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Let $X_i$ be i.i.d $U(-\theta,2\theta)$ for i=1,2,...n.

$f(x)=\frac{1}{3\theta}$ and $L(\theta)=(3\theta)^{-n}\mathbb{1}_{[-\theta<X_{(1)}<X_{(2)}<...<X_{(n)}<2\theta]}$.

I don't know if the Maximum Likelihood Estimator is $X_{(1)}$ or $\frac{1}{2}X_{(n)}$. Don't know which one.

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    $\begingroup$ This looks like a homework problem. If it is, please add the self study tag and try to tell us why you think $X_{(1)}$ or $\frac{1}{2}X_{(n)}$ might be an answer. What else have you tried? $\endgroup$ – StatsStudent Aug 26 at 23:39
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    $\begingroup$ @pfmr1995 Clearly $X_{(1)}$ cannot itself be ML for $\theta$ in any case. $\endgroup$ – Glen_b Aug 27 at 0:35
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    $\begingroup$ If you do not provide (more) input on what you tried, this question is likely to get closed. $\endgroup$ – Xi'an Aug 27 at 5:04
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    $\begingroup$ Hints: (a) before looking for the MLE, you should identify the sufficient statistic (b) if hesitating between two values, compute and compare the likelihoods at both values. $\endgroup$ – Xi'an Aug 27 at 5:25
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This is an important problem because it illustrates a situation in which estimation is not entirely straightforward. However, it is a little messier than it needs to be in order for you to see the main points. In order to help you get started without ruining the chance for you to work the problem on your own, I will make some comments and show some simulations for a related problem that is a little simpler because of its symmetry.

Suppose you have a random sample of size $n = 10$ from $\mathsf{Unif}(-\theta,\theta).$ and you want to estimate $\theta.$

By looking at the likelihood function you have already guessed that the minimum sample value $X_{(1)}$ and the maximum sample value $X_{(10)}$ may be useful (also see @Xi'an's comment on sufficient statistics). So it makes sense that both $X_{(1)}$ and $X_{(10)}$ might be useful in finding the best estimate of $\theta.$

By simulation, we can illustrate that three unbiased estimators of $\theta$ are $-\frac{11}{9}X_{(1)},$ $\frac{11}{9}X_{(10)},$ and $\frac{11}{18}(X_{(10)}-X_{(1)}).$ Also, the latter of these three estimators has the smallest variability of the three. One intuitive reason why the range is best is that it cannot ever have the wrong sign. [Derivations of the exact distributions (related to four-parameter beta distributions) for each estimator are not difficult.]

The following simulation in R uses 10 million iterations with samples of size $n=10$ and $\theta = 5.$

set.seed(2020)
th = 5; n = 10
m = 10^7; min = max = numeric(m)
for(i in 1:m) {
 x = runif(n,-th,th)
 min[i] = min(x); max[i] = max(x)
 }
mean(min); mean(max)
[1] -4.090951
[1] 4.091096
rng = max-min; mean(rng)
[1] 8.182047
mean(-min*11/9);  mean(max*11/9);  mean(rng*11/18)
[1] 5.000051
[1] 5.000229
[1] 5.00014
sd(-min*11/9); sd(max*11/9); sd(rng*11/18)
[1] 1.014797
[1] 1.014481
[1] 0.6806108   # smallest variability of the three

enter image description here

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