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Let X be a random variable having a probability density function $f \in (f_o,f_1)$, where

$f_0(x) = \biggl(1 \ \ \ \ if \ \ 0\leq x \leq 1\,, \ 0 \ otherwise\biggr )$

and

$f_1(x) = \biggl(\dfrac{1}{2} \ \ \ \ if \ \ 0\leq x \leq 2\,, \ 0 \ otherwise\biggr )$

For testing the null hypothesis $H_0:f=f_0$ against $H_1:f=f_1$ based on a single observation on $X$, the power of the most powerful test of size $\alpha = 0.05$ equals ?

My inpuit : I applied NP lemma. Since we have a single observation. Test is to reject $H_0$ if $Pr$$\left\{\dfrac{f_1}{f_0}\geq K \right\}$ = $.05$

I am having problem in finding the critical region. I divided these densities and i got $\dfrac{f_1}{f_0} = \dfrac{1}{2} \ \ \ \ \ \ \ 0\leq x \leq 1\ $

$\dfrac{1}{2} \geq K$

What do i depict from it ?

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  • $\begingroup$ How about finding a most powerful test of size $\alpha = 0.05.$ $\endgroup$
    – BruceET
    Jan 10, 2018 at 9:01
  • $\begingroup$ Meaning? I am having problem in finding critical region. $\endgroup$
    – Daman
    Jan 10, 2018 at 9:03
  • $\begingroup$ You have two intervals to consider $(0,1)$ and $(1,2).$ You have some arbitrary choices to make in order to have a test of size $\alpha$ The criterion involves $\ge K$ not $> K.$ Use part of an interval. $\endgroup$
    – BruceET
    Jan 10, 2018 at 9:08
  • $\begingroup$ Can i show you 4 options i had in this question. Could you please give a try coz i tried it tons of time. $\endgroup$
    – Daman
    Jan 10, 2018 at 9:09
  • $\begingroup$ If you reject just for $x$ in the interval $(1,2),$ then what is $\alpha$ and what is the power? $\endgroup$
    – BruceET
    Jan 10, 2018 at 9:17

1 Answer 1

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Let $\Lambda(x) = \frac{f_0(x)}{f_1(x)}$ and reject if $\Lambda(x) < \tau$. Note that because the likelihood ratio has the Monotone Likelihood Ratio Property we can instead reject if $x>\tau'$.

Letting $\phi(x) = I_{x>.95}$ be our test. We can see that $\alpha = E_0[\phi(x)] = P_0[X>.95] = .05$.

To get the power of the test: $E_{1}[\phi(x)] = P_1[X>.95] = \frac{1}{2}[2-.95] = \frac{21}{40}$.

Because the test was constructed using the Neyman-Pearson Lemma, and because there is only one parameter in the alternative set, it is a UMP test.

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  • $\begingroup$ So today is the day I understood this topic. I was weak in Stas. inference that's why didn't understand a thing last year. I just wanted to say thanks. I understood your answer now. Last year it went over my head. $\endgroup$
    – Daman
    Nov 27, 2018 at 17:01

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