Consider $\theta^\ast = X_{(1)}+X_{(n)}$. I have the following questions:
Is $\theta^\ast$ unbiased?
I would say so. I need to show that $\mathbb E[\theta^\ast] = \mathbb E[X_{(1)}+X_{(n)}] = \theta$. The expectation of $\mathbb E[X_{(n)}]$ is
\begin{align*} F_{X_{n}}(x) &= \mathbb P(X_{(n)}\leqslant x)\\ &= \mathbb P(X_{(1)}\leqslant x,\dots, X_{(n)}\leqslant x)\\ &= \left(\frac x\theta\right)^n\\ f_{X_{(n)}}(x) &= F'_{X_{(n)}}(x)\\ &=n\theta^{n-1}x^{n-1}\\ \mathbb E[X_{(n)}] &= \int\limits_0^\theta xf_{X_{(n)}}(x)\\ &=\frac {n}{n+1}\theta. \end{align*}
The expectation of $\mathbb E[X_{(1)}]$ is \begin{align*} F_{X_{(1)}}(x) &= 1-[1-F_{X}]^n\\ &=1-[1-\frac x\theta ]^n\\ f_{X_{(1)}}(x)&= F'_{X_{(1)}}(x)\\ &=n\theta^{-1} \left(1-\frac x \theta\right)^{n-1}\\ \mathbb E[X_{(1)}] &= \int\limits_{0}^\theta xf_{X_{(1)}}(x)\\ &=\frac 1{n+1}\theta. \end{align*} And therefore we have $$\mathbb E[X_{(1)}+X_{(n)}] = \frac{n}{n+1}\theta + \frac1 {n+1}\theta = \theta.$$
Is $\theta^\ast$ unbiased consistent?
I need to show that $\mathbb P(|\theta-\theta^\ast|\geqslant \varepsilon)\to0$ for all $\varepsilon >0, n\to\infty$. I am not sure how to do this or even if it is true. I believe I have to use some central limit theorem.
To see that indeed $\Pr [|X_{(1)}+X_{(n)}-\theta |\geqslant \epsilon ]\to 0$ as $n\to +\infty $ it is enough to show that $X_{(1)}\to 0$ and $X_{(n)}\to \theta $ in probability, because
$$ \Pr [|X_{(1)}+X_{(n)}-\theta |\geqslant \epsilon ]\leqslant \Pr [|X_{(1)}|\geqslant \epsilon /2]+\Pr [|X_{(n)}-\theta |\geqslant \epsilon /2] $$