MLE for a uniform distribution

Question

I have trouble with finding a conclusion or possible $θ_{MLE}$ on this problem:

Let $Y_1, Y_2, \ldots, Y_n$ be a random sample from a $U(θ, θ + 1)$ distribution.
(a) Obtain an unbiased estimator $θ_\mathrm{mle}$ based on the MLE of $θ$.
(b) Show that $θ_2 = Y-\frac{1}{2}$ is another unbiased estimator.
(c) Find the efficiency of $θ_2$ relative to $θ_\mathrm{MLE}$.

The pdf I got is:

\begin{equation*} \ f_X(x)= \begin{cases} 1, & \text{if } θ\lt y_1 \lt y_2 <\cdots \lt y_n \lt θ+1\\ 0, & \text{otherwise} \end{cases} \ \end{equation*}

but maximizing $L(θ)^*$ will result to $0$. \begin{equation*} \ \text{*} L(θ) = \prod_{i=1}^n p_X(k_i;0) \end{equation*}

Any ideas, thoughts, solutions are accepted.

Answer 1

I will do my best to make sense of your Question. Please make edits or leave comments to resolve ambiguities and errors.

1) Based on @HansEngler's Comment, I choose the midrange $M$ as $an$ MLE. (That is, the average of the max and min.) Then $\hat \theta = M - 1/2$ is an unbiased estimator based on $an$ MLE, if not $the$ MLE. (@Did'a objections to the total lack of rigorous language, duly noted.)

2) I will assume that 'Y' means $\bar Y$ so that another unbiased estimator is $\tilde \theta = \bar Y - 1/2$.

3) In R, I will use simulation of 100,000 samples in the case $n = 10$ and $\theta = 4$ to illustrate the unbiasedness of both proposed estimators, and check which one has the smaller standard deviation.

Then if any of this useful, you might use analytic methods to derive some of the relevant facts to finish your assignment.

 m = 10^5;  n = 10;  th = 5
 x = runif(m*n, th, th+1)
 DTA = matrix(x, nrow=m)  # each row a sample of 5.
 mn = apply(DTA, 1, min)
 mx = apply(DTA, 1, max)
 hat = (mn + mx)/2 - 1/2
 mean(hat); sd(hat)
 ## 5.000074    # 'MLE' unbiased
 ## 0.0619935   # SD of 'MLE'
 tilde = rowMeans(DTA) - 1/2
 mean(tilde); sd(tilde)
 ## 4.999757    # Alt est unbiased
 ## 0.0916963   # noticeably more variable

Indications are that both $\hat \theta$ and $\tilde \theta$ are unbiased, but that the former is a 'better' estimator because it has a smaller standard deviation.