain higher-order terms. The Edgeworth series corrects a sampling distribution through its cumulants and Fourier inversion; Laplace's method corrects an integral through local derivatives at its dominant point. Neither should be confused with simply replacing a likelihood by a normalized posterior density.

Use Minkowski spacetime with signature $(+,-,\dots ,-)$ and $\hslash =1$, with a vacuum time-ordering prescription. At a regulator, define the normalized generating functionalThe normalized vacuum correlation functions of a time-ordered product areThe connected generating functional isThe logarithm selects connected source diagrams, or equivalently cumulants. With the assumed vanishing one-point function, derivatives of $Z=e^{iW}$ give the centered four-point cumulant decomposition:In detail, four differentiations either act on one connected block or form one of three pair partitions. Contributions with blocks of size one vanish. If $W_{ij}$ and $W_{1234}$ denote source derivatives at zero, the same relation is $G_{1234}=i{W}_{1234}-W_{12}{W}_{34}-W_{13}{W}_{24}-W_{14}{W}_{23}$, since $C_2=-i{W}_2$ and $C_4=i{W}_4$.

Let $\phi (x)=\delta W/\delta J(x)$ be the source-dependent mean field. On a locally invertible source-to-field branch, define the quantum effective action by the Minkowski Legendre transformIts variation is $\delta \Gamma =-\int J\delta \phi$, so $\delta \Gamma /\delta \phi =-J$. Differentiating the two inverse source-field maps yields the Minkowski inverse-Hessian relation for an effective action:Thus ${\Gamma }^{(2)}=-(W^{(2)}{)}^{-1}$. The inverse is understood at the regulator and on a nonsingular fluctuation sector; it is not an assertion that every source-field map is globally invertible.

For a free real scalar field, integrate by parts to writewhere repeated spacetime variables are integrated. Specify the propagator convention explicitly:Completing the Gaussian path integral around $\varphi =-A_{ϵ}^{-1}J$ cancels the source-independent determinant between numerator and denominator. ThereforeThe positive imaginary term in $A_{ϵ}$ damps the oscillatory Gaussian. Together with vacuum projection at the time boundaries, it selects the Feynman poles and the time-ordered, in-out correlation function rather than a retarded inverse. It can also be obtained by continuation from the Euclidean vacuum integral. A denominator convention without the numerator $i$ redistributes the displayed factors of $i$.

Finally $\phi =-A_{ϵ}^{-1}J$, hence $J=-A_{ϵ}\phi$. Substitution into the Legendre transform gives the vacuum-normalized effective action of a free scalar field:All connected functions beyond order two vanish and all proper vertices beyond the classical quadratic kernel vanish. Vacuum normalization fixes the additive constant; without it, the quantum effective action equals the classical action up to a field-independent determinant contribution.

Use the Probabilists' Hermite polynomials, rather than the differently scaled physicists' convention:Thus $H_2=x^2-1$, $H_3=x^3-3x$, and $H_5=x^5-10x^3+15x$. The cumulant-generating function of an exponential variable of mean $\theta$ is $-\log (1-\theta t)$ near zero, so its $r$th cumulant is $(r-1)!{\theta }^r$. Hence the standardized third and fourth cumulants are ${\rho }_3=2$ and ${\rho }_4=6$; the fourth is a cumulant, not the fourth raw central moment.

Write $F_n$ for the cumulative distribution function of the pivot in part (i). Its Edgeworth expansion, valid here because the exponential law is smooth, nonlattice and has exponential moments, isat the fixed endpoints $t=\pm z$. The coverage is exactly $F_n(z)-F_n(-z)$. Since $H_2$ and $\varphi$ are even, the $n^{-1/2}$ terms cancel; the two odd Hermite terms double. Consequently the symmetric Edgeworth coverage cancellation givesIn particular its coverage probability error is $\boxed{O(n^{-1})}$. This is a fixed-confidence-level expansion; it is not a uniform statement for normal cutoffs increasing with $n$.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

Let the summands have mean mu, variance ${\sigma }^2>0$, and cumulants ${\kappa }_j$. For the standardized sum $Z_n=(S_n-n\mu )/(\sigma \sqrt{n})$, expansion of its log-characteristic function gives a normal leading term and successively smaller cumulant terms. Fourier inversion turns these terms into Probabilists' Hermite polynomials. Under sufficient moments and a smooth nonlattice condition, the Edgeworth expansion beginsfor bounded z. Higher orders include the fourth cumulant and products of lower cumulants. This is a central expansion in powers of $n^{-1/2}$; it need not remain positive, normalized, or accurate in far tails after truncation.

For a saddlepoint density approximation, require the summand cumulant-generating function K to be finite on an interval about zero. At a sum value $s=ny$, find the interior saddle $\hat{t}$ satisfying $K^{\prime }(\hat{t})=y$. Exponential tilting makes the tilted mean y, so the tilted sum is approximated locally by its normal probability density at its mean. Undoing the tilt givesEquivalently this follows by deforming the transform-inversion contour through its stationary point. Suitable analytic and nonlattice regularity gives relative error $O(n^{-1})$ at fixed interior y. This approximation incorporates a nonquadratic tail exponent and is positive, while an Edgeworth truncation uses a normal probability density times a polynomial. It uses more information from the distribution and stronger exponential-moment assumptions. Its integral may differ from one by a higher-order amount, so positivity does not imply exact normalization.

