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With unknown independent covariate and centered error distributions, take the regular mean-preserving error tangent space, let $h={\partial }_{\theta }{g}_{\theta }$, $\rho =-f^{\prime }/f$ and ${\tau }^2=E_f{\epsilon }^2>0$. Under $E_f\rho =0$, $E_f(\epsilon \rho )=1$ and finite second moments, orthogonal projection removes $(E_{v}h)(\rho -\epsilon /{\tau }^2)$ from $h\rho$. Thus the efficient score and efficient information are $\tilde{\ell }=(h-E_{v}h)\rho +(E_{v}h)\epsilon /{\tau }^2$ and $\tilde{I}={\mathrm{Var}}_v(h)E_f{\rho }^2+(E_{v}h{)}^2/{\tau }^2$. The reduction to $h\epsilon /{\tau }^2$ is valid for a normal distribution of errors or for constant $h$, but need not hold otherwise. Centered covariates and logistic distribution errors give the counterexample $\tilde{\ell }=X\tanh (\epsilon /2)$.

For a real location parameter $\mu$ and positive scale parameter $s$, the logistic distribution has probability density function and cumulative distribution functionDifferentiating $F$ recovers $f$, and its limits at infinity show normalization of a probability distribution. The density is symmetric about $\mu$ with exponentially decaying tails, so its expected value exists and equals $\mu$. Allowing $s<0$ in the displayed expression makes $f$ negative and its integral $-1$; a signed scale must instead be replaced by $|s|$ in the prefactor to define the same valid density.

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.

Now $p_{\theta ,\eta }(x,y)=v(x)f(y-g_{\theta }(x))$. With $h_{\theta }={\partial }_{\theta }{g}_{\theta }$ and the location score $\rho (e)=-f^{\prime }(e)/f(e)$, differentiation of the log-likelihood givesTake the usual regularity conditions $h_{\theta }\in L^2(v)$ and $\rho \in L^2(f)$ so that this score function belongs to L2 space.

For the specified nuisance statistical path, differentiating $\log (1+t\gamma (e))$ at zero gives the score function $\gamma (e)$. Normalization requires $E_f\gamma =0$. There is a second constraint: the statistical path must preserve the model's zero error expected value. ThusThis constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric logistic distribution, for example, $\gamma (e)=\tanh (e)$ is bounded and has zero expected value, but $E[\epsilon \tanh (\epsilon )]>0$; it would not preserve the mean.

For any $\psi \in L^2(v)$, independence impliesAll these products are integrable by the Cauchy-Schwarz inequality, since $\epsilon \psi (X)$ is square-integrable. Thus every such error-density score function is orthogonal to every $e\psi (x)$. Covariate-density nuisance score functions $b(X)$ are also orthogonal to these functions, since $E\epsilon =0$. This is the mean-preserving error tangent space orthogonality used in the final calculation.

First use the additional admitted form of the efficient score. Write $\tilde{\ell }=e\zeta (X)$ and let ${\tau }^2=E_f{\epsilon }^2>0$. The difference $\dot{\ell }-\tilde{\ell }$ is the orthogonal projection onto the nuisance tangent space. Part (c) makes every $e\psi (X)$ orthogonal to that space. Therefore, for all $\psi \in L^2(v)$,The function in braces belongs to L2 space; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction isUnder the regular tail condition $sf(s)\to 0$ at both infinities, integration by parts gives $\int s{f}^{\prime }(s)ds=-1$, and the formula becomes $e{h}_{\theta }(x)/{\tau }^2$. This also confirms the sign.

The admitted product form is an extra restriction; it does not follow for every independent-error regression model. To locate the restriction precisely, suppose the error-density nuisance statistical paths preserve both normalization and mean to first order. Their closed mean-preserving error tangent space is $\{\gamma :E_f\gamma =E_f(\epsilon \gamma )=0\}$. The full nuisance tangent space is the orthogonal sum of this space and the centered functions of $X$.

For completeness, bounded functions satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its expected value, and then subtract a multiple of a fixed bounded centered function $q$ with $E_f(\epsilon q)\ne 0$. Such a $q$ exists by truncating $\epsilon$, since ${\tau }^2>0$. The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.

Let $m=E_v{h}_{\theta }(X)$. With $E_f\rho =0$ and $E_f(\epsilon \rho )=1$, the orthogonal projection of $h_{\theta }(X)\rho (\epsilon )$ onto the error nuisance space is $m\{\rho (\epsilon )-\epsilon /{\tau }^2\}$; its projection onto the covariate nuisance space is zero. Thus the general efficient score in independent-error regression isThe orthogonality of the two terms gives the displayed efficient information. For a normal distribution of the error, $\rho (e)=e/{\tau }^2$, so this reduces to the admitted formula. It also does so when $h_{\theta }$ is constant.

A concrete counterexample to the generality of the admission is $g_{\theta }(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard logistic distribution error. Here $m=0$ and $\rho (e)=\tanh (e/2)$, so the actual efficient score is $x\tanh (e/2)$. It cannot equal $e\zeta (x)$ because $\tanh (e/2)/e$ is not constant. This score function is already orthogonal to every nuisance score function: its factor $X$ is centered against error-only directions, and its factor $\rho (\epsilon )$ is centered against covariate-only directions. The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.

Let ${\epsilon }_i$ be independent standard logistic random variables and define the latent variableSet $Y_i=1$ exactly when $Z_i>0$. If $F$ is the logistic cumulative distribution function, its symmetry and the identity $F^{-1}(p)=\mathrm{logit}(p)$ givewhich is precisely the fitted logistic regression.

For ${\theta }_1>{\theta }_0$, the likelihood ratio iswhich is increasing in $x$. Thus the logistic location family has a monotone likelihood ratio in $X$.

The upper-tail rejection probability is increasing in the location parameter, since $X=\theta +Z$ with $Z$ having the standard logistic distribution. Hence for the critical value from part (i),The test has size $\alpha$ for the composite null.

Now fix any alternative ${\theta }_1>0$. The Neyman-Pearson lemma applied to the simple hypotheses $\theta =0$ and $\theta ={\theta }_1$ says that this same upper-tail test is most powerful among all tests whose rejection probability at $0$ is at most $\alpha$. Every test of size at most $\alpha$ for the composite null satisfies that restriction. The upper-tail test is therefore most powerful at every ${\theta }_1>0$, so it is the uniformly most powerful upper-tail test for a logistic location.

The likelihood ratio of the simple alternative to the simple null isThis is a strictly increasing function of $x$. By the Neyman-Pearson lemma, the most powerful test therefore rejects for $X>c$. The logistic distribution hasThe size condition ${\mathbb{P}}_0(X>c)=\alpha$ givesBecause the distribution is continuous, no boundary randomization is needed. The most powerful size-$\alpha$ test is therefore

The same integration by substitution gives the expected valueThe remaining integrand is an odd function. Its integral is absolutely convergent, since ${\mathrm{sech}}^{2}t\le 4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. ConsequentlyEquivalently, the logistic distribution is symmetric about $\mu$, and its exponentially decaying tails ensure that its expected value exists.

For one observation from the logistic distribution, the size-$\alpha$ test of $H_0:\theta \le 0$ against $H_1:\theta >0$ rejects when