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?

  • $\begingroup$ 1. You have not correctly applied the lemma. 2. If this is an exercise for some subject please see the help center under "Homework". $\endgroup$
    – Glen_b
    Commented Sep 2, 2018 at 6:32
  • $\begingroup$ What is the correct way of applying the lemma? $\endgroup$
    – Kazusa
    Commented Sep 2, 2018 at 6:33
  • $\begingroup$ Sorry, I am not quite understanding what you mean by bringing in a condition that's not part of the lemma, can you please expand on it, thanks $\endgroup$
    – Kazusa
    Commented Sep 2, 2018 at 6:41
  • $\begingroup$ The only conditions I know is the Neyman Pearson Lemma, which apply to only simple null and simple alternative, and the likelihood ratio test, which applies for more general alternatives, where the alternative can be composite etc. If you mean by conditions are when the samples are independent, then that was my fault, i forgot to add that. From what I understand, the Neyman Pearson Lemma is comparing the likelihood of the alternate vs the simple at the values specified by the hypothesis. And if it is large enough, one reject H-0. $\endgroup$
    – Kazusa
    Commented Sep 2, 2018 at 7:10
  • $\begingroup$ And in this case, where I take the log of the LR and try and get the L.H.S. into some form of test statistic, isolated from the theta_A, and couldn't do so, therefore I conclude (which my reason seems fishy at best) that since the LHS is dependent on the value of theta_A, therefore there is no most powerful test in this case. $\endgroup$
    – Kazusa
    Commented Sep 2, 2018 at 7:12

1 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.

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.

But when you look for a range

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

then the test will compare with the maximum likelihood estimate

$$\sum[\log((1+\hat{\theta}\, x_i)/(1+\theta_0 \, x_i))]\geq C$$

and this may not need to coincide with the most powerful for every potential $\theta_b < \theta_0$, in which case one should use

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

I came across this old answer of my and it has been formulated a bit too quickly. There is some nuance possible.

In the case that the likelihood is a monotonous function of the parameter. Then the power may not be the same for every of the alternative hypotheses but it may still be the best powererfull test that we can do.

This is related to the Karlin-Rubin theorem (see also Understanding the proof of Karlin–Rubin Theorem). If the likelihood function is not a monotonous function (in the sufficient statistic, which is a bit tricky to visualise in your case which has no sufficient statistic except the entire sample) then there may be two boundaries for the hypothesis test, and we can make a tradeoff between the two sides which makes that a test is not most powerful everywhere (e.g. compare the power of one-sided and two-sided t-tests).

For a small sample $x_1,x_2$ and null hypothesis $\theta_0 = 0$ we may compute the likelihood ratio as function of $x_1,x_2$.

The likelihood function is a polynomial function

$$\mathcal{L}(\theta) = (1+x_1\theta)(1+x_2\theta) = x_1x_2 \theta^2 + (x_1+x_2) \theta + 1$$

and then likelihood ratio statistic is

$$\frac{\mathcal{L}(\theta_a)}{\mathcal{L}(0)} = (1+x_1\theta_a)(1+x_2\theta_a)$$

The most powerful tests for $\theta_0=0$ with alternative $\theta_a$ will consider critical regions that are constrained by a hyperbola, the points $x_1,x_2$ where the likelihood $(1-x_1\theta_a)(1-x_2\theta_a) = constant$. For different $\theta_a$ that will be different hyperbolas and different regions for different $\theta_a$ (the hyperbolas will intersect instead of overlap).

So there is no way to select a critical region that will be most powerful for all multiple point hypotheses at once. (because the most powerful critical regions of the individual hypotheses do not coincide).


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