Let X be a random variable having the probability density function f which belongs to $f_{0}$,$f_{1}$. The $f_{0}$ is from uniform distribution, $U(0,1)$ while $f_{1}$ is also uniform with $U(0,2)$. For testing the null hypothesis $H_{0}:f=f_{0}$ versus the alternative hypothesis $H_{1}:f=f_{1}$ based on a single observation on X, the power of the most powerful test of size $\alpha =0.05$ will be?

I know, from the Neyman Pearson lemma, that the most powerful test is the ratio of likelihood functions under the alternative and null hypothesis. But, all the problems that I solved, the critical region used to come as a function of sample values. But in this problem, I am not able to get a critical region. If someone help me in the critical region, I am able to find the power of the test. Thanks in advance.


1 Answer 1


The most reasonable test has power 50% and significance 0. Not a formal proof, but:

  • If the null hypothesis is true, the single observation X will lie in $(0,1)$
  • If the null hypothesis is false, the single observation X will lie in $(0,2)$
  • Therefore, the only reasonable test (for any $\alpha$ below 50%) is to accept the null hypothesis if X lies in $(0,1)$ and reject it if it lies in $(1,2)$.
  • If the nul hypothesis is false, the probability of getting X in $(1,2)$ is 50%, and then power of the test is 50%.

However, you could increase the power by using $\alpha=0.05$. You just need to reject the null hypothesis for a 5% of observations falling within $(0, 1)$, and since both under the null and the alternative hypothesis all points in $(0, 1)$ have the same probability, it doesn't matter how you decide for which 5% of points you reject the null hypothesis. For example, the following methods would be fine:

  • Rejecting the null hypothesis for all values over 0.95
  • Randomly deciding whether to reject with probability 5% in $(0, 1)$ and always rejecting in $(1, 2)$ (as suggested in Björn's comment).

This way, if the null hypothesis is false the, probability of rejecting it (aka power) is $0.5·0,05+0.5=0.525$, although you wouldn't use this test for a practical purpose.

  • $\begingroup$ I got your point. It makes sense also. But what if this type of question is asked as a objective problem in which we have two option as 0.535 and 0.465. The actual answer is 0.535. $\endgroup$
    – userNoOne
    Commented Jan 3, 2018 at 10:24
  • 1
    $\begingroup$ The test in my answer actually gets alpha=0; by expanding the cut point below 1, you can get a larger alpha and power - although a less reasonable test. Then, a likelihood ratio test would get a power slightly over 0.5. $\endgroup$
    – Pere
    Commented Jan 3, 2018 at 10:45
  • 2
    $\begingroup$ You could also randomly reject 5% of the time for values between 0 and 1. Nobody really does that in practice, but it’s also a more powerful test. But then the power should be 0.525, not 0.535. $\endgroup$
    – Björn
    Commented Jan 3, 2018 at 13:01
  • $\begingroup$ But how you got power as $0.525$. $\endgroup$
    – userNoOne
    Commented Jan 13, 2018 at 13:24
  • $\begingroup$ @RAHUlJHa Please see my edit. $\endgroup$
    – Pere
    Commented Jan 13, 2018 at 19:23

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