[proofplan] We compare the two tests through their local noncentrality parameters under alternatives \begin{align*} \theta_n=\frac{h}{\sqrt n}. \end{align*} The paired $t$-statistic has local noncentrality \begin{align*} \frac{h}{\sigma}. \end{align*} For the Wilcoxon signed-rank statistic, the assumed Hájek projection reduces the calculation to differentiating a limiting one-observation signed-rank score functional; an integration-by-parts computation identifies that derivative as $2\int_{\mathbb R}f^2\,d\mathcal L^1$. The finite Fisher information, contiguity, local asymptotic normality, and projection assumptions are used here as black-box hypotheses, not re-proved from the density assumptions. After standardising by the null variance of the signed-rank score, the Wilcoxon local noncentrality is $h\sqrt{12}\int_{\mathbb R}f^2\,d\mathcal L^1$, and the asserted efficiency is the ratio of squared noncentralities. [/proofplan] [step:Identify the local noncentrality of the paired $t$-statistic] Let \begin{align*} \bar Y_n:=\frac{1}{n}\sum_{i=1}^n Y_i \end{align*} denote the sample mean, and let \begin{align*} T_n:=\frac{\sqrt n\,\bar Y_n}{\sigma} \end{align*} denote the population-variance normalisation of the paired $t$-statistic. Let $S_n$ denote the usual sample standard deviation of $Y_1,\dots,Y_n$, and define the usual paired $t$-statistic by \begin{align*} \widetilde T_n:=\frac{\sqrt n\,\bar Y_n}{S_n}. \end{align*} Because the errors have finite positive variance $\sigma^2$, the law of large numbers gives $S_n\to\sigma$ in probability under the null and under the contiguous local shifts. Hence Slutsky's theorem implies that $\widetilde T_n$ and $T_n$ have the same local limiting noncentrality. It is therefore enough to compute the limit for $T_n$. Under \begin{align*} \theta_n=\frac{h}{\sqrt n}, \end{align*} we have \begin{align*} \sqrt n\,\bar Y_n = \sqrt n\,\theta_n+\frac{1}{\sqrt n}\sum_{i=1}^n \varepsilon_i = h+\frac{1}{\sqrt n}\sum_{i=1}^n \varepsilon_i. \end{align*} Since the $\varepsilon_i$ are independent and identically distributed with mean $0$ by evenness of $f$ and variance $\sigma^2\in(0,\infty)$, the [central limit theorem](/theorems/521) gives \begin{align*} T_n \Longrightarrow N\left(\frac{h}{\sigma},1\right). \end{align*} Thus the paired $t$-test has local noncentrality \begin{align*} \delta_t=\frac{h}{\sigma}. \end{align*} [/step] [step:Write the limiting signed-rank score functional] Define the null distribution function of $|\varepsilon_1|$ as the map $H:[0,\infty)\to[0,1]$ given by \begin{align*} H(t)=\mathbb P(|\varepsilon_1|\le t)=2\int_0^t f(x)\,d\mathcal L^1(x). \end{align*} Define the signed-rank score function as the map $g:\mathbb R\to[-1,1]$ given by \begin{align*} g(y)=\operatorname{sgn}(y)\,H(|y|), \end{align*} where $\operatorname{sgn}(y)=1$ for $y>0$, $\operatorname{sgn}(0)=0$, and $\operatorname{sgn}(y)=-1$ for $y<0$. Define the limiting signed-rank mean functional as the map $m:\mathbb R\to\mathbb R$ given by \begin{align*} m(\theta):=\int_{\mathbb R} g(y) f(y-\theta)\,d\mathcal L^1(y). \end{align*} The stated Hájek projection assumption says that, under \begin{align*} \theta_n=\frac{h}{\sqrt n}, \end{align*} the local mean shift of the