Confidence measure/statistic for inferring a category with random independent sampling

Question

Suppose there's a group of dogs and a set of people A,B,C,D. If I know that, in this hypothetical scenario, the person who owns the dog is the person who walks it the most I can try to infer who the owner of a dog x is by sampling observations of the dog being walked.

Suppose I sample (1,1,98,0) after taking N=100 independent observations of x being walked (where the entries in the vector correspond to person A,B,C,D). I can say, with "confidence", that person C is the owner of x.

Statistically, I am thinking of this as a categorical distribution and, it is my understanding, that with a Dirichlet prior the maximum likelihood estimate corresponds to the proportion observed. For example, the maximum likelihood estimate for person C being the owner of x based on the 100 observations is .98. However, suppose my sample size were smaller and I had observed (0, 0, 1, 0). The estimate for the proportion/probability here would be 1 but I would be less confident. What i'm looking at is which test statistic/s or p-value is/are the relevant one/s to this model. Is it a Chi squared test? And if so, would I want to test (1/100,1/100,98/100,0) against (0,0,1,0) (because I know the dog can only have one owner)?

Answer 1

Here is one possible approach. Suppose the counts are $(n_1, n_2, n_3, n_4)$ and let $j$ denote the index where the count is largest. You can think of this as a hypothesis testing problem of $$ H_0 : \max(p_1, p_2, p_3, p_4) \neq p_j $$ against $$ H_1 : \max(p_1, p_2, p_3, p_4) = p_j $$ where $p_i$ is the probability of individual $i$ walking the dog. An idea then would be to determine the maximum likelihood of the sample under the two hypotheses and reject $H_0$ when the data are much more likely under $H_1$. In either case the likelihood is $$ L = p_1^{n_1} p_2^{n_2} p_3^{n_3} p_4^{n_4} $$ which is maximized under $H_1$ when each probability equals the sample proportion for that category. Under $H_0$ it's not quite as obvious since we want to maximize the likelihood under the somewhat unusual constraint that the largest estimated probability not be the category with the largest count, but my guess is that we can do this by averaging the sample proportions of the most common and second most common categories. Even though $H_0$ would not strictly be true in this circumstance, we can view the resulting likelihood as a supremum over $H_0$.

Having obtained these two likelihoods we can then calculate a ratio and reject $H_0$ when this ratio is unduly favorable to $H_1$. As far as computing a $p$-value goes it's possible that minus two times the likelihood ratio would be approximately $\chi^2$ but I'm not sure about the degrees of freedom. An alternative might be to devise a Monte Carlo simulation to come up with an approximate $p$-value.