question about Neyman-Pearson lemma vs the likelyhood ratio. From my textbook it says that if you want to test:

  • $H_0: \theta \in \Theta_0$ vs $H_1: \theta \in \Theta_0 ^ c$, then you can use a LRT
  • $H_0: \theta = \theta_0$, vs $H_1: \theta = \theta_1$, then you can use UMP

And for the record, I am defining LRT and UMP from casella berger as follows:

LRT is a test for $H_0: \theta \in \Theta_0$ vs $H_1: \theta \in \Theta_0 ^ c$ that has a rejection region of the form: $$\frac{\sup_{\Theta_0} L(\theta|X)}{\sup_{\Theta}L(\theta|X)} \leq c $$ Where c is a constant $\Theta$ is the parameter space, $\Theta$ is the parameter space for the null, and $L(\theta|X)$ is the likelihood.

UMP/Neyman-Pearson is a test for $H_0: \theta = \theta_0$, vs $H_1: \theta = \theta_1$that is of the form:

  • Reject if $f(x|\theta_1) > k f(x|\theta_0)$

Where k is some constant and f is the pdf

Now it seems to me that these seem to specify overlapping situations and I would just like to be sure I understand which tests to use for which situations:

  • $H_0: \theta= \theta_0$, $H_1: \theta= \theta_1$, UMP
  • $H_0: \theta= \theta_0$, $H_1: \theta\neq \theta_0$, LRT or UMP as you can write $H_1 : \theta = \theta_1 \neq \theta_0$
  • $H_0: \theta= \theta_0$, $H_1: \theta > \theta_0$, UMP by writing $H_1: \theta = \theta_1 > \theta_0$

Now if this is right, it seems that when it comes time to invert these for confidence intervals:

  • 2 sided confidence interval => invert a $H_0: \theta= \theta_0$, $H_1: \theta\neq \theta_0$ (which could come from LRT or UMP)
  • 1 sided confidence interval => invert a $H_0: \theta= \theta_0$, $H_1: \theta > \theta_0$ (which you can only get by inverting a UMP test).

Is this correct?

  • $\begingroup$ Can you explain precisely what tests you're referring to by 'UMP' & 'LRT'? - else I fear differences in terminology may lead to confusion. $\endgroup$ – Scortchi - Reinstate Monica Aug 8 '18 at 21:14
  • $\begingroup$ Updated to define UMP and LRT. $\endgroup$ – Anonymous Emu Aug 8 '18 at 21:51
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    $\begingroup$ I am not sure whether I understand you correctly, but as far as I know, you cannot simply write the Alternative Hypothesis from $H_0:\theta=\theta_0 \quad vs: H_1: \theta \neq \theta_0$ as $H_1: \theta=\theta_1$ you can only do this if $\Theta=\{\theta_0, \theta_1\}$ $\endgroup$ – Sebastian Aug 9 '18 at 12:26
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    $\begingroup$ @0rangetree: Nor in general for testing $H_0:\theta=\theta_0$ vs $ H_1: \theta \geq \theta_0$. Though when the likelihood ratio is a monotone function of the minimal sufficient statistic, you end up with the UMP test for any $\theta_1$ in $H_1$ (the Karlin-Rubin theorem). $\endgroup$ – Scortchi - Reinstate Monica Aug 9 '18 at 14:49
  • $\begingroup$ So to address @Scortchi , and 0rangetree at the same time I actually disagree, or at least PSU does according to (onlinecourses.science.psu.edu/stat414/node/308), they test the hypothesis $H_0:\mu = 10$ vs $H_1: \mu > 10$, by saying that a test: $H_0: \mu = 10$ is UMP for every $\mu_1 > 10$, and thus we can apply Neyman Pearson to get a UMP test. $\endgroup$ – Anonymous Emu Aug 9 '18 at 17:08

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