projected Wilcoxon statistic is governed by $h\,m'(0)$. [/step] [step:Differentiate the signed-rank functional and reduce it to $\int_{\mathbb R}f^2\,d\mathcal L^1$] Because $f$ is absolutely continuous and the local signed-rank differentiation is assumed valid, the LAN score derivative formula applies to the score functional $g$. In this notation, the black-box hypothesis asserts that $g f'$ is integrable in the improper score sense and that \begin{align*} m'(0) = -\int_{\mathbb R} g(y) f'(y)\,d\mathcal L^1(y). \end{align*} Equivalently, the location score $\ell:\mathbb R\to\mathbb R$ given by $\ell(y)=f'(y)/f(y)$ on $\{y\in\mathbb R:f(y)>0\}$ is square-integrable with respect to the null law, and the derivative of $\theta\mapsto \int_{\mathbb R}g(y)f(y-\theta)\,d\mathcal L^1(y)$ at $0$ is obtained by differentiating the location density. This improper integral is therefore well-defined by the differentiability hypothesis; the integration-by-parts identity below identifies the same value through an absolutely convergent $L^2$ integral. On $(0,\infty)$, $g(y)=H(y)$, and on $(-\infty,0)$, $g(y)=-H(-y)$. Since $f$ is even, \begin{align*} H'(y)=2f(y)\quad\text{for }\mathcal L^1\text{-a.e. }y>0, \end{align*} and \begin{align*} \frac{d}{dy}\bigl[-H(-y)\bigr]=H'(-y)=2f(-y)=2f(y) \quad\text{for }\mathcal L^1\text{-a.e. }y<0. \end{align*} Thus $g'(y)=2f(y)$ for $\mathcal L^1$-a.e. $y\ne0$. Fix $a<0<b$. Since $f$ and $g$ are absolutely continuous on the compact intervals $[a,0]$ and $[0,b]$, ordinary [integration by parts](/theorems/210) on these two intervals gives \begin{align*} -\int_a^b g(y)f'(y)\,d\mathcal L^1(y)= -g(b)f(b)+g(a)f(a)+\int_a^b g'(y)f(y)\,d\mathcal L^1(y), \end{align*} where the boundary contribution at $0$ is zero because $H(0)=0$ and hence $g(0)=0$. Letting $a\to-\infty$ and $b\to\infty$, the endpoint terms vanish because $|g|\le1$ and $f(y)\to0$ as $|y|\to\infty$. The integral involving $g'f$ converges absolutely, since $g'(y)=2f(y)$ for $\mathcal L^1$-a.e. $y\ne0$ and $f\in L^2(\mathbb R)$. By the absolute convergence of $g'f$ and the [Dominated Convergence Theorem](/theorems/4), applied to the functions $\mathbb{1}_{[a,b]}(y)g'(y)f(y)$ dominated by $|g'(y)f(y)|$, the integrals of $g'f$ over $[a,b]$ converge to the full-line integral as $a\to-\infty$ and $b\to\infty$. Together with the defining differentiability assumption for $m'(0)$, this justifies passage to the limit in the finite-interval integration-by-parts identity. Therefore \begin{align*} m'(0)=\int_{\mathbb R} g'(y)f(y)\,d\mathcal L^1(y)=2\int_{\mathbb R} f(y)^2\,d\mathcal L^1(y). \end{align*} [guided] The purpose of this step is to compute the one quantity that controls the local power of the Wilcoxon signed-rank statistic: the derivative at the null of its limiting signed-rank mean. The limiting signed-rank score is the map $g:\mathbb R\to[-1,1]$ given by \begin{align*} g(y)=\operatorname{sgn}(y)\,H(|y|), \end{align*} where $H:[0,\infty)\to[0,1]$ is given by \begin{align*} H(t)=2\int_0^t f(x)\,d\mathcal L^1(x) \end{align*} and is the distribution function of $|\varepsilon_1|$ under the null law. Under a location shift $\theta$, the density of $Y_1$ is $y\mapsto f(y-\theta)$, so the limiting signed-rank