Consider a continuous-time Markov process, described by:

$$ \dot p = A_\theta\, p \tag{1}$$

where $p \in \mathbb R^n$ is a probability vector, and $A_\theta \in \mathbb R^{n\times n}$ is a transition matrix, of rank $n-1$, parameterized by $\theta$.

Given some initial distribution $p_0$, the probability vector in time is given by:

$$ p(t,\theta) = e^{A_\theta t}\,p_0 \tag{2}$$

As $t$ increases, $p(t,\theta)$ converges to some stationary distribution $p_\theta^\infty$, the unique eigenvector of $A_\theta$ with null eigenvalue.

I want to study the Fisher information of $p(t,\theta)$ about $\theta$, so I look for its derivative:

$$\begin{align} \partial_\theta p(t,\theta) &= \partial_\theta (e^{A_\theta t}\,p_0) \\ &= t \int\limits_0^1 e^{\alpha A_\theta t} (\partial_\theta A_\theta) e^{-\alpha A_\theta t} d\alpha\ e^{A_\theta t}\, p_0 \\ &\equiv t\, B(t,\theta)\, p(t,\theta) \end{align} \tag{3}$$

I expected (¿maybe incorrectly?) the derivative $\partial_\theta p(t,\theta)$ to converge to some vector independent of $t$:

$$\begin{align} \lim_{t\rightarrow \infty} \partial_\theta p(t,\theta) &\stackrel{?}= \partial_\theta \lim_{t\rightarrow \infty} p(t,\theta) \tag{4}\\ \lim_{t\rightarrow \infty} t\, B(t,\theta)\, p(t,\theta) &\stackrel{?}= \partial_\theta p_\theta^\infty \tag{5} \end{align}$$

I know $\partial_\theta p_\theta^\infty$ is a non-zero finite vector.

Question: Are Eqs. $(4\text{ - }5)$ correct?

If YES, how is the term $t\, B(t,\theta)$ not exploding as $t\rightarrow \infty$? I fail to see how $\lim_{t\rightarrow \infty} B(t,\theta)\, p(t,\theta) \propto \frac{1}{t}$.

If NO, where did I go wrong in my reasoning?

Thank you very much


1 Answer 1


This question made me think this problem from a different perspective, to try and work with the eigendecomposition of $A_\theta$.

Let $\{p(\theta),v_2(\theta),\cdots,v_n(\theta)\}$ be the eigenvectors of $A_\theta$, associated with the eigenvalues $\{0,\lambda_2(\theta),\cdots,\lambda_n(\theta)\}$, where $\mathrm{Re}\{\lambda_j\}<0 \quad\forall_{j=2,\cdots,n}$.

Write the initial distribution in the above eigenbasis, $p_0 = p(\theta)+\sum_{j=2}^n c_j(\theta) v_j(\theta)$, to obtain:

$$ p(t,\theta)= e^{A_\theta t}p_0= \underbrace{\ p(\theta)\ }_{t-\mathrm{indep.}}+ \underbrace{\sum_{j=2}^n c_j(\theta) e^{\lambda_j(\theta)t} v_j(\theta)}_{t-\mathrm{dependent}} $$

As expected, $p(t,\theta)$ is converging to the stationary distribution $p(\theta)$. Turning to the derivative we again find that the $t$-dependent part vanishes as $t\rightarrow \infty$:

$$ \partial p(t,\theta)= \underbrace{\ \partial_\theta p(\theta)\ }_{t-\mathrm{indep.}}+ \underbrace{\sum_{j=2}^n e^{\lambda_jt} \big(t\, \partial_\theta\lambda_j\, c_j\, v_j+ \partial_\theta c_j\, v_j + c_j\, \partial_\theta v_j\big)}_{t-\mathrm{dependent}} $$

Thus, $$ \lim_{t\rightarrow \infty} \partial_\theta p(t,\theta) = \partial_\theta p(\theta) = \partial_\theta \lim_{t\rightarrow \infty} p(t,\theta) $$ Which means Eq.$(4)$ of the question is correct. I still don't know what to think of the term $tB(t,\theta)$ from Eq.$(5)$, but I guess I found a better way to frame the derivative, a way I am more comfortable with.


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