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Let $X_{1}, \dots, X_{n}$ be independent random sample with cumulative distribution function $F_{\theta}(x) = 1 - 2^{-(x-\theta)}$ for $x > \theta$ and $0$ elsewhere, where $\theta$ is unknown parameter. Let us consider the uniformly most powerful test of $H_{0}: \theta = 0$, against $H_{1}: \theta >0$ at significance level $0.01$. For what $n$ at $\theta > 0$ power is more than $0.64$.

I know that I am supposed to use Karlin-Rubin theorem and i showed that it has monotone likelihood function, I just don't know how to manipulate this distribution to find critical region. Any help?

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  • $\begingroup$ Why do you uses $\theta_1$ instead of $\theta$? You need to add the self-study tag to your question. $\endgroup$ – Michael R. Chernick Nov 4 '19 at 18:26
  • $\begingroup$ Added and changed $\endgroup$ – Berto Nov 4 '19 at 18:30
  • $\begingroup$ Okay, then notice that it is just a shifted exponential distribution with location/shift $\theta$ and scale $1/\ln 2$ (i.e. $X-\theta$ is Exp with mean $1/\ln 2$). $\endgroup$ – StubbornAtom Nov 4 '19 at 19:20
  • $\begingroup$ What critical region do you get from Karlin-Rubin? (Although you don't need this theorem for testing simple versus composite hypotheses.) $\endgroup$ – StubbornAtom Nov 4 '19 at 19:39
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    $\begingroup$ Power is relative to a specific alternative, and you haven't provided one. For example, consider $\theta = 100$ and $n=100$. In that case, your power is going to be $> 0.64$. Consider $\theta = 0.00000001$ and $n=100$. Your power is is going to be roughly equal to your significance level $0.01$. So the question, as stated, can't be answered; for any $n$, you can find a $\theta$ close enough to zero such that the power is arbitrarily close to your significance level. $\endgroup$ – jbowman Nov 4 '19 at 20:31
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The pdf of your distribution is $$f(x)=(\ln 2)2^{-(x-\theta)}1_{x>\theta}\quad,\,\theta\in\mathbb R$$

So pdf of the sample $X=(X_1,\ldots,X_n)$ is $$f_{\theta}(x_1,\ldots,x_n)=(2^\theta\ln 2)^n 2^{-\sum x_i}1_{x_{(1)}>\theta}\,,$$ where $x_{(1)}=\min\{x_1,x_2,\ldots,x_n\}$.

In general, any reasonable test involving the unknown quantity is based on a sufficient statistic. And here it is clear from the joint pdf that a sufficient statistic for $\theta$ is $X_{(1)}$.

If one wishes to use Karlin-Rubin theorem, one has to show that $f_{\theta}$ has the monotone likelihood ratio (MLR) property. And here you would actually find that $f_{\theta}$ has MLR in $X_{(1)}$. That would give you the critical region $\{X: X_{(1)}>c\}$ for testing $H_0:\theta=\theta_0$ against $H_1:\theta>\theta_0$ for any specified $\theta_0$.

To find $c$ subject to a level/size restriction, you need the probability $P_{\theta_0}(X_{(1)}>c)$. And this can be found directly from the CDF you have without referring to any sampling distribution.

Just notice that $P(X_{(1)}>c)=P(X_1>c,X_2>c,\ldots,X_n>c)=(P(X_1>c))^n$

As for power of the test at some $\theta=\theta_1(>0)$, it is given by $P_{\theta_1}(X_{(1)}>c)$. You can now establish a relationship between this function of $\theta_1$ and the sample size $n$.

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  • $\begingroup$ So the power of this test will be $P_{H_1}(X_{(1)}>c)$? Does that mean that power is independent from $n$? $\endgroup$ – Berto Nov 4 '19 at 20:24
  • $\begingroup$ Okay, finally i got it, thank you. $\endgroup$ – Berto Nov 4 '19 at 20:55
  • $\begingroup$ As @jbowman mentioned in a comment, note that power is usually defined for a specific alternative. Power at $\theta=\theta_1(>0)$ is $P_{\theta_1}(X_{(1)}>c)$. You cannot find $P_{H_1}(X_{(1)}>c)$ exactly when $H_1$ is composite. $\endgroup$ – StubbornAtom Nov 4 '19 at 21:23

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