quick question.
I need to find the rejection region of H_0, meaning that I need to find the constant value for which if X=x belongs to the region H_0 is rejected.
The pdf is c(1-x)^(c-1) with x belonging to (0,1) , the size of the test is 0.1 and under H_0: c=2, H_1: c=3.
Solving with the NPL formula I get: $x> 1-(2/3)k$
How do I find the value of k?
I tried integrating the pdf on the interval 0 to $1-(2/3)k$ and equating to the size 0.1, but I seems to get it wrong.
Am I doing silly calculation mistakes? How should I find k ( and so the rejection region ):
Thank you so much guys.
$X$ has a $\mathsf{Beta}(1,c)$ distribution, so you can verify that $1-X\sim \mathsf{Beta}(c,1)$ and hence $$-c\ln(1-X)\sim \mathsf{Exp}(1)\,.$$
By NP lemma, a most powerful test for testing $H_0:c=2$ versus $H_1:c=3$ rejects $H_0$ for large values of the likelihood ratio $r(x)=f_{H_1}(x)/f_{H_0}(x)$ where $f(\cdot)$ is the pdf of $X$.
Now $r(x)=\frac32(1-x)$ for $x\in(0,1)$, hence we reject $H_0$ if $1-X>k$ for some constant $k\in(0,1)$. Size of the test is $0.1$, which means
$$P_{H_0}(1-X>k)=P_{H_0}(-2\ln(1-X)<-2\ln k)=0.1$$
That is, $$\int_0^{-2\ln k}e^{-x}\,dx=1-e^{2\ln k}=0.1$$
Solving for $k$ you get the rejection region $\{X:X<1-k\}$.