This post continues a series on possible remedies to a counterexample to the Law of Likelihood based on Borel’s paradox. In this post I discuss the possibility of avoiding the counterexample by using symmetry principles to restrict the conditional probabilities that appear in the Law of Likelihood. **This remedy is prima facie appealing, but it runs afoul of an ingenious version of Borel’s paradox due to Kenny Easwaran** (2008, Ch. 8).

### Recap of the Counterexample

The counterexample under consideration runs as follows. Take a sphere of circumference 360 miles equipped with an arbitrary system of latitudes and longitudes. Let $E$ be the datum that a point $P$ randomly selected from a uniform distribution on the surface of the sphere lies within one mile of the intersection of the equator and the prime meridian. Let $H_1$ be the hypothesis that $P$ lies on the equator. Let $H_2$ be the hypothesis that $P$ lies on the union of the prime meridian and its opposite (the “prime meridional circle”).^{1}^{
}

The Law of Likelihood^{2} implies that $E$ favors $H_2$ over $H_1$ when the relevant conditional probabilities are computed in the standard way. **This result is wrong** because the distinction between the equator and the prime meridional circle is (by stipulation) arbitrary. Thus, the Law of Likelihood needs to be abandoned, restricted, or revised.

Abandoning the Law of Likelihood would be a rather drastic response to a problem that only arises for probability-zero hypotheses, so I have been considering possible restrictions and revisions.

### Remedy 4

There are many ways to apply Kolmogorov’s theory of regular conditional distributions in the problem above. A naive approach that the theory permits yields a likelihood ratio of 1.6, but a different approach that the theory also permits would give yield the desired likelihood ratio of 1.^{3}

What I’m calling Remedy 4 to Borel’s paradox would be to argue for a principle that places **additional restrictions on Kolmogorov’s theory**, at least as it is applied in the context of the Law of Likelihood, so that **only approaches that yield the likelihood ratio of 1 would be permitted.**

An obvious place to look for such a restriction is to symmetry considerations. A coordinate system that treats the red circle in the figure above as an equator and the blue circle as a meridional circle, for instance, introduces an asymmetry that is not present in the setup of the problem. It seems right to say that conditional probabilities that reflect that asymmetry are inappropriate and should not be used.

### Why Remedy 4 Fails

Avoiding easy variants on the example above requires adopting the following restriction on the conditional probabilities that appear in the Law of Likelihood when it is applied to that example:

**Rotation Preservation:**For any hypothesis $H$ according to which $P$ lies in a region that is invariant under rotations of the sphere about some axis, it is required that $\mbox{Pr}(H|G)$ have a common value $a$ for any hypothesis $G$ according to which $P$ lies on a great circle that passes through the poles of that axis.

Unfortunately for Remedy 4, Easwaran (2008, Ch. 8) shows that **Rotation Preservation leads to a violation of the essentially universally accepted principle of Finite Additivity**^{4} in the presence of the following additional assumptions:

**Weak Conglomerability:**If there is a constant $a$ such that $\mbox{Pr}(A|E)=a$ for all $E\in\mathcal{E}$, where $\mathcal{E}$ is a partition, then it is required that $\mbox{Pr}(A)=a$.**Conjunction Identity**(my term)**:**If $A \mbox{&} E=B \mbox{&} E$, then it is required that $\mbox{Pr}(A|E)=\mbox{Pr}(B|E)$.

Denying Finite Additivity is not an option. Denying Weak Conglomerability would cause Remedy 4 to collapse into Remedy 3. The only remotely plausible way to deny Conjunction Identity is to maintain that it holds only relative to some structure on the hypothesis space such as a $\sigma$-field or a partition. Adopting that claim would cause Remedy 4 to collapse into Remedy 2. Thus, **Remedy 4 is not an adequate alternative to other remedies I have presented.**

### Proof

Here’s how Rotation Preservation, Weak Conglomerability, and Conjunction Identity lead to a violation of Finite Additivity.

Let $A$ be the hypothesis that $P$ is in the union of the circular regions of points that are within some number of miles $0<d<90$ of the poles in some system of latitudes and longitudes. Let $\mathcal{G}_1$ be a partition^{5} of the sphere into great circles that pass through those poles. Rotation Preservation requires that $\mbox{Pr}(A|G)$ be invariant under rotations of that great circle around those poles, i.e. that $\mbox{Pr}(A|G)=a$ for some $a$ and all $G \in \mathcal{G}_1$. It follows by Weak Conglomerability that $\mbox{Pr}(A)=a$.

