1
$\begingroup$

enter image description here

My Attempt

Comparing $f(x;\theta)$ with the form $a(\theta)b(x)exp[c(\theta)d(x)]$ ,

we get $d(x) = log (1-x)$ and $ c(\theta ) = \theta -1 $ as monotone , increasing function in $\theta$ and therefore , critical region C =[$(x) :log(1-x) > k$ ] gives uniformly most powerful size - $\alpha $ test for testing

$H_0 :\theta \le \theta_0 $ V/S $ H_1 : \theta > \theta_0 $

So for testing given hypothesis, my critical region is C =[$(x) :log(1-x) > k$ ] and then we can find that answer will be option A.

Is my attempt correct?

$\endgroup$

1 Answer 1

3
$\begingroup$

Your attempt is correct, but for this particular multiple choice question we can find the cut-off point $c$ directly given that the critical region is of the form $-\sum\limits_{i=1}^n \ln(1-X_i)^2 <c$.

As you might have already noticed, $$X_i\sim \mathsf{Beta}(1,\theta)\implies 1-X_i\sim \mathsf{Beta}(\theta,1)\implies-2\theta\ln (1-X_i)\sim \chi^2_2$$

Since size of the test is $\alpha$, we have

$$P_{H_0}\left[-\sum_{i=1}^n \ln(1-X_i)^2 <c\right]=P_{H_0}\left[-2\sum_{i=1}^n \ln(1-X_i)< c\right]=P\left[\chi^2_{2n}< c\right]=\alpha$$

Or, $$P\left[\chi^2_{2n}> c\right]=1-\alpha\quad,\,\text{i.e.}\quad c=\chi^2_{2n,1-\alpha}$$

$\endgroup$

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.