Let $A$ be the generator matrix of a continuous-time Markov chain. This means that $A$ has positive off-diagonal elements $A_{ij} > 0$, $i \ne j$, and row sums $\sum_j A_{ij}$ equal to $0$. For example, $A$ could be $$ A = \left( \begin{matrix} -7 & 4 & 3 \\ 1 & -2 & 1 \\ 3 & 5 & -8 \end{matrix} \right). $$ I am interested in proving the following claim about the matrix $B = x(x I - A)^{-1}$ for some $x > 0$.

Claim. For the matrix $B = x(x I - A)^{-1}$, it holds that the diagonal elements of $B - B^2$ are non-negative.

Using numerical simulations I have convinced myself that this claim is likely true; however, I have not been able to make much progress toward proving it.

It is straightforward to show that the matrix $B$ is stochastic. However, the claim above is not true for all stochastic matrices $B$; there is something special about stochastic matrices of this particular form.

Any ideas?

  • $\begingroup$ $x$ is an arbitrary positive real number. Thanks, edited $\endgroup$
    – user133281
    Mar 1 at 18:04
  • 4
    $\begingroup$ I suggest you just write $B = ( I - A)^{-1}$ as the $x$ doesn't contribute anything. I.e. you are asking about a claim for $B = x(x I - A)^{-1}$ to hold for arbitrary $x\gt 0$, for abittrary generator matrix $A$ which is equivalent to asking about $B = ( I - x^{-1}A)^{-1}$ but this is equivalent to asking about $B = ( I - A)^{-1}$ since rescaling $A$ by a positive number does not change the rows summing to 0 or the off diagonals of $A$ being $\gt 0$ [i.e. re-scaling by $x^{-1}$ maps from one arbitrary generator to another]. $\endgroup$ Mar 2 at 2:18

4 Answers 4


Let me continue from @user1551's idea and show that indeed $\mathbf{B} - \mathbf{B}^2$ has non-negative diagonals.

Step 1. We recap @user1551's reduction.

Let $\mathbf{A}$ be a transition-rate matrix as in OP, and let $\mathbf{B} = ( \mathbf{I} - \mathbf{A})^{-1}$. (Here, we are assuming $x=1$ without losing the generality.) Choose $d > 0$ sufficiently large so that

$$ \mathbf{P} = \mathbf{I} + \frac{1}{d} \mathbf{A} $$

is a stochastic matrix. Solving this for $\mathbf{A}$ gives $\mathbf{A} = d(\mathbf{P} - \mathbf{I})$, hence

\begin{align*} \mathbf{B} - \mathbf{B}^2 &= -\mathbf{A}(\mathbf{I} - \mathbf{A})^{-2} \\ &= \frac{d}{(d+1)^2} (\mathbf{I} - \mathbf{P})(\mathbf{I} - c \mathbf{P})^{-2} \end{align*}

where $c = \frac{d}{d+1} \in (0, 1)$. In light of this, it suffices to prove:

Claim. Let $\mathbf{P}$ be a stochastic matrix. Then for any $c \in (0, 1)$, the diagonal entries of $$ (\mathbf{I} - \mathbf{P})(\mathbf{I} - c\mathbf{P})^{-2} $$ are non-negative.

Step 2. Let $X = (X_n)_{n\geq 0}$ denote a Markov chain with the transition matrix $\mathbf{P}$. Fix a state $j$, and define $(\tau_k)_{k\geq 0}$ as the sequence of return times to state $j$. More precisely,

$$ \tau_0 = 0 \qquad \text{and} \qquad \tau_{k+1} = \inf\{ n > \tau_k : X_n = j \}. $$

If $\mathbb{P}_j$ denotes the law of $X$ started at $j$, then

\begin{align*} \mathbf{e}_j^{\top} (\mathbf{I} - c \mathbf{P})^{-1} \mathbf{e}_j &= \sum_{n=0}^{\infty} c^n (\mathbf{e}_j^{\top} \mathbf{P}^n \mathbf{e}_j) = \sum_{n=0}^{\infty} c^n \mathbb{P}_j(X_n = j) \\ &= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c^n \mathbb{P}_j(\tau_k = n) = \sum_{k=0}^{\infty} \mathbb{E}_j[c^{\tau_k}] = \sum_{k=0}^{\infty} \mathbb{E}_j[c^{\tau_1}]^k \\ &= \frac{1}{1 - \mathbb{E}_j[c^{\tau_1}]}. \end{align*}

This computation is related to the original problem as follows:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}c} \frac{1 - c}{1 - \mathbb{E}_j[c^{\tau_1}]} &= \frac{\mathrm{d}}{\mathrm{d}c} \mathbf{e}_j^{\top} (1 - c)(\mathbf{I} - c \mathbf{P})^{-1} \mathbf{e}_j \\ &= - \mathbf{e}_j^{\top} (\mathbf{I} - \mathbf{P})(\mathbf{I} - c\mathbf{P})^{-2} \mathbf{e}_j. \end{align*}

Step 3. The above relation shows that the claim is equivalent to showing:

Claim 2. The function $$ f(c) = \frac{1 - c}{1 - \mathbb{E}_j[c^{\tau_1}]} $$ is non-increasing for $c \in (0, 1)$.

