Let $(X_1,\dots,X_n)$ be a random sample from an arbitrary statistical model $X\sim f_X(x;\theta)$ such that $\mathbb{E}|X|<\infty$, $\mathbb{E}(X)=\theta$. Consider the statistic $\frac{X_{(1)}+X_{(n)}}{2}$. Show that this is unbiased for $\theta$.
I tried to do the exercise in this way:
Let's compute the distribution of the minimum:
\begin{align} F_{X_{(1)}}(x) & =\mathbb{P}(X_{(1)}\leq x)\\ & = \mathbb{P}(\min(X_1,\dots,X_n)\leq x)\\ & =1-\mathbb{P}(\min(X_1,\dots,X_n)\geq x)\\ & = 1-\mathbb{P}(X_1\geq x,\dots,X_n\geq x)\\ & =1-[1-\mathbb{P}(X\leq x)]^n=1-[1-F_{X}(x)]^n. \end{align}
So that its density function is:
\begin{equation} f_{X_{(1)}}(x)= n[1-F_{X}(x)]^{n-1}f_X(x). \end{equation}
Let's now compute the distribution of the maximum:
\begin{align} F_{X_{(n)}}(x)&=\mathbb{P}(X_{(n)}\leq x)\\ & = \mathbb{P}(\max(X_1,\dots,X_n)\leq x)\\ & =\mathbb{P}(X_1\leq x,\dots,X_n\leq x)\\ & =[\mathbb{P}(X\leq x)]^n\\ & =[F_{X}(x)]^n. \end{align}
So that its density function is:
\begin{equation} f_{X_{(n)}}(x)= n[F_{X}(x)]^{n-1}f_X(x) \end{equation}
Hence:
\begin{align} \mathbb{E}\biggl(\frac{X_{(1)}+X_{(n)}}{2}\biggr) & =\frac{1}{2}\int\biggl\{xf_{X_{(1)}}(x)+xf_{X_{(n)}}(x) \biggr\}dx \\ & =\frac{1}{2}\int\biggl\{nx[1-F_{X}(x)]^{n-1}f_X(x)+nx[F_{X}(x)]^{n-1}f_X(x)\biggr\}dx \\ & = \frac{1}{2}\int\biggl\{nxf_X(x)\biggl[[1-F_{X}(x)]^{n-1}+[F_{X}(x)]^{n-1} \biggr]\biggr\}dx. \end{align}
And here I get stucked! I don't know how to solve this integral in order to get $\theta$! If I manage to prove that $[1-F_{X}(x)]^{n-1}+[F_{X}(x)]^{n-1} =\frac{2}{n}$ I obtain my result but I don't know how to prove it! Someone can help me? Many thanks!
The estimator $\dfrac{X_{(1)}+X_{(n)}}{2}$ is not an unbiased estimator of $\theta$. For example, if $f(x;\theta) = \frac{1}{\theta}e^{-x/\theta}$ for $x \ge 0$ then $$E[X_{(1)}] = \frac{\theta}{n},$$ and $$E[X_{(n)}] = \theta\sum_{i=1}^{n}\frac{1}{i}.$$ It is easy to verify that, for $n >2$, $$\frac{E[X_{(1)}]+E[X_{(n)}]}{2} \neq \theta.$$