mean is \begin{align*} m(\theta)=\int_{\mathbb R} g(y)f(y-\theta)\,d\mathcal L^1(y). \end{align*} By the differentiability assumption in the theorem, we may differentiate this functional at $\theta=0$. The finite Fisher information hypothesis supplies the standard location-score integrability control behind this differentiation, while the theorem explicitly assumes the local signed-rank differentiation itself. Since differentiating $f(y-\theta)$ with respect to $\theta$ gives $-f'(y-\theta)$, we obtain \begin{align*} m'(0) = -\int_{\mathbb R} g(y)f'(y)\,d\mathcal L^1(y). \end{align*} Now we transfer the derivative from $f$ onto $g$. This is where the evenness of $f$ enters. For $y>0$, the score is $g(y)=H(y)$, hence \begin{align*} g'(y)=H'(y)=2f(y) \quad\text{for }\mathcal L^1\text{-a.e. }y>0. \end{align*} For $y<0$, the score is $g(y)=-H(-y)$. Differentiating this expression gives \begin{align*} g'(y)=H'(-y)=2f(-y). \end{align*} Since $f$ is even, $f(-y)=f(y)$, and therefore \begin{align*} g'(y)=2f(y) \quad\text{for }\mathcal L^1\text{-a.e. }y<0. \end{align*} The boundary terms in [integration by parts](/theorems/2098) vanish. At $y=0$, we have $H(0)=0$, so $g(0)=0$. At infinity, $|g(y)|\le1$ and $f(y)\to0$ as $|y|\to\infty$, so the endpoint contribution $g(y)f(y)$ tends to $0$ along both half-lines. To pass from local integration by parts to the full line, fix $a<0<b$. On $[a,0]$ and $[0,b]$, the functions are absolutely continuous, so integration by parts gives \begin{align*} -\int_a^b g(y)f'(y)\,d\mathcal L^1(y)= -g(b)f(b)+g(a)f(a)+\int_a^b g'(y)f(y)\,d\mathcal L^1(y), \end{align*} with no boundary contribution at $0$ because $g(0)=0$. Now let $a\to-\infty$ and $b\to\infty$. The endpoint terms tend to $0$ since $|g(y)|\le1$ and $f(y)\to0$ as $|y|\to\infty$. The integral containing $g'f$ has a finite full-line limit because $g'(y)f(y)=2f(y)^2$ for $\mathcal L^1$-a.e. $y\ne0$ and $f\in L^2(\mathbb R)$. Thus $|g'f|\in L^1(\mathbb R)$. By the [Dominated Convergence Theorem](/theorems/4), applied to $\mathbb{1}_{[a,b]}(y)g'(y)f(y)$ dominated by $|g'(y)f(y)|$, the finite-interval integrals of $g'f$ converge to the full-line integral as $a\to-\infty$ and $b\to\infty$. The left-hand side is interpreted through the assumed differentiability formula for $m'(0)$, whose score-integrability requirements are part of the finite-Fisher-information and local asymptotic framework in the statement. Hence the finite-interval identity passes to the full line and gives \begin{align*} -\int_{\mathbb R} g(y)f'(y)\,d\mathcal L^1(y)=\int_{\mathbb R} g'(y)f(y)\,d\mathcal L^1(y). \end{align*} Substituting $g'(y)=2f(y)$ for $\mathcal L^1$-a.e. $y\ne0$ yields \begin{align*} m'(0) = 2\int_{\mathbb R} f(y)^2\,d\mathcal L^1(y). \end{align*} This is the analytic core of the proof: the local signed-rank drift is determined by the $L^2$ mass of the density. [/guided] [/step] [step:Standardise the Wilcoxon projection under local alternatives] Let $U:\mathbb R\to[-1,1]$ denote the one-observation limiting signed-rank score under the null law, given by \begin{align*} U(y)=g(y). \end{align*} Under $\theta=0$, the distribution function $H$ is continuous because $f$ is a density and $H(t)=2\int_0^t f(x)\,d\mathcal