On the information content of discrete phylogenetic characters

Abstract

Phylogenetic inference aims to reconstruct the evolutionary relationships of different species based on genetic (or other) data. Discrete characters are a particular type of data, which contain information on how the species should be grouped together. However, it has long been known that some characters contain more information than others. For instance, a character that assigns the same state to each species groups all of them together and so provides no insight into the relationships of the species considered. At the other extreme, a character that assigns a different state to each species also conveys no phylogenetic signal. In this manuscript, we study a natural combinatorial measure of the information content of an individual character and analyse properties of characters that provide the maximum phylogenetic information, particularly, the number of states such a character uses and how the different states have to be distributed among the species or taxa of the phylogenetic tree.

Appendix

Proof of Lemma 1

We first consider the case $m=2$. In this case, we have $b(m+s)=b(2+s)$ and $b(m)=b(2)=1$ as well as $b(m+s-1)=b(s+1)$ and $b(m+1)=b(3)=1$. In total, we have $b(m+s) \cdot b(m) = b(s+2) > b(s+1) = b(m+1)\cdot b(m+s-1)$, which is true for all $s\ge 2$.

We now consider the case $m\ge 3$. As $s\ge 2$, we have:

$$\begin{aligned}&2m + 2s> 2m +2 \\&\quad \Rightarrow 2(m+s)-5> 2m-3 \\&\quad \Rightarrow (2(m+s)-5) \cdot (2(m+s)-7)!! \cdot (2m-5)!! \\&\qquad> (2m-3) \cdot (2(m+s)-7)!! \cdot (2m-5)!! \\&\quad \Rightarrow (2(m+s)-5)!! \cdot (2m-5)!!> (2m-3)!! \cdot (2(m+s)-7)!! \\&\quad \Rightarrow (2(m+s)-5)!! \cdot (2m-5)!!> (2(m+1)-5)!! \cdot (2(m+s-1)-5)!! \\&\quad \Rightarrow b(m+s) \cdot b(m) > b(m+1)\cdot b(m+s-1). \end{aligned}$$

The last line uses the fact that $b(m)=(2m-5)!!$ for all $m\ge 3$. This completes the proof. $\square$

Note that Lemma 1 is only stated for $m \ge 2$. If $m=1$, the lemma only holds for $s\ge 3$. To see this, consider the case $m=1$ and $s=2$. Then, $b(m+s)\cdot b(m)= b(1+2)\cdot b(1) = b(1+1)\cdot b(1+2-1)= b(m+1)\cdot b(m+s-1)$, as $b(1)=b(2)=b(3)=1$. Therefore the strict inequality stated in the lemma no longer holds.