For a generator $G$ of a Markov process in continuous time and finite state space, how would one prove that the entries of $e^{tG}$ are non-negative?

Question

I have a generator matrix G for a Markov chain in continuous time and finite state space and I am looking to prove that the entries of $e^{tG} \geq 0 $

By definition $G = P'(0)$ with entries $g_{ij} = p_{ij} '(0)$

Where $p_{ij}(t) = Prob[X_j = t | X_i = 0]$

There are 2 cases to take into account:

$i≠j$:

$$g_{ij} = \lim_{t\to 0} \frac{p_{ij}(t) - p_{ij}(0)}{t} = \lim_{t\to 0} \frac{p_{ij}(t) - 0}{t}$$

$$\Rightarrow g_{ij}\geq 0, (p_{ij}(t) \in [0,1])$$

$i=j:$

$$g_{ii} = \lim_{t\to 0} \frac{p_{ii}(t) - p_{ii}(0)}{t} = \lim_{t\to 0} \frac{p_{ii}(t) - 1}{t}$$

$$\Rightarrow g_{ii} \leq 0, (p_{ii}(t) \in [0,1])$$

Now I know that the diagonal entries of G are $\leq 0$ and the off-diagonal entries are $\geq 0$.

I know that $e^{tG}= \sum_{n=0}^\infty \frac{(tG)^n}{n!}$

I also know that I could let $R$ be the matrix of eigenvectors of $G.$

$\Rightarrow R^{-1}GR = D $, where $D$ is a diagonal matrix with eigenvalues of $G$ as its entries.

$\Rightarrow G = RDR^{-1}$ $\Rightarrow e^{tG} = Re^{tD}R^{-1}$

And $e^{tD}$ is going to be a diagonal matrix with non-negative entries.

What I'm having a problem with is now proving that $e^{tG}_{ij} \geq 0$ , for all $i,j,$ knowing the above.

Answer 1

Pick some $a$ such that $a\geqslant-g_{ii}$ for every $i$ (this assumes the state space is finite). The matrix $G_a=G+aI$ has only nonnegative entries hence, for every $n\geqslant0$, so has $(G_a)^n$.

Now, $\mathrm e^{tG_a}=\sum\limits_{n\geqslant0}(t^n/n!)(G_a)^n$ is a linear combination with nonnegative coefficients of matrices with nonnegative entries, hence $\mathrm e^{tG_a}$ has only nonnegative entries, and $\mathrm e^{tG}=\mathrm e^{-ta}\mathrm e^{tG_a}$ hence $\mathrm e^{tG}$ has only nonnegative entries.