L^1(x)$. By the probability integral transform, the [random variable](/page/Random%20Variable) $H(|\varepsilon_1|)$ is uniformly distributed on $[0,1]$, and by evenness of $f$ the sign $\operatorname{sgn}(\varepsilon_1)$ is independent of $|\varepsilon_1|$ and takes the values $1$ and $-1$ with probability $1/2$ each. Hence \begin{align*} \mathbb E[U(\varepsilon_1)]=0. \end{align*} Also, \begin{align*} \operatorname{Var}(U(\varepsilon_1))=\mathbb E[H(|\varepsilon_1|)^2]. \end{align*} Since $H(|\varepsilon_1|)$ is uniformly distributed on $[0,1]$, \begin{align*} \mathbb E[H(|\varepsilon_1|)^2]=\int_0^1 u^2\,d\mathcal L^1(u)=\frac{1}{3}. \end{align*} Therefore the standardised Hájek projection has null variance $1$. Since the local mean shift of the unstandardised projected signed-rank score is $h\,m'(0)$, the standardised Wilcoxon local noncentrality is \begin{align*} \delta_W=\frac{h\,m'(0)}{\sqrt{1/3}}=h\sqrt{3}\,m'(0)=h\sqrt{12}\int_{\mathbb R} f(y)^2\,d\mathcal L^1(y). \end{align*} Equivalently, for the conventional positive-rank normalisation of the Wilcoxon statistic, the null rank-scale variance is $1/12$ and the same standardised noncentrality is obtained. [/step] [step:Take the ratio of squared local noncentralities] Define the Pitman efficiency $e_W(f)$ of the Wilcoxon signed-rank test relative to the paired $t$-test as the limiting sample-size ratio \begin{align*} \frac{n_t}{n_W}, \end{align*} where $n_t$ denotes the sample size for the paired $t$-test and $n_W$ denotes the sample size for the Wilcoxon signed-rank test, required for the two tests to attain the same asymptotic local power against the same fixed local direction $h\ne0$. In a local asymptotic normal experiment, if the corresponding standardised statistics have local noncentralities $\delta_W$ and $\delta_t$, equality of limiting local power is equality of the normal mean shifts after rescaling sample size. Therefore the sample-size ratio is the ratio of squared local noncentralities: \begin{align*} e_W(f)=\frac{\delta_W^2}{\delta_t^2}. \end{align*} The assumed contiguity, LAN expansion, and Hájek projection hypotheses are exactly the hypotheses that justify using these local noncentralities for the two statistics. Substituting the two local noncentralities gives \begin{align*} e_W(f)=\frac{\left(h\sqrt{12}\int_{\mathbb R}f(y)^2\,d\mathcal L^1(y)\right)^2}{(h/\sigma)^2}=12\sigma^2\left(\int_{\mathbb R}f(y)^2\,d\mathcal L^1(y)\right)^2. \end{align*} The factor $h^2$ cancels, so the efficiency is independent of the particular nonzero local direction $h$. [/step] [step:Evaluate the formula for Gaussian errors] Assume \begin{align*} f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{x^2}{2\sigma^2}\right). \end{align*} Then \begin{align*} \int_{\mathbb R} f(x)^2\,d\mathcal L^1(x)=\int_{\mathbb R}\frac{1}{2\pi\sigma^2}\exp\left(-\frac{x^2}{\sigma^2}\right)\,d\mathcal L^1(x)=\frac{1}{2\pi\sigma^2}\cdot \sqrt{\pi}\sigma=\frac{1}{2\sqrt{\pi}\sigma}. \end{align*} Substituting this value into the general efficiency formula gives \begin{align*} e_W(f)=12\sigma^2\left(\frac{1}{2\sqrt{\pi}\sigma}\right)^2=\frac{3}{\pi}. \end{align*} This proves both the general formula and the Gaussian special case. [/step]