For the centered Laplace distribution, $K(t)=t^2+t^4/2+\cdots$, hence ${\kappa }_2=2$ and ${\kappa }_3=0$. Its skewness vanishes, so the order-$n^{-1/2}$ Edgeworth correction is zero. With $z=s/\sqrt{2n}$, the requested expansion, omitting terms of order $n^{-1}$ in the standardized correction, isFor bounded z the exact probability density function is this expression times $1+O(n^{-1})$. The prefactor is $n^{-1/2}$ because this is the probability density function of the unstandardized sum; that change of units should not be confused with the orders inside the Edgeworth expansion.

For the saddlepoint, $K^{\prime }(t)=2t/(1-t^2)$. Write $y=s/n$ and $d=\sqrt{1+y^2}$. The root in $(-1,1)$ isThese formulas include $s=0$ without a division by y. Substitution yieldsThis is the leading saddlepoint approximation, without its order-$n^{-1}$ relative correction. Expanding d near one recovers the same central normal approximation, while retaining d captures the Laplace-sum large-deviation exponent away from the center.

An Edgeworth expansion refines the central limit theorem by incorporating higher cumulants. For iid centred observations with variance ${\sigma }^2$, standardized sum $Z_n={\sum }_i{X}_i/(\sigma \sqrt{n})$, standardized third cumulant ${\gamma }_1={\kappa }_3/{\sigma }^3$ and fourth cumulant ${\gamma }_2={\kappa }_4/{\sigma }^4$, its central-region density expansion isunder sufficient moment and smoothness/nonlattice conditions. Here $H_3=z^3-3z$, $H_4=z^4-6z^2+3$, $H_6=z^6-15z^4+45z^2-15$ are probabilists' Hermite polynomials. Expanding the characteristic exponent gives the cubic and quartic terms, while squaring the cubic term in the exponential gives the $H_6$ correction. Integrating the first correction, using $(\varphi H_{k-1}{)}^{\prime }=-\varphi H_k$, givesThus skewness determines the first departure from the normal approximation; inversion gives a Cornish-Fisher expansion for quantiles. These are asymptotic expansions, not guaranteed nonnegative densities in extreme tails. Lattice data require continuity/lattice corrections, and moment assumptions alone do not justify a uniform smooth-density expansion.

Let $W=2[\ell (\hat{\theta })-\ell (\tilde{\theta })]$ be a regular likelihood-ratio test statistic for $r$ smooth restrictions, so its leading null law is ${\chi }_r^2$. Suppose a higher-order calculation gives ${\mathbb{E}}_{0}W=r+a/n+O(n^{-2})$. The Bartlett correction rescales the statistic:Its mean is $r+O(n^{-2})$, eliminating the first-order mean error. In regular likelihood models satisfying the usual Bartlett expansion, the same scaling removes the leading $n^{-1}$ distributional error against the chi-square reference law, leaving an $O(n^{-2})$ error for fixed tail probabilities. Matching a mean alone would not establish this distributional conclusion for an arbitrary statistic.

The coefficient depends on the model, null restrictions and nuisance parameters; it can be obtained from higher derivatives and cumulants of the log-likelihood, or estimated by null simulation when an analytic expression is inconvenient. Nuisance parameters require their appropriate null values or estimates. A correction improves finite-sample calibration within regular asymptotic theory; it does not turn a boundary-parameter or unidentifiable problem into a regular chi-square problem.

Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.

Let the summands be independent and identically distributed random variables with mean $\mu$, variance ${\sigma }^2>0$, and standardized third cumulant ${\gamma }_1=\mathbb{E}[(Y_1-\mu {)}^3]/{\sigma }^3$. DefineAn Edgeworth expansion refines the central limit theorem using the higher cumulants of one summand. Some smoothness and moment assumptions are essential: finite variance alone does not supply a density expansion. A convenient sufficient set for the expansions here is a smooth density whose derivatives decay faster than every inverse power, together with $\mathbb{E}{e}^{\delta |Y_1|}<\infty$ for some $\delta >0$. These assumptions give all required moments, Fourier smoothing and the nonlattice characteristic-function condition. Less restrictive sufficient conditions are possible, but this set covers the logistic model below.

Write $\varphi$ and $\Phi$ for the standard normal density and standard normal distribution function. The first-order density Edgeworth expansion isThe remainder is uniform on every fixed compact interval under the stated assumptions. To see the origin of the correction, for $Z=(Y_1-\mu )/\sigma$ the cumulant expansion of its characteristic function givesExponentiating and using Fourier inversion yields the displayed correction, since the inverse transform of $(it{)}^3{e}^{-t^2/2}$ is $-{\varphi }^{(3)}(x)=(x^3-3x)\varphi (x)$. The Probabilists' Hermite polynomial convention is being used: $H_2(x)=x^2-1$ and $H_3(x)=x^3-3x$.