$\newcommand{\szdp}[1]{\!\left(#1\right)}\newcommand{\eff}{\operatorname{eff}}$ Problem Statement: Let $Y_1, Y_2, \dots, Y_n$ denote a random sample from the uniform distribution on the interval $(\theta, \theta+1).$ Let $$\hat\theta_1=\overline{Y}-\frac12\qquad\text{and}\qquad \hat\theta_2=Y_{(n)}-\frac{n}{n+1}.$$ Find the efficiency of $\hat\theta_1$ relative to $\hat\theta_2.$ [Note: this is Problem 9.3b in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer.]
My Attempt: It is fairly straight-forward to show that both $\hat\theta_1$ and $\hat\theta_2$ are unbiased estimators of $\theta.$ Note that the distribution of $Y_i$ is given by \begin{align*} f(y)&= \begin{cases} 1,&\theta<y<\theta+1\\ 0,&\text{elsewhere,} \end{cases}\\ F(y)&= \begin{cases} 0,&y\le\theta\\ y-\theta,&\theta<y<\theta+1\\ 1,&\theta+1\le y, \end{cases} \end{align*} so that the distribution of the maximum order statistic $Y_{(n)}:=\max(Y_1,Y_2,\dots,Y_n)$ is given by \begin{align*} g_{(n)}(y) &=n[F(y)]^{n-1}\,f(y)\\ &= \begin{cases} n(y-\theta)^{n-1},&\theta<y<\theta+1\\ 0,&\text{elsewhere.} \end{cases} \end{align*} Now we must compute variances: \begin{align*} V\szdp{\hat\theta_1} &=\frac{1}{n^2}\,\sum_{i=1}^n V(Y_i)\\ &=\frac{1}{12n}\\ V\szdp{\hat\theta_2} &=V\szdp{Y_{(n)}}\\ &=n\int_\theta^{\theta+1}y^2\,(y-\theta)^{n-1}\,dy-\theta^2\\ &=n\left[\frac{1}{n+2}+\frac{2\theta}{n+1}+\frac{\theta^2}{n}\right]-\theta^2\\ &=\frac{n}{n+2}+\frac{2n\theta}{n+1}\\ &=\frac{n(n+1)+2n(n+2)\theta}{(n+1)(n+2)}\\ &=\frac{n(n+1+2n\theta+4\theta)}{(n+1)(n+2)}. \end{align*} Finally, we compute the efficiency as \begin{align*} \eff\szdp{\hat\theta_1,\hat\theta_2} &=\frac{V\szdp{\hat\theta_2}}{V\szdp{\hat\theta_1}}\\ &=\frac{12n^2(n+1+2n\theta+4\theta)}{(n+1)(n+2)}. \end{align*}
My Question: This is not the book's answer, which is $$\frac{12n^2}{(n+1)^2(n+2)}.$$ Am I doing something wrong, or is the book wrong, or are we both wrong?
