consistently under maximum likelihood. Proof of Theorem 4.2. This theorem follows from the proofs of Theorems 3.2 and 4.1. The only difficulty is to check whether there might exist a Q which is part of the model θ and a corresponding Q$_{0}$ which is a part of the model θ$_{0}$ such that P$_{θ}$ = P$_{θ}$$_{0}$ but Q = Q$_{0}$ . This could only occur if the transition probability matrix P from the same edge and model θ mapped to multiple valid Markov generators. As for the proof of Theorem 4.1, all transition probability matrices are uniquely determined by P$_{θ}$ and as P$_{θ}$ = P$_{θ}$$_{0}$ we must have that P = P$_{0}$ for the corresponding P$_{0}$ in θ$_{0}$ , so by assumption Q must be unique and Q = Q$_{0}$ . Assumption A of Corollary 3.1 implies that the process under consideration is non- stationary and we subsequently find that we can recover the root. We further show in example 3.1 that in the stationary, non-reversible case it is not possible to recover the root. This seems to fit with the collective empirical evidence of Yang and Roberts (1995), Huelsenbeck et al. (2002), and Yap and Speed (2005). That is, in some cases a non-stationary process can recover the root, whereas a stationary, non-reversible process cannot. Assumption A is also the source of a subtle and interesting con- tradiction. It can be shown under mild assumptions that the product of arbitrary stochastic matrices will eventually have almost identical rows as the number of terms in the product increases (Seneta 2006, Section 4.3). It is not difficult to show that the rows must be the stationary distribution of the product, and that if the terms are strand symmetric then the product must be strand symmetric. So for any rea- sonably long history of strand-symmetric processes, we must have that the marginal probabilities are also strand-symmetric. The conclusion is that for our model to work, the process should probably have been strand asymmetric prior to the root of the tree, and strand symmetric thereafter. An alternative interpretation is that 23 5 Discussion