# Doubt in neyman pearson lemma problem while considering the critical region

The given problem is X follows $$Rectangular(\theta,\theta^2)$$ To test $$H_0:\theta=2$$ against $$H_1:\theta=3$$ They asked to calculate the power and size if given crictical region is $$X\geq4.5$$ Now they have calculated Size=$$P_{H_0}(X\geq4.5)$$ But my question is if we follow Neyman Pearson Lemma then $$L(x|H_0)=0$$ when $$4\leq X\leq9$$ Since under $$H_0$$ distribution is $$Rectangular(2,4)$$ and also $$L(x|H_1)=\frac{1}{6}$$ when $$4\leq X\leq9$$ Since under $$H_1$$ distribution is $$Rectangular(3,9)$$. Now Under $$H_0$$ there is probability 0 in $$4\leq X\leq9$$ hence $$\frac{L(x|H_0)}{L(x|H_1)}=0$$ in $$4\leq X\leq9$$. hence since neyman pearson lemma says $$\frac{L(x|H_0)}{L(x|H_1)} \leq K$$ is the critical region for any positive K, then in $$4\leq X\leq9$$ always $$\frac{L(x|H_0)}{L(x|H_1)}=0\leq K$$ since K is positive. Hence $$4\leq X\leq9$$ is always a critical region.So while calculation of size we must consider $$(4,4.5)$$ also as a crtical region apart from the given critical region $$X\geq4.5$$. Is my explanation correct???

The Neyman-Pearson Lemma is about what the optimal rejection region is (which is apparently what you call "critical region", chances are both terms are in use), but you can define non-optimal tests with other rejection regions. If I understand your question right, you got $$X\ge 4.5$$ given as a rejection region, so this is a test different from the Neyman-Pearson optimal one, and if you are asked to analyse that one, you can leave the Neyman-Pearson Lemma alone.