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Suppose that $X_1, \ldots, X_n$ come from the distribution

$$ f(x|\theta) = e^{-(x-\theta)} \ \ \text{for} \ \ x \geq \theta \ \ \text{and} \ \ 0 \ \ \text{otherwise} $$

Define $X_{(1)} = \min_{1 \leq i \leq n}X_i$ as the smallest value among the $X_i$.

I want to find the distribution of the minimum of $X_i$ under the constrained parameter space $-\infty < \theta\leq \theta_0$ where $\theta_0$ is some constant. Normally, this is taken as $-\infty < \theta\leq \infty$.

In my book, it states that:

$$ P(X_{(1)} \geq c) = e^{-n(c-\theta_0)} $$

However, I cannot find this answer on my own. It seems that I am always left with $e^{-n(c-\theta)}$ instead and don't know how to incorporate the $\theta_0$. Does anyone have any ideas or tips?

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  • $\begingroup$ Does the textbook give any hints of a prior distribution on $\theta$, or just the upper bound? If not, is there a chance it's asking for an upper bound on this distribution? $e^{-n(c-\theta)}$ is correct, if you've fixed $\theta$, and would be maximised at $\theta=\theta_0$. $\endgroup$ – πr8 Dec 29 '16 at 23:29
  • $\begingroup$ It's just an upper bound, it's actually a hypothesis testing question and under the null, $\theta \leq \theta_0$ $\endgroup$ – user321627 Dec 29 '16 at 23:36
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    $\begingroup$ $$\sup_{\theta\leqslant\theta_0}e^{-n(c-\theta)}=e^{-n(c-\theta_0)}$$ $\endgroup$ – Did Dec 29 '16 at 23:48
  • $\begingroup$ @Did If this is a hypothesis test, then why is it taken to be the supremum? Normally we don't take the supremum over to infinity right? $\endgroup$ – user321627 Dec 30 '16 at 0:26
  • $\begingroup$ The only way to have $$P_\theta(X_{(1)}\geqslant c)\leqslant A$$ for every $\theta\leqslant\theta_0$ simultaneously is to have $$A\geqslant\sup_{\theta\leqslant\theta_0}P_\theta(X_{(1)}\geqslant c)$$ To sum up, your misunderstanding is that you still believe your book stated $$P_\theta(X_{(1)}\geqslant c)=e^{-n(c-\theta_0)}$$ for every $\theta\leqslant\theta_0$, which is absurd, while your book stated $$P_\theta(X_{(1)}\geqslant c)\leqslant e^{-n(c-\theta_0)}$$ for every $\theta\leqslant\theta_0$, which is correct and optimal. $\endgroup$ – Did Dec 30 '16 at 7:50

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