Let us suppose that we are trying to analyze a given starfish. We would like to know which species does the starfish belong to. We have a list of 1000 starfish species, but we know that there is an unneglectable probability that the list is incomplete.

We know that $p\%$ of all starfish belong to species $A_1$. Next, we email the pictures of the starfish to an expert on marine biology and he says that the starfish can be $A_1$, but it is also possible that it is $A_2$ or $A_3$.

Let us denote by $A_1, A_2, A_3$ the statement that the starfish is $A_1, A_2, \text{or } A_3$. Let us denote $A^{+}_2, A^{+}_3$ the statement that the marine biologist adds $A^{+}_2, A^{+}_3$ to the list of candidate species. Finally $N_n$ means that the starfish does not belong to any of the species $A_1, \dots, A_n$. We would then update our probabilities as follows:

$$ P(A_1|A^{+}_{2}A^{+}_{3}) = P(A_1)P(A^{+}_{2}A^{+}_{3}|A_1)\frac{1}{P(A^{+}_{2}A^{+}_{3})} \\ P(A_2|A^{+}_{2}A^{+}_{3}) = P(A_2)P(A^{+}_{2}A^{+}_{3}|A_2)\frac{1}{P(A^{+}_{2}A^{+}_{3})} \\ P(A_3|A^{+}_{2}A^{+}_{3}) = P(A_3)P(A^{+}_{2}A^{+}_{3}|A_3)\frac{1}{P(A^{+}_{2}A^{+}_{3})} \\ P(N_3|A^{+}_{2}A^{+}_{3}) = P(N_3)P(A^{+}_{2}A^{+}_{3}|N_3)\frac{1}{P(A^{+}_{2}A^{+}_{3})} \\ $$ Now, here is the thing. We have no reason to believe that the microbiologist is any better at identifying $A_2$ rather than $A_3$. So we would expect the posterior for both $A_2$ and $A_3$ to be equal: $$ P(A_2|A^{+}_{2}A^{+}_{3}) = P(A_3|A^{+}_{2}A^{+}_{3}) $$ From the abovementioned equations, using the chain rule we see that is the case if and only if $$ \frac{P(A^{+}_{3}|A^{+}_{2}A_3)}{P(A^{+}_{3}|A^{+}_{2}A_2)} = \frac{P(A^{+}_{2}|A_2)}{P(A^{+}_{2}|A_3)} $$ Now, this still leaves open the question of priors on $A_2$ and $A_3$. Intuition tells us, that the prior probability should be a small fraction of the probability that the starfish does not belong to any of the species in consideration so far: $$ P(A_2|X) = fP(N_1|X) = f(1-p) $$ However, this leads to a strange situation when we try to update the probabilities sequentially, by first updating on $A^{+}_2$: $$ P(A_2|A^{+}_2) = P(A_2)\frac{P(A^{+}_2|A_2)}{P(A^{+}_2)} = f(1-p)\frac{P(A^{+}_2|A_2)}{P(A^{+}_2)} \\ P(N_2|A^{+}_2) = P(N_2)\frac{P(A^{+}_2|N_2)}{P(A^{+}_2)} = (1-f)(1-p)\frac{P(A^{+}_2|N_2)}{P(A^{+}_2)} $$ and then $A^{+}_3$: $$ P(A_2|A^{+}_2A^{+}_3) = P(A_2|A^{+}_2)\frac{P(A^{+}_3|A_2A^{+}_2)}{P(A^{+}_3|A^{+}_2)} = f(1-p)\frac{P(A^{+}_2|A_2)}{P(A^{+}_2)}\frac{P(A^{+}_3|A_2A^{+}_2)}{P(A^{+}_3|A^{+}_2)} \\ P(A_3|A^{+}_2A^{+}_3) = P(A_3|A^{+}_2)\frac{P(A^{+}_3|A_3A^{+}_2)}{P(A^{+}_3|A^{+}_2)} = f(1-f)(1-p)\frac{P(A^{+}_2|N_2)}{P(A^{+}_2)}\frac{P(A^{+}_3|A_3A^{+}_2)}{P(A^{+}_3|A^{+}_2)} $$ where we used the assumption that $P(A_3|A^{+}_2) = f P(N_2|A^{+}_2)$. Putting these equations together produces: $$ f(1-p)\frac{P(A^{+}_2|A_2)}{P(A^{+}_2)}\frac{P(A^{+}_3|A_2A^{+}_2)}{P(A^{+}_3|A^{+}_2)} = f(1-f)(1-p)\frac{P(A^{+}_2|N_2)}{P(A^{+}_2)}\frac{P(A^{+}_3|A_3A^{+}_2)}{P(A^{+}_3|A^{+}_2)} $$ which simplifies to: $$ P(A^{+}_2|A_2)P(A^{+}_3|A_2A^{+}_2) = (1-f)P(A^{+}_2|N_2)P(A^{+}_3|A_3A^{+}_2) $$ and using our necessary and sufficient condition derived from the equality of the posteriors: $$ \frac{P(A^{+}_2|A_3)}{P(A^{+}_2|A_3)}\frac{P(A^{+}_2|A_2)}{P(A^{+}_2|N_2)} = (1-f)\frac{P(A^{+}_3|A_3A^{+}_2)}{P(A^{+}_3|A_2A^{+}_2)} \\ P(A^{+}_2|N_2) = \frac{P(A^{+}_2|A_3)}{1-f} $$ Therefore unless we set $P(A^{+}_2|N_2)$ as equal to a multiple of $P(A^{+}_2|A_3)$ we will get the unintuitive result that the posteriors of $A_2$ and $A_3$ are not equal. This would be very odd. It would be as if it mattered in which order $A_2$ and $A_3$ appear in the biologist's email.

This model is quite inflexible, since we cannot set $P(A^{+}_2|N_2)$ at liberty, but must always fix it as a multiple of $P(A^{+}_2|A_3)$, which limits our modeling capacities. Furthermore we have to be careful with setting $f$, since the probabilities have to be smaller than 1.

Given that this model suffers from these restrictions, what would be a better way of setting a prior on a finite, very large, but unknown set?

  • 2
    $\begingroup$ There is a whole Bayesian literature on estimating the number of species, see, e.g., Lijoi et al. $\endgroup$ – Xi'an Apr 15 at 15:54

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