BenB52058
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I am a bit stuck on the following question: We have a p.d.f for a r.v X

f(x:\theta)=\theta x^{\theta-1} if (0<x<1) (theta >1) and 0 otherwise.

I want to find the most powerful test for H0: \theta=\theta_0 against HA: \theta=\theta_1.

So I use the Neyman-Pearson lemma. To say the MP test is the likelihood ratio test.

I want to take the likelihood ratio but I don't really know how since the null hypothesis is not simple if it was simple I would just get

\Lambda=(\theta_0/\theta_1)x^{\theta_0-\theta_1} and then reject for small values of this i.e. (\Lambda &lt;c ) and use the p.d.f. to find what value of c we need to use here to get the correct level we desire.

However I really don't know what the ratio would be here as I don't know what the values of \theta_0

and \theta_1 are surely the values we reject on would be depending on this yet the question makes no mention. (I.e. if \theta_1&gt;\theta_0 then the exponent would be negative so we would look for larger values of x rather than smaller to make the ratio smaller for instance.)

I have seen sometimes where you use the supremum on the numerator (taken over a restriction of values ) and then the supremum (of all possible values) on the denominator but I don't know how to do that here? Should I be looking at MLE's? (maximum likelihood estimates) to proceed or what.

I think when I get this confusion cleared about how to deal with the composite null hypothesis rather than simple and how that effects the ratio test I will be able to complete this question so any help is appreciated.

Thanks

(hopefully I explained myself well enough for someone to understand what I mean.)
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Gregorius
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(Original post by BenB52058)
I am a bit stuck on the following question: We have a p.d.f for a r.v X

f(x:\theta)=\theta x^{\theta-1} if (0<x<1) (theta >1) and 0 otherwise.

I want to find the most powerful test for H0: \theta=\theta_0 against HA: \theta=\theta_1.

So I use the Neyman-Pearson lemma. To say the MP test is the likelihood ratio test.

I want to take the likelihood ratio but I don't really know how since the null hypothesis is not simple.
Why do you think that this null hypothesis is not simple?
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BenB52058
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(Original post by Gregorius)
Why do you think that this null hypothesis is not simple?
I thought because theta 0 is not known that it wouldn't be simple but I guess from your reply that it would be.

Okay so maybe I've just confused a definition but suppose that it is simple how would I proceed because it would still depend on the vales of theta 0 and 1 what the rejection region is wouldn't it I mean since 0<x<1 then if the exponent is negative we would reject for large x but if the exponent was positive we would reject for small x.

Am I supposed to break this into cases I.e theta0<theta1 and vice versa?

Thanks!
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Gregorius
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(Original post by BenB52058)
I thought because theta 0 is not known that it wouldn't be simple but I guess from your reply that it would be.

Okay so maybe I've just confused a definition but suppose that it is simple how would I proceed because it would still depend on the vales of theta 0 and 1 what the rejection region is wouldn't it I mean since 0<x<1 then if the exponent is negative we would reject for large x but if the exponent was positive we would reject for small x.

Am I supposed to break this into cases I.e theta0<theta1 and vice versa?

Thanks!
A simple null hypothesis is one that specifies completely the distribution concerned.

Splitting your test according to the sign of theta 0 minus theta 1 would be fine.
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BenB52058
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(Original post by Gregorius)
A simple null hypothesis is one that specifies completely the distribution concerned.

Splitting your test according to the sign of theta 0 minus theta 1 would be fine.
thanks!!!
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