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?

  • $\begingroup$ Why do you uses $\theta_1$ instead of $\theta$? You need to add the self-study tag to your question. $\endgroup$ Commented Nov 4, 2019 at 18:26
  • $\begingroup$ Added and changed $\endgroup$
    – Berto
    Commented Nov 4, 2019 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 $1/\ln 2$). $\endgroup$ Commented Nov 4, 2019 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.) $\endgroup$ Commented Nov 4, 2019 at 19:39
  • 1
    $\begingroup$ 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
    Commented Nov 4, 2019 at 20:31

1 Answer 1


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 $n$.

  • $\begingroup$ 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
    Commented Nov 4, 2019 at 20:24
  • $\begingroup$ Okay, finally i got it, thank you. $\endgroup$
    – Berto
    Commented Nov 4, 2019 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. $\endgroup$ Commented Nov 4, 2019 at 21:23

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.