# 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?

• MLE is not unique, as shown here and here for example. Apr 4, 2020 at 10:29
• Does this answer your question? Likelihood Function for the Uniform Density $(\theta, \theta+1)$ Apr 4, 2020 at 12:33
• Only partially, I need to calculate bias and check whether this MLE is consistent. Apr 4, 2020 at 12:38
• If you take $\alpha(X_{(n)}-1)+(1-\alpha)X_{(1)}$ as your MLE for any constant $\alpha\in(0,1)$ free of $\theta$, then unbiasedness would not hold in general. But consistency would, as is shown in the answer by @joriki. Apr 4, 2020 at 18:37

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.