Let $X_i$, $i=1,...,n$ be iid with $f(x,\theta) = \frac{1}{2\theta}$ for $-\theta<x<\theta$. Find the best unbiased estimator of $\theta$ if one exists.
So I first tried $T(X)=X_{(n)}$, which gave me $$\mathbb{E}[T]=\frac{n-1}{n+1}\theta.$$ I thus concluded that $T(X)=\frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator. Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states $$\operatorname{Var}(T)\geq \frac{1}{I_X(\theta)},$$ so I first calculated the Fisher information for a single random variable. The calculation yields $$I_X(\theta)=\mathbb{E}_\theta[V(\theta,x)^2] = \int_{-\theta}^\theta\left(\partial_\theta\log\frac{1}{2\theta}\right)^{\!2}\frac{1}{2\theta}\,dx = \frac{1}{\theta^2}.$$ Thus the lower bound is $\dfrac{\theta^2}{n}$. Next, I computed the variance of the estimator $$\operatorname{Var}[T] = \int_{-\theta}^\theta \left(\frac{n+1}{n-1}\right)^{\!2}x^2n\frac{(x+\theta)^{n-1}}{2^n\theta^n}\,dx-\theta^2 = ... = \theta^2\frac{4n}{(n-1)^2(n-2)},$$ which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.