methods to overcome the identifiability issue. Instead of working with the pairwise co-occurrence probabilities, they start by estimating the joint probabilities of three consecutive observations $\Pr[Y_{t-1},Y_t,Y_{t+1}]$. Noticing that these three random variables are conditionally independent given $X_t$, the triple-occurrence probability factors into $$\begin{aligned}
\Pr[Y_{t-1},Y_t,Y_{t+1}] = \sum_{k=1}^{K}\Pr[X_t=x_k]\Pr[Y_{t-1}|X_t=x_k]\Pr[Y_{t}|X_t=x_k]\Pr[Y_{t+1}|X_t=x_k],\end{aligned}$$ which admits a tensor *canonical polyadic decomposition* (CPD) model [@Hsu2009; @Anandkumar2012; @Anandkumar2014]. Assuming $K\leq N$, the CPD is essentially unique if two of the three factor matrices have full column rank, and the other one is not rank one [@harshman1970foundations]; in the context of HMMs, this is equivalent to assuming ${\boldsymbol{M}}$ and ${\boldsymbol{\varTheta}}$ both have linearly independent columns, which is a relatively mild condition. The CPD is known to be unique under much more relaxed conditions [@sidiropoulos2017tensor], but in order to uniquely retrieve the transition probability using the relationship $$\Pr[Y_{t+1}|X_{t}] = \sum_{j=1}^{K}
\Pr[Y_{t+1}|X_{t+1}\!=\!x_j]\Pr[X_{t+1}\!=\!x_j|X_t],$$ $K\leq N$ is actually the best we can achieve using triple-occurrences without making further assumptions. [^1] A salient feature in this case is that if the triple-occurrence probability $\Pr[Y_{t-1},Y_t,Y_{t+1}]$ is exactly given (meaning the rank of the triple-occurrence tensor is indeed smaller than $N$), the CPD can be efficiently calculated using generalized eigendecomposition and related algebraic methods [@sanchez1990tensorial; @leurgans1993decomposition; @DeLathauwer2004a]. These 