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?

  • 2
    $\begingroup$ MLE is not unique, as shown here and here for example. $\endgroup$ Apr 4, 2020 at 10:29
  • $\begingroup$ Does this answer your question? Likelihood Function for the Uniform Density $(\theta, \theta+1)$ $\endgroup$ Apr 4, 2020 at 12:33
  • $\begingroup$ Only partially, I need to calculate bias and check whether this MLE is consistent. $\endgroup$
    – blahblah
    Apr 4, 2020 at 12:38
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 4, 2020 at 18:37

1 Answer 1


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.


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