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Let $S=\{1,...,h\}$ be a finite state space and $X(t)$ an irreducible Markov chain fully described by a generator matrix $Q$ with a transition probability matrix $P(t)=e^{Qt}$ on time horizon $[0,T]$. I have noticed in literature that identifiability of $Q$ is always assumed; however, I can hardly imagine that this cannot be proven. Unfortunately, I get stuck on where to look using the definition of identifiability \begin{equation} P_{Q_{1}}=P_{Q_{2}}\implies Q_{1}=Q_{2} \end{equation} and the MLE \begin{equation} q_{ij}=\frac{N_{ij}(T)}{R_{i}(T)} \end{equation} with $N_{ij}(t)$ the number of $i\rightarrow j$ transition up to time and $R_{i}(t)=\int_{0}^{t}1_{\{X(u)=i\}}du$ for $t\in[0,T]$ and any $i,j\in S$ with $j\neq i$ and \begin{equation} q_{ii}=-\sum_{j=1,j\neq i}q_{ij} \end{equation}

Does anyone have some tips on where to start on or literature proving identifiability?

  • Do you want to know whether $P$ determines $Q$ uniquely? – nejimban Jan 04 '22 at 17:15
  • No, that is the case as by definition $P$ is a real matrix with $\det(P)\neq0$ and thus $\log(P)=Q$ uniquely. However; identifiability of any statistical model implies (for as far as I understand) that $L(Q_{1}|X)=L(Q_{2}|X)\implies Q_{1}=Q_{2}$ for any $X$. I am not sure if my understanding of identifiability is correct, whether it is simply not true or where to begin with the proof. –  Jan 04 '22 at 17:19
  • If $P_{Q_{1}}=P_{Q_{2}}\implies Q_{1}=Q_{2}$ since $Q=\log(P)$ one-to-one, then I am already done..., but does that capture identifiability? –  Jan 04 '22 at 17:32

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If $Q_1\not=Q_2$ then $P_1(t)\not=P_2(t)$ for all $t>0$, and the law of a Markov chain with generator $Q_1$ will be different from the law of a Markov chain with generator $Q_2$. If you are thinking of this as a statistical model with $Q$ as parameter, then the model is identifiable.

In fact, as you seem to be suggesting, a single sample path of the chain will almost surely allow you to recover $Q$. First, $R_i(t)/t$ converges a.s. to $\pi_i$ (the stationary probability for state $i$); second, the diagonal entry, $q_i=-q_{ii}$, is the a.s. limit $\lim_{t\to\infty} N_i(t)/(t\pi_i)$, and finally, for $i\not=j$, $q_{ij} =\lim_{t\to\infty} N_{ij}(t)/(t\pi_i/q_i)$.

John Dawkins
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  • Thank you; however, is identifiability always considered in the limits, since the model is defined on finite $[0,T]$, a single finite Markov chain might have an outcome with $P_{Q_{1}}(X)=P_{Q_{2}}(X)$? –  Jan 08 '22 at 18:40
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    If $P_{Q_1}(t) = P_{Q_2}(t)$ for all $t\in[0,T$ (some $T>0$) then certainly $Q_1=Q_2$. Is this the sort of identifiability you are looking for? – John Dawkins Jan 08 '22 at 18:53
  • Forgot to answer, but yes thank you! If it indeed holds for all $t\in[0,T]$ the model must be identifiable. –  Jan 24 '22 at 14:52
  • I forgot that the estimator does not allow for unused $q_{ij}$ to differ as these are set to $0$ by definition. –  Jan 24 '22 at 15:00