Let $A’$ be the hypothesis that $P$ is in the band the circles just discussed sweep out when they are rotated around some axis perpendicular to the axis that passes through the poles at their centers. Let $\mathcal{G}_2$ be a partition^{6} of the sphere into great circles that pass through the poles of that axis. Take the unique great circle $G^*$ that is in both $\mathcal{G}_1$ and $\mathcal{G}_2$. $A \mbox{&}G^*=A’ \mbox{&}G^*$, so Conjunction Identity requires that $\mbox{Pr}(A|G^*)=\mbox{Pr}(A’|G^*)$. But by the same Rotation Preservation and Weak Conglomerability argument given above, $\mbox{Pr}(A’)=\mbox{Pr}(A’|G^*)$. It follows that $\mbox{Pr}(A)=\mbox{Pr}(A’)$

But Finite Additivity implies $\mbox{Pr}(A’)>\mbox{Pr}(A)$:

**Finite Additivity:**If $A\mbox{&}B=\varnothing$, then it is required that $\mbox{Pr}(A\cup B)=\mbox{Pr}(A)+\mbox{Pr}(B)$.

Finite Additivity requires $\mbox{Pr}(A’)=\mbox{Pr}(A\cup (A’\setminus A))=\mbox{Pr}(A)+\mbox{Pr}(A’\setminus A)>\mbox{Pr}(A)$.

### Conclusion

Remedy 4 requires Rotation Preservation, which leads to a violation of Finite Additivity on pain of denying either Weak Conglomerability or Conjunction Identity. Denying Weak Conglomerability would cause Remedy 4 to collapse into Remedy 3, while denying Conjunction Identity would cause it to collapse into Remedy 2. Thus, Remedy 4 is not a viable alternative to those other remedies to Borel’s paradox as a counterexample to the Law of Likelihood.

Thanks to Kenny Easwaran for permission to use images.

To **share your thoughts about this post**, comment below or send me an email. Comments support $\LaTeX$: surround mathematical expressions with single dollar signs for inline mode or double dollar signs for display mode (just as you would in writing a $\LaTeX$ document).

- As in “Borel’s Paradox as a Counterexample to the Law of Likelihood,” one should remove from each hypothesis one of the two points at which the two circles intersect so that one cannot avoid the counterexample by restricting the Law of Likelihood to mutually exclusive hypotheses. I omit this point here for ease of exposition. ↩
- The Law of Likelihood says that $E$ favors $H_1$ over $H_2$ if and only if $\mbox{Pr}(E|H_1)/\mbox{Pr}(E|H_2)=k>1$, and $k$ measures the degree of that favoring. ↩
- Thanks to Michael Lew and to Adam Brodie for emphasizing this point in personal correspondence. ↩
- I write “
*essentially*universally accepted” because some formal theories that resemble probability theory violate additivity, such as Dempster-Shafer theory and John Norton’s approach to representing ignorance. Additivity is appropriate in probability theory, in which degrees of belief are assumed to span belief and disbelief, as Norton puts it. It is not appropriate in alternative theories in which degrees of belief are assumed to span something like knowledge and ignorance. ↩ - $\mathcal{E}$ is not technically a partition because the two poles themselves are in each of its members. Easwaran (2008, 84, fn. 3) suggests several ways around this nuisance. Each one requires a bit of fudging, but that fudging seems excusable given that the problem involves only two of the uncountably many points on the surface of the sphere and on the great circles that pass through those points. I’ll address this problem explicitly in a more formal treatment of this topic. ↩

Michael Lew says

Greg, on my reading you have misrepresented our correspondence. My suggestion was that the scope of the law of likelihood is naturally restricted to circumstances where the likelihoods in the ratio come from the same probability model so that the proportionality constants are known to be identical. Your version here seems to apply a restriction to Kolmogorov’s theory of probability rather than to the law of likelihood.

Greg Gandenberger says

Thanks, Michael. The only item in this post that I said came out of our correspondence was the point that Kolmogorov’s theory can be applied to a pair of great circles in a way that yields a likelihood ratio of one. Is that a misrepresentation? If so, I apologize and will be happy to fix the problem!