MLE of Uniform on $(\theta, \theta +1)$ and consistency/bias I see there were a few questions on SE about MLE of Uniform already but none of them helped me with this one:
We are to compute MLE of $U(\theta, \theta +1)$ and check if it is biased and consistent. 
I tried by making a spin-off of an example with $U(0, \theta)$ but I am not sure if it is correct. Suppose there's $X_1, X_2, \dots, X_n$ i.i.d with $U(\theta, \theta +1)$, $T(X_1, \dots, X_n)$ is the statistic and $(x_1, \dots, x_n)$ a sample from that statistic.
I start of with computing $L(\theta)$
$$
L(\theta)=\prod_{i=1}^n\mathbb{1}_{[\theta, \theta +1]}(x_i) = \mathbb{1}_{(-\infty, X(1)]}(\theta)\cdot\mathbb{1}_{[X(n),\infty)}(\theta+1)
$$
Since $P(x_i \geq \theta) = 1$ this is just
$$
L(\theta)=\mathbb{1}_{[X(n),\infty)}(\theta+1) = \begin{cases}
      1, & \text{if}\ \theta + 1 \geq X(n) \\
      0, & \text{otherwise}
    \end{cases}
$$
The smallest value of $\theta = 1$ is then $\frac{X(n) - 1 + X(1)}{2}$ and this is our MLE. As @StubbornAtom pointed out in comments, this is not the only MLE possible.
How can I calculate bias and consistency of the $\hat{\theta}^{MLE}$ of my choosing?
 A: Your argument “Since $P(x_i\ge\theta)=1$” is incorrect; the resulting likelihood function is $1$ for arbitrarily large $\theta$. The correct simplified form is $\mathbb 1_{[X_{(n)}-1,X_{(1)}]}$. But your estimator lies in this interval, so it’s one of the possible maximum-likelihood estimators.
By symmetry, the expected values of $X_{(1)}$ and $X_{(n)}$ are symmetric about $\theta+\frac12$, so the expected value of your estimator is $\theta$, so it’s unbiased.
For consistency, note that by symmetry $X_{(1)}$ and $X_{(n)}$ have the same variance, so
\begin{eqnarray}
\operatorname{Var}\hat\theta
&=&
\operatorname{Var}\left(\frac{X_{(n)}-1+X_{(1)}}2\right)
\\
&=&
\frac14\operatorname{Var}\left(X_{(n)}+X_{(1)}\right)
\\
&=&
\frac14\left(\operatorname{Var}X_{(n)}+\operatorname{Var}X_{(1)}+2\operatorname{Cov}(X_{(n)},X_{(1)})\right)
\\
&\le&
\frac14\left(\operatorname{Var}X_{(n)}+\operatorname{Var}X_{(1)}+2\sqrt{\operatorname{Var}X_{(n)}\operatorname{Var}X_{(1)})}\right)
\\
&=&
\operatorname{Var}X_{(1)}\;.
\end{eqnarray}
The order statistic $X_{(1)}$ of $n$ random variables uniformly distributed on $[0,1]$ has distribution $\mathsf{Beta}(1,n)$ (see Wikipedia) and the shift by $\theta$ doesn’t change the variance, so the variance is that of $\mathsf{Beta}(1,n)$ (see Wikipedia):
$$
\operatorname{Var}\hat\theta\le\frac n{(n+1)^2(n+2)}\;.
$$
Thus the estimator is consistent.
