trees on a taxon set $\mathcal
X$, let ${\mathbb P}(T)$ denote the probability of gene tree $T$. Then for each subset $\mathcal C_g\subseteq \mathcal X_g$, the probability that $\mathcal C_g$ is a clade on a gene tree is $${\mathbb P}(\mathcal C_g)= \sum_T {\mathbb P}(\mathcal C_g | T) {\mathbb P}(T) = \sum_T I(\mathcal C_g\in \mathcal H(T))\, {\mathbb P}(T),$$ where $I$ is the indicator function with values of 1 or 0. Note that the probability of any trivial clade is therefore 1.

We emphasize that the clade probabilities for an $n$-taxon species tree $\sigma$ do *not* form a probability distribution. The presence of different clades may not be mutually exclusive events (for instance, if $\mathcal C_g \subset \mathcal C'_g$), and their probabilities do not sum to 1.

\[prop:trivinv\] If $|\mathcal X|=n$, then for any distribution of binary gene trees the sum of the probabilities of all non-trivial clades is $n-2$.

Denoting $n$-taxon gene trees by $T$, $$\sum_{\mathcal C_g\subset \mathcal X_g \atop\text{non-trivial}} \mathbb P(\mathcal C_g)= \sum_{\mathcal C_g\subset \mathcal X_g \atop\text{non-trivial}}  \sum_{T} I(\mathcal C_g \in \mathcal H(T))\, \mathbb P(T)=\sum _{T} \sum_{\mathcal C_g \subset \mathcal X_g\atop\text{non-trivial}} I(\mathcal C_g\in \mathcal H(T))\,  \mathbb P(T).$$ But since each binary gene tree has $n-2$ non-trivial clades, this shows that 