# Power of uniformly most powerful test

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?

• Why do you uses $\theta_1$ instead of $\theta$? You need to add the self-study tag to your question. Commented Nov 4, 2019 at 18:26
• Added and changed Commented Nov 4, 2019 at 18:30
• 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 $1/\ln 2$). Commented Nov 4, 2019 at 19:20
• What critical region do you get from Karlin-Rubin? (Although you don't need this theorem for testing simple versus composite hypotheses.) Commented Nov 4, 2019 at 19:39
• 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. Commented Nov 4, 2019 at 20:31

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 $$n$$.

• So the power of this test will be $P_{H_1}(X_{(1)}>c)$? Does that mean that power is independent from $n$? Commented Nov 4, 2019 at 20:24
• Okay, finally i got it, thank you. Commented Nov 4, 2019 at 20:55
• 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. Commented Nov 4, 2019 at 21:23