# What distribution has the following likelihood function?

I'm working with a model that uses the Beta-Binomial natural conjugate family. In other words, the prior over the variable of interest $$\theta \sim Beta(\alpha_0,\beta_0)$$ distribution over $$[0,1]$$ and the signal follows a Binomial distribution $$s \sim Binom(\theta,n)$$ over $$\{0, \dots, n\}$$, resulting in a posterior that follows a $$(\theta|s) \sim Beta(\alpha_0+s,\beta_0+n-s)$$ distribution which is easy to update given the prior parameters and the signal). See here for more details. This setup has a nice interpretation (loosely speaking) of being able to say that the prior is equivalent to having seen $$\alpha_0$$ successes out of $$\alpha_0+\beta_0$$ Bernoulli trials with an underlying probability of $$\theta$$, and that the posterior is equivalent to having seen $$\alpha_0+s$$ successes out of $$\alpha_0+\beta_0+n$$ trials.

What I have found is that $$\alpha_0+\beta_0$$ is relatively small (around 10) for the Beta prior distribution that best fits my data, meaning that the posterior is quickly overwhelmed by the signal as the number $$n$$ increases. This is no problem in of itself, but it constrains the set of signals and corresponding posterior means to be too coarse given the strength of the information (I want $$s$$ to be equivalent to only 3 or 4 trials in order to preserve much of the prior belief, but that leaves only 4 or 5 possible signals that could be observed). What I want to do is increase the fineness of the signal space to some other, larger size of $$m$$ trials while keeping the total strength of the signal as "equivalent to" $$n$$ trials. What I came up with as a solution is to use the following likelihood for the signal: $$\ell(s|\theta) \propto \theta^{\frac{n}{m}s}(1-\theta)^{\frac{n}{m}(m-s)} \; \mathrm{ over } \; s \in {0, 1, \dots, m}$$ Applying Bayes' rule, the posterior would then still follow a $$(\theta|s) \sim Beta(\alpha_0+\frac{n}{m}s,\beta_0+n-\frac{n}{m}s)$$ distribution so that we can think of the normalized signal $$\tilde{s} \equiv \frac{s}{m}$$ has exactly the strength relative to the prior that I originally wanted (being "equivalent to" a sample from $$n$$ trials). The difference being that $$\tilde{s}$$ is no longer restricted to the set $$\{0, \frac{1}{n}, \dots, 1\}$$ as it would be if $$m=n$$, but instead can finely cover the entire interval $$\{0, \frac{1}{m}, \dots, 1\}$$ by making $$m$$ large.

That seems to be the needed adjustment to the classic Beta-Binomial model, but I'm struggling to describe the sampling distribution that generates that likelihood $$\ell(\tilde{s}|\theta)\propto \theta^{n\tilde{s}}(1-\theta)^{n-n\tilde{s}} \; \mathrm{ over } \; \tilde{s} \in {0, \frac{1}{m}, \dots, 1}.$$

Is it some form of correlated Binomial distribution? Does anyone recognize this likelihood as belonging to a specific distribution family?

Many thanks!

• It comes across as bizarre that you want to favor the prior belief over the data. This issue seems like the heart of the matter. What is the reason for discrediting your data to that degree?
– whuber
Sep 17 at 17:37
• @whuber Good question. It's not that I want to discredit the data per se, it's that this signal-generating process is part of a stylized model of a particular decision problem, and the practical details suggest that the amount of information that will be learned in the given time period will be about 1/3 of the amount of information that we already have about $\theta$. My challenge is that I want an almost-continuum of such signals rather than the very coarse set that a $Binomial(\theta,3)$ would offer. Sep 17 at 19:47
• If the additional information truly is that little, then the use of a correct Bayesian update will reflect that. That is more likely to be effective and defensible than the seemingly ad hoc approach you describe.
– whuber
Sep 17 at 21:17
• @whuber I agree, I don't want an ad hoc model, and that is why I am trying to find the specific type of signal that will give the type of correct Bayesian update that my model needs. I am hoping for a much more fine-grained signal space than just the set {0, 1, 2, 3}. Sep 19 at 17:59
• I am unable to reconcile that comment with the background information in your question, which states "the signal follows a Binomial distribution." If that's the case, there's nothing you can do about the "signal space:" it is what it is.
– whuber
Sep 19 at 18:48