The authors wish to acknowledge an error that was overseen in the proof of Result 1 in [1]. By carefully inspecting the authors’ proof, it becomes clear that an MLE cannot exist if N_k = N for at least one k. This is easily seen from the authors’ formula for in Result 1, where N_k = N implies , a contradiction since then . All results however remain valid as stated if their Assumption 1 is replaced by the following version.

Assumption 1 Assume that the sum over the lineages’ prevalences is larger than one, but no alleles is 100% prevalent. In other words, more than one lineage is found in at least one infection, i.e., and all lineages are not found in every infection, i.e., N_k ≠ N for all k.

By replacing Assumption 1 with the version above in [1], the results hold without modifications. All other modifications that need to be made in the article are minor and obvious. However, the case N_k = N for at least one k was not properly addressed. This occurred because it was overseen that the proof of in Result 1 is not applicable then. What goes wrong in this case? The answer is somewhat subtle. Heuristically, this contradiction occurs because no point in the parameter space is a critical point, i.e., a point at which all derivatives of L vanish. However, for any fixed λ, L(λ, p|x) attains a maximum for some , with . The reason is that L(λ, p|x) = −∞ for (where denotes the n−1-dimensional simplex). For λ → 0, . Hence, is necessarily monotonically increasing in λ, implying that no MLE exists. In mathematical terms this can be formulated as follows:

Remark 1 Assume that at least one lineage is found in every sample, i.e.,N_k = N for at least one k, but not all are found in every sample, i.e.,N_k ≠ N for at least one j. Then, the log-likelihood function does not attain a maximum. However, its smallest upper bound is

The supremum is reached in the limit of any sequence (λ_t, p_t) with , if N_k ≠ N and if N_k = N.

Proof. Because L(λ, p|x) is bounded by 0, the supremum exists. Furthermore, a sequence (λ_t, p_t) exists with .

Without loss of generality let N₁,…,N_m < N and N_m+1 = … = N_n = N. Hence, (1)

Let (λ_t) be any monotone sequence with . Moreover, let c_k >0 for k = 1,…,m. Now let p_t be a sequence satisfying for k = 1,…,m and for k = m + 1,…,n. Without loss of generality let for k = 1,…,m and for k = m + 1,…,n. For sufficiently large t this sequence is defined and . Hence,

Next define . Note that this definition is independent of the sequence (λ_t, p_t), with λ_tp_t →(c₁,…,c_m, ∞,…,∞) for t → ∞.

The next aim, is to identify potential maxima of f. Clearly, . Equating the partial derivatives to zero gives . The Hessian matrix is given by and clearly negative definite. Thus, f attains a global maximum at . Therefore .

If (λ_t, p_t) is any sequence with for a k with 1 ≤ k ≤ m, it is easily seen from (1) that . Moreover, if for 1 ≤ k ≤ m and at least one k with m + 1 ≤ k ≤ n, without loss of generality for , (1) implies implying that this limit is less than the maximum of f. The above considerations imply that the supremum of the log-likelihood function must be the maximum of f. Deriving finishes the proof.

The case that N_k = N for all k is treated in [1]. Moreover, obviously in Remark 1 of [1] a misprint occurred. The expression needs to be replaced by , while the same expression needs to be replaced by in the paragraph below Result 2.