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 $\theta$?

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    $\begingroup$ 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 '15 at 19:46
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    $\begingroup$ It doesn't make much sense if you have only one observation and say that one is sufficient for $\theta$. $\endgroup$ – Zhanxiong Aug 7 '15 at 19:49
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    $\begingroup$ 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 '15 at 19:51

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.


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