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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 $\tilde{\ell }=X\tanh (\epsilon /2)$.

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

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