Neyman Pearson Lemma and most powerful test

Question

This is a homework question. I was given a random sample of independent and identically distributed $X_i$'s and wish to test the hypotheses:

$$H_0: \theta = \theta_0$$ $\text{vs}$
$$H_A: \theta = \theta_A \quad (\theta_A < \theta_0)$$

$f(x;\theta)=0.5(1+\theta \, x)\,$ where $-1\leq \theta \leq 1$ and $-1\leq x \leq1$

In this case, I should apply the Neyman Pearson Lemma, as we are testing for a simple null and simple alternate

However, the question asked to show that there is no most powerful test in this case, and explain why

What I did was applying the N-P Lemma, so

NP: $[(\frac{1}{2^n})\,\prod (1+\theta_A \, x_i)]/[(\frac{1}{2^n})\, \prod (1+\theta_0 \, x_i)]>k$

and if i take the log of it i get

$\sum[\log((1+\theta_A\, x_i)/(1+\theta_0 \, x_i))]\geq \ln k$

And then I concluded, since there is no simple closed way to arrange the LHS to be in the form of a test statistic, of only the data, and the test statistic depends on $\theta_A$, therefore there is no most powerful test in this case, is this a correct assessment of the situation?

Also, please correct my misunderstanding, I was under the assumption that if we are testing for simple null and simple alternate, there is always a most powerful test as the Neyman-Pearson lemma suggests, as evident by this question, my understanding was wrong, but I am not sure in what way does the Neyman-Pearson lemma give the most powerful test?

Answer 1

You are probably supposed to answer something in the direction of https://en.wikipedia.org/wiki/Uniformly_most_powerful_test The NP lemma tells that the likelihood ratio test is most powerful for point hypotheses.

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For instance if $$H_0:\theta = \theta_0 \quad \text{and} \quad H_a:\theta=\theta_a$$ then the test: $$\sum[\log((1+\theta_a\, x_i)/(1+\theta_0 \, x_i))]\geq \ln k$$ is most powerful.

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But when you look for a range

$$\quad H_a:\theta < \theta_0$$

then the test:

$$\sum[\log((1+\theta_a\, x_i)/(1+\theta_0 \, x_i))]\geq \ln k_a$$

is not anymore most powerful for every $\theta < \theta_0$, since it is only most powerful for $\theta_a$. In the case of some other $\theta_b \neq \theta_a$ you should use

$$\sum[\log((1+\theta_b\, x_i)/(1+\theta_0 \, x_i))]\geq \ln k_b$$