Let $X = (X_1, \dots, X_n)$ be a random sample from the Uniform($-2 \theta, 5 \theta$) distribution with $\theta > 0$ unknown. Find the maximum likelihood estimator (MLE) for $\theta.$ Furthermore, determine whether the MLE $\hat{\theta}$ is a function of a one-dimensional sufficient statistic for $\theta.$

Let $M = \max{ \{X_1, \dots, X_n \}}$ and $L = \min{ \{X_1, \dots, X_n \}}.$ Consider the likelihood function of $\theta$ $$L(\theta; x) = \prod_{k=1}^{n} f(x_k ; \theta) = \prod_{k=1}^{n} \frac{1}{7 \theta} \cdot \mathbf{1}_{(-2 \theta, 5 \theta)}(x_k) = \frac{1}{(7 \theta)^n} \cdot \mathbf{1}_{(-2 \theta, 5 \theta)}(m) \cdot \mathbf{1}_{(-2 \theta, 5 \theta)}(\ell) \cdot \prod_{k=1}^{n} \mathbf{1}_{\mathbf{R}}(x_k).$$ By the Factorization Theorem, it follows that $(M, L)$ is sufficient for $\theta,$ and in fact, it is easy to show that $(M, L)$ is minimal sufficient for $\theta.$ Our candidates for the MLE include $M,$ $L,$ and functions of $M$ and $L,$ e.g., the midrange $\frac{M-L}{2};$ however, I am running into difficulty finding the MLE and establishing that it gives a maximum.

On first glance, it appeared that $\hat{\theta} = L$ because $m \geq \ell$ implies that $\frac{1}{(7m)^n} \leq \frac{1}{(7 \ell)^n};$ however, this is only true if $m \geq \ell > 0.$

Reading a few other posts on here, I considered the possibility that the midrange $\frac{M-L}{2}$ is the MLE for $\theta.$ Of course, the difficulty arises out of the fact that there are many possibilities for $L$ and $M$: $m \geq \ell > 0,$ $m \geq 0 > \ell,$ and $0 \geq m > \ell,$ to name a few.

Can anyone offer any helpful insight or tips?


The likelihood function is $(7\theta)^{-n}$ for $-2\theta <L$ and $5\theta>M,$ otherwise zero. So the maximum likelihood value of theta is the smallest value of $\theta$ satisfying $-\theta <L/2$ and $\theta>M/5.$ Thus it is $\max\{M/5,-L/2\}.$

  • $\begingroup$ I agree that the maximum likelihood estimate -- and therefore, the maximum likelihood estimator -- is the smallest value of $\theta$ such that $\theta > M/5$ and $\theta > -L/2;$ however, I do not understand why this gives that the maximum of $M/5$ and $-L/2$ is the MLE. By our initial argument, should it not be the minimum? Moreover, what can be said about the question of whether the MLE is a function of a one-dimensional sufficient statistic? It does not seem to be the case. $\endgroup$ – Dylan_Carlo_Beck Aug 13 '17 at 6:04
  • 2
    $\begingroup$ Well if $\theta < M/5$ or $\theta < -L/2$ then the likelihood is zero (since we are outside the support of the PDF) so it must be the smallest thing not less than either of them, i.e. the max. The MLE is certainly a function of a one-dimensional statistic, since $\max\{M/5,-L/2\}$ is a one dimensional statistic. (It's a somewhat ugly function of $X_1,\ldots,X_n$ but it's still just one number computed from them, not unlike, say, the sample mean). And it is a sufficient statistic since the likelihood can be written $L(\theta) = (7\theta)^{-n}I(\theta \ge \max\{M/5,-L/2\})$. $\endgroup$ – spaceisdarkgreen Aug 13 '17 at 6:22
  • $\begingroup$ Can it not be shown that $(M,L)$ is minimal sufficient? Furthermore, does it not make sense that $\max{ \{M/5, -L/2 \}}$ is a function of $(M,L)?$ $\endgroup$ – Dylan_Carlo_Beck Aug 13 '17 at 15:02
  • $\begingroup$ What do you means, does it not make sense? Not sure if you're asserting it does or doesn't make sense and what is the significance? It obviously is a function of $(M,L)$ so I'd say it makes sense but again I'm not sure what you're trying to say. As for your first question I've already said that I think $\max\{M/5,-L/2\}$ is sufficient. $\endgroup$ – spaceisdarkgreen Aug 13 '17 at 15:31
  • $\begingroup$ I apologize if I am being unclear. Certainly, $g(M,L) = \max{ \{M/5, -L/2 \}}$ is sufficient: it is a function of the minimal sufficient statistic $(M,L);$ however, to me, that implies that $g(M,L)$ is NOT a function of one-dimensional sufficient statistic. Furthermore, it is my understanding that there are no one-dimensional sufficient statistics, as $(M,L)$ is minimal sufficient. $\endgroup$ – Dylan_Carlo_Beck Aug 13 '17 at 16:10

The support of $L(\theta; x)$ is given by $L\ge -2 \theta$ , $M\le 5\theta$; or, equivalently $\theta \ge M/5$ and $\theta \ge -L/2$. Or $$\theta \ge T \triangleq \max(M/5,-L/2)$$

Because over its support $L(\theta; x)$ (for $n>1$) is decreasing, then $ \theta_{ML}=T$

Regarding $(M,L)$ being or not minimal sufficient:

You say " it is easy to show that $(M,L)$ is minimal sufficient for $\theta$" but I don't think that's true.

$$L(\theta; x)=\frac{1}{(7\theta)^n}\mathbf{1}_{[M \le 5 \theta]} \mathbf{1}_{[L \ge -2 \theta]}=\frac{1}{(7\theta)^n} \mathbf{1}_{[T\le\theta]} $$ tells us that both $(M,L)$ and $T$ are sufficient. $T$ is clearly minimal. Because $T=f(M,L)$ (but not the reverse) then $(M,L)$ cannot be minimal.

Put in other way, consider some $x_1$ with $(M_1,L_1)=(100,-2)$ and some $x_2$ with $(M_2,L_2)=(100,-4)$, so that $T_1=T_2=20$

Then $\frac{L(\theta; x_1)}{L(\theta; x_2)}=1$ (doesn't depend on $\theta$), but $(M_1,L_1)\ne (M_2,L_2)$, hence $(M,L)$ is not minimal.

  • $\begingroup$ Can you justify your claim that $L(\theta ; x)$ is decreasing over its support? I do not see why this is true. Furthermore, can you confirm my claim that $(M,L)$ is minimal sufficient for $\theta?$ What can be said about the question of whether $\theta_{ML}$ is a function of a one-dimensional sufficient statistic? It does not seem to be the case. $\endgroup$ – Dylan_Carlo_Beck Aug 13 '17 at 6:01
  • $\begingroup$ Over its support, we have the function $(7\theta)^{-n}$ which is decreasing for $n\ge 1$ $\endgroup$ – leonbloy Aug 13 '17 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.