This is the same as showing $1/f(1-x)$ is non-increasing for $x \in (0, 1)$. However,

\begin{align*} \frac{1}{f(1-x)} &= \frac{1 - \mathbb{E}_j[(1-x)^{\tau_1}]}{x} = \int_{0}^{1} \mathbb{E}_j[ \tau_1 (1 - xt)^{\tau_1 - 1} ] \, \mathrm{d}t \end{align*}

It is clear that $x \mapsto \tau_1 (1 - xt)^{\tau_1 - 1}$ is non-increasing for each $t \in [0, 1]$, hence the claim is proved. $\square$


Not an answer, but some observations. Let $d$ be the maximum diagonal entry of $-\frac{1}{x}A$. Then $S=I+\frac{1}{xd}A$ is a stochastic matrix whose diagonal elements are less than $1$. Conversely, given any $d>0$ and any stochastic matrix $S$ whose diagonal elements are less than $1$, $A=-xd(I-S)$ is a generator of a CTMC. Therefore, while the stochastic matrix $B$ in the OP takes a very special form and the statement that $B-B^2$ has a nonnegative diagonal is not true for a general stochastic matrix $B$, if we express $B$ in terms of $S$, we may reformulate the statement in question as one about an almost general stochastic matrix $S$.

More specifically, let $c=\frac{d}{d+1}$. Then $0<c<1$ and \begin{align*} B-B^2 &=(B^{-1}-I)B^2\\ &=-\frac{A}{x}\left(I-\frac{A}{x}\right)^{-2}\\ &=d(I-S)\big(I+d(I-S)\big)^{-2}\\ &=d(d+1)^{-2}(I-S)(I-cS)^{-2}.\\ \end{align*} Hence the OP is essentially asking whether $(I-S)(I-cS)^{-2}$ has a nonnegative diagonal for any $c\in(0,1)$ and any stochastic matrix $S$ whose diagonal elements are less than $1$.

When $c$ is small, the answer is clearly affirmative, because $(I-cS)^{-2}$ is close to $I$ and $e_j^T(I-S)(I-cS)^{-2}e_j\approx e_j^T(I-S)e_j=(1-s_{jj})\ge0$. In fact, for any $i\ne j$, we have \begin{align*} \left((I-cS)^{-2}\right)_{jj} &=\left(I+2cS+3c^2S^2+4c^3S^3+\cdots\right)_{jj}\ge1,\\ \left((I-cS)^{-2}\right)_{ij} &=\left(I+2cS+3c^2S^2+4c^3S^3+\cdots\right)_{ij}\\ &\le2c+3c^2+4c^3+\cdots\\ &=(1-c)^{-2}-1. \end{align*} Therefore, when $(1-c)^{-2}-1\le1$, i.e., when $c\le1-\frac{1}{\sqrt{2}}\approx0.2929$ or $d\le\sqrt{2}-1$, the largest element in the nonnegative vector $v=(v_1,v_2,\ldots,v_n)^T=(I-cS)^{-2}e_j$ will be $v_j$. Consequently, \begin{align*} e_j^T(I-S)(I-cS)^{-2}e_j &=e_j^T(I-S)v\\ &=(1-s_{jj})v_j-\sum_{k\ne j}s_{jk}v_k\\ &\ge(1-s_{jj})v_j-\sum_{k\ne j}s_{jk}v_j\\ &=\left(1-\sum_{k=1}^ns_{jk}\right)v_j\\ &=0. \end{align*} This settles the question for small $c$. However, numerical simulation suggests that the answer is affirmative for actually any $c\in(0,1)$.


Although a solution has been posted using analytical means, I note that it is possible to see the nonnegativity of the diagonal elements of $B - B^2$ purely algebraically, via combinatorial interpretation of the entries of the matrix $B$.

Fix $n$ and regard $A$ as a function of its off-diagonal entries $a_{ij}$, $i \neq j$, with its diagonal entries determined hy the requirement that the row sums of $A$ are all zero. A monomial in the variables $a_{ij}$, $i \neq j$, can be thought of as a directed graph on the vertices $\{1,2,\dots,n\}$ without "self-loops": the power of $a_{ij}$ in the monomial records how many directed edges go from $i$ to $j$ in the graph (the distinctness of the indices $i$ and $j$ in the variables $a_{ij}$ means that no vertex is ever connected to itself by a single edge). When $G$ is such a directed graph I will write $a_G = \prod_{(i,j) \in G} a_{ij}$ for the corresponding monomial (this product being interpreted "with multiplicity" whenever $G$ has more than one edge from $i$ to $j$ - although such $G$ will not arise in the cases of interest below). A forest on the vertices $\{1,2,\dots,n\}$ is an undirected graph on these vertices whose connected components contain no loops (i.e., are trees), and a rooted forest is a forest whose connected components each have a distinguished vertex. A rooted forest can be regarded as a directed graph by orienting each edge toward the root of its component. I will write $\mathcal{F}_k$ for the set of all rooted forests on $\{1,\dots,n\}$ having exactly $k$ components (equivalently, having exactly $n-k$ edges), and for $1 \leq j \leq n$ I will write $\mathcal{F}_k(j)$ for the set of rooted forests on $\{1,\dots,n\}$ in which $j$ is a root. Note that there is exactly one element $F$ of $\mathcal{F}_n$ (the graph having vertices $\{1,2,\dots,n\}$ and no edges), and in this case the corresponding monomial $a_F$ is $1$ and the sets $\mathcal{F}_n(j)$ are all equal to $\mathcal{F}_n$.

