The problem is to show that the largest order statistic $x_{(n)}$ is sufficient for $\theta$ where $X \sim \frac{1}{\theta}\mathbb{I}_{x \in (0, \theta)}$ is a uniform distribution.

I believe I have proved the result, but my argument cannot make sense, due to the consequences. Here's my argument:

The likelihood of a random sample of size $n$ is, for $\theta >0$, $$L(\theta) = \bigg(\frac{1}{\theta} \bigg)\bigg(\frac{1}{\theta} \bigg) \cdots \bigg(\frac{1}{\theta} \bigg) = \frac{1}{\theta^n},$$ which factors into a product of a function of $\theta$ and $\hat\theta$, and a function of the sample $x_i$ for any statistic $\hat\theta$. Therefore, by the Fisher-Neyman Factorization Theorem, any statistic is sufficient for $\theta$. In particular, the largest order statistic $x_{(n)}$ is sufficient for $\theta$.

It seems to me like any statistic should be sufficient for $\theta$. Of course, "any" statistic includes a constant, so my argument must be wrong. I'm not sure why, though. Can anyone enlighten me?


  • 1
    $\begingroup$ The likelihood, when applied to the parameter $\theta$, is supposed to be a number, but on the right hand side of your equation you have written a function (namely, a multiple of $\mathbb{I}_{(0,\theta)}$). That can't possibly be correct, can it? Has there perhaps been a typographical omission? $\endgroup$ – whuber Mar 31 '15 at 22:46
  • $\begingroup$ Hey, I'm a big fan of some of your posts! Of course, you're right. I edited $\mathbb{I}_{(0, \theta)}$ into the post after I posted it, thinking I needed some condition on $\theta$. I should just write that $\theta > 0$. Thanks! $\endgroup$ – user795305 Mar 31 '15 at 22:52
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    $\begingroup$ OK, but why aren't you evaluating $\mathbb{I}_{(0,\theta)}(x_i)$ for each observation $x_i$? Why do the $x_i$ simply disappear? I believe this is where your respondent Julius has been trying to direct your attention. $\endgroup$ – whuber Mar 31 '15 at 22:53
  • $\begingroup$ Juluis' recent edit and your post has made the problem clear. I was ignoring the domain of the pdf of the uniform distribution $X$! Thank you for the help. $\endgroup$ – user795305 Mar 31 '15 at 22:55

The problem is in the way you wrote the likelihood function. Suppose that we observe $x_{(n)}=2$, where $x_{(n)}$ is the $n$th ordered value of the sample $x_1,\dots,x_n$. Then what is $L(1)$? According to your formula it would be 1, but $\theta$ being 1 is not likely at all since we observe a draw from $U(0,\theta)$ that is higher than 1. Therefore, $L(1)=0$.

That is, for $\theta$ to be feasible, i.e. to have a positive likelihood, we cannot observe $x_{(n)}>\theta$ drawn from $U(0,\theta)$. In other words, by recalling that the density function of $U(0,\theta)$ actually is $f(x)=\frac1\theta 1_{(0,\theta)}(x)$, we get

$$L(\theta)=\frac1\theta1_{(0,\theta)}(x_1)\dots \frac1\theta1_{(0,\theta)}(x_n)=\frac{1}{\theta^n}1_{(0,\theta)}(\max\{x_1,\dots,x_2\})=\frac{1}{\theta^n}1_{(0,\theta)}(x_{(n)}),$$

in which case not any statistic is sufficient.

  • $\begingroup$ Ah, it seems like my likelihood definitely is wrong. I don't see why yours is right, though. Where does the $x_{(n)}$ term come from? $\endgroup$ – user795305 Mar 31 '15 at 22:37
  • $\begingroup$ Let me update the answer. $\endgroup$ – Julius Vainora Mar 31 '15 at 22:39

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