If $X_1\sim U(0,\theta)$ then $X_1$ is a sufficient statistic for $\theta$. Also when $X_2\sim U(0,\theta + 1)$ then $X_2$ is a sufficient statistic for $\theta$. Is that right?

Now if $X_1, X_2$ are independent then $\max(X_1,X_2)$ should give a sufficient statistic. Isn't it?

Also I read a theorem that if $T$ is a sufficient statistic of say $\omega$ then any monotone function of $T$ will be a sufficient statistic. So can I apply this above and say that: $\max(X_1 +1 ,X_2)$, $\max(X_1, X_2-1)$ are sufficient statistics of $\theta$?

  • 3
    $\begingroup$ Neither $\max(X_1+1, X_2)$ nor $\max(X_1, X_2-1)$ are even well-defined functions of $\max(X_1, X_2)$! $\endgroup$
    – whuber
    Aug 7, 2015 at 19:46
  • 1
    $\begingroup$ It doesn't make much sense if you have only one observation and say that one is sufficient for $\theta$. $\endgroup$
    – Zhanxiong
    Aug 7, 2015 at 19:49
  • 1
    $\begingroup$ If $T(X) = X$ then we have that $f(x|\theta) = f(T(x)|\theta)$ so $T$ is always sufficient for $\theta$ by the Neyman factorization theorem. $\endgroup$
    – jld
    Aug 7, 2015 at 19:51

1 Answer 1


Suppose $$X_i \sim \operatorname{Uniform}(0,\theta+i-1), \quad i = 1, 2, \ldots, n$$ are independent random variables from which the sample $\boldsymbol x = (x_1, \ldots, x_n)$ is drawn. Then the joint density, and thus the likelihood, is $$\mathcal{L}(\theta \mid \boldsymbol x) = f(\boldsymbol x \mid \theta) = \prod_{i=1}^n (\theta + i - 1)^{-1} \mathbb{1}(0 \le x_i \le \theta+i-1).$$ The support of $\mathcal L$ is clearly $$\theta \ge \max_i (x_i - i + 1).$$ On this interval, it is equally clear that $\mathcal L$ is a strictly decreasing function of $\theta$; thus the likelihood is maximized for $\hat \theta = \max_i(x_i - i + 1)$.

This also suggests that $x_{(n)} = \max_i x_i$ is not sufficient for $\theta$, because the knowledge of the maximum order statistic alone is not sufficient to ascertain the value of $\hat\theta$. This makes sense: you don't know which numbered observation generated the maximum observation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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