 Solution

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 $H_3(x)=x^3-3x$.

The corresponding distribution function expansion isuniformly on fixed compact intervals. The sign follows from $\frac{d}{dx}[H_2(x)\varphi (x)]=-H_3(x)\varphi (x)$. These are expansions of both the density and distribution function; a uniform density error alone would not justify integrating an error bound over the entire real line without the accompanying tail control.

For a fixed $\alpha \in (0,1)$, write $z=z_{\alpha }={\Phi }^{-1}(\alpha )$. At leading order the proposed quantile expansion gives $\Phi (p_0(z))=\Phi (z)$, and strict monotonicity gives $p_0(z)=z$. Substitute $y_{\alpha }=z+p_1(z)n^{-1/2}+O(n^{-1})$ into the distribution function expansion. A Taylor expansion at $z$ givesSince $\varphi (z)>0$, the coefficient of $n^{-1/2}$ must vanish. Thus the first Cornish-Fisher expansion coefficients are

For the logistic distribution, set $Z_i=Y_i-\theta$. Its density $q(z)=e^z/(1+e^z{)}^2$ satisfies $q(-z)=q(z)$ and has exponentially decreasing tails. Symmetry and integrability give $\mathbb{E}{Z}_i=0$, and the supplied integral givesTherefore $\mathbb{E}{Y}_i=\theta$, $\mathrm{Var}(Y_i)={\pi }^2/3$, and the central limit theorem givesWriting $z_{-}=z_{\alpha /2}$ and $z_{+}=z_{1-\alpha /2}$, the condition $z_{-}\le T_n\le z_{+}$ is equivalent toConsequently the confidence intervalhas limiting coverage $\Phi (z_{+})-\Phi (z_{-})=1-\alpha$. Whether the finite endpoints are included makes no probability difference because the sample mean has a density.

The centered logistic distribution is symmetric, so ${\gamma }_1=0$. Its smooth density and exponentially decreasing derivatives satisfy the conditions stated above; in particular all moments exist and Fourier smoothing is available. The first correction in the Edgeworth expansion vanishes, leaving $F_{T_n}(z)=\Phi (z)+O(n^{-1})$ at each of the two fixed quantiles. Subtracting gives the stronger conclusionThe standardized law is independent of $\theta$, so this coverage-error conclusion holds with the same bound for every location parameter.