Likelihood of Ordered Binomials

Question

I have four numbers ($k_1 \dots k_4$) that come from 4 binomial trials with parameters $n$ and $\theta_1 \dots \theta_4$. I want to test (via likelihood ratio test) the hypotheses

$$H_0: \theta_1 = \theta_2 = \theta_3 = \theta_4$$ versus $$H_1: \theta_1 > \theta_2 > \theta_3 > \theta_4$$

I tried to calculate

$$P(k_1,k_2,k_3,k_4|\theta_1 =\dots = \theta_4 = \theta) =\int_0^1 \left(\prod_{i=1}^4\binom{n}{k_1}\theta^{k_i}(1-\theta)^{n-k_i}\right)d\theta$$

and

$$P(k_1,k_2,k_3,k_4|\theta_1 > \theta_2 > \theta_3 > \theta_4) = \int_0^1\binom{n}{k_1}\theta_1^{k_1}(1-\theta_1)^{n-k_1}\int_0^{\theta_1}\binom{n}{k_2}\theta_2^{k_2}(1-\theta_2)^{n-k_2}\int_0^{\theta_2}\binom{n}{k_3}\theta_3^{k_3}(1-\theta_3)^{n-k_3}\int_0^{\theta_3}\binom{n}{k_4}\theta_4^{k_4}(1-\theta_4)^{n-k_4}d\theta_4\dots d\theta_1$$

I.e., since the $\theta$s are ordered under the alternative hypothesis, I only consider each $\theta_i$ up to the value of its predecessor.

Using the values of $k_i = {8,7,5,4}$ and $n=12$, evaluating these expressions numerically seems to give me that the equality case has a higher probability. I would not expect this since the $k$s are ordered. Is there a problem with my expressions?

Answer 1

The integrals you described are not frequentist likelihood ratios, which involve maximizing over the unknown parameters given the constraints in the two models, but rather are akin to the Bayesian marginal likelihood $$P(x|\mathcal{M}_i) = \int P(x|\theta,\mathcal{M}_i) \pi(\theta|\mathcal{M}_i) d\theta$$ where $\mathcal{M}_i$, $i=0,1$ are the null and alternative models In the null case, implicitly there is a prior $\pi(\theta|\mathcal{M_0}) = 1$ hidden in the integral, which does indeed integrate to 1 over the unit hypercube $[0,1]^4$.

In the alternate case, the same implicit prior is at work, now there is a normalization constant missing, equal to the integral over the polytope defined by the inequality constraints, the value of which is less than 1. Once that has been included, you could compare the two marginal likelihoods via the Bayes factor.