Letting $B := x(xI - A)^{-1}$ and denoting the row $i$, column $j$ entry of $B$ by $B_{ij}$, and letting $d(xI - A) := \det(xI - A)/x$ (note that it is clear a priori that $x$ divides $\det(xI - A)$), the combinatorial interpretations of the entries of $B$ promised up above are that $$ d(xI-A) = \sum_{k=1}^n \left(\sum_{F \in \mathcal{F}_k} a_F\right) x^{k-1}, $$ and that $$ d(xI-A) B_{ij} = \sum_{k=1}^n \left(\sum_{F \in \mathcal{F}_k(j), i \sim_F j} a_F\right) x^{k-1}, \qquad 1 \leq i, j \leq n, $$ where $i \sim_F j$ means that $i$ and $j$ belong to the same connected component of $F$ (a condition that is vacuous when $i=j$). In this last formula note that when $i \neq j$ the inside sum corresponding to the index value $k=n$ is empty (because the one element of $\mathcal{F}_n$ has no edges) and does not contribute to the sum.

As the OP observed, it is straightforward to show that $B$ is (right) stochastic, i.e., that it has all its row sums equal to $1$. The formulas above are a refinement of that observation. That the rows of $B$ sum to $1$, or equivalently that the rows of $d(xI-A) B$ sum to $d(xI - A)$, reflects the fact that for any row index $i$, the rooted forests on the vertices $\{1,\dots,n\}$ can be partitioned according to the root of the component that contains $i$, with $d(xI - A) B_{ij}$ accounting for those forests in which $i$ lies in a component in which $j$ is a root, and with $d(xI-A)$ accounting for all of these forests.

To see how these formulas imply the nonnegativity of the diagonal of $B - B^2$, note that for any stochastic matrix $B$ (i.e. not just $B = x(xI - A)^{-1}$ with $A$ as above) we have $$ B_{ii} = \left(\sum_{j=1}^n B_{ij}\right) B_{ii} = \sum_{j=1}^n B_{ij} (B_{ji} + B_{ii} - B_{ji}) = [B^2]_{ii} + \sum_{j \neq i} B_{ij} (B_{ii} - B_{ji}). $$ For $B$ of the above specific type (but not for all stochastic matrices), our formulas will show that $B_{ii} - B_{ji} \geq 0$ for all $j \neq i$, which then (by the formula displayed immediately above) implies $B_{ii} \geq [B^2]_{ii}$ for all $i$.

Indeed, while $d(xI - A) B_{ii}$ is a sum over all rooted forests in which $i$ is a root, for $i \neq j$ the term $d(xI-A) B_{ji}$ accounts for only those forests in which $i$ is a root and $j$ is in the same component as $i$. For $i \neq j$ the difference $p_{ji} := d(xI - A) B_{ii} - d(xI - A) B_{ji}$ is thus the sum $$ p_{ji} = \sum_{k=1}^n \left(\sum_{F \in \mathcal{F}_k(i), j \not \sim_F i} a_F\right) x^{k-1}; $$ which is plainly nonnegative whenever $x$ and the $a_{ij}$ are because it is a sum of monomials in these variables with nonnegative integer coefficients (indeed, any positive coefficient is $1$). Since $d(xI - A)$ is also such a sum (and is positive for positive $x$ and positive values of the $a_{ij}$), the nonnegativity of $B_{ii} - B_{ij}$ for $i \neq j$, and hence the nonnegativity of $B_{ii} - [B^2]_{ii}$ for all $1 \leq i \leq n$ follows.

The formula for $d(xI-A) = \det(xI-A)/x$ and the formulas for the matrix entries $B_{ij}$ (which are themselves determinant formulas by e.g. Cramer's rule) are slightly nontrivial to verify. Note, for example, that while it is straightforward to see that $d(xI - A)$ is a sum of monomials in $x$ and $a_{ij}$ of degree $n-1$, it is perhaps less clear which monomials of this form occur in the expansion and which do not. When $n=4$, for example, the formula immediately shows that the monomial $a_{12} a_{24} a_{41}$ does not appear with nonzero coefficient in $\det(xI - A)/x$ (the corresponding directed graph has a loop), although the monomial $a_{12} a_{24} a_{41} x$ does indeed appear with nonzero coefficient in the product of the diagonal entries of $xI - A$ (as do all degree $4$ products of $x$ with three "$a$" variables taken from distinct rows). So the formula above for $d(xI-A)$ implies that a substantial amount of cancellation would take place were one to simply evaluate the determinant via Laplace expansion (for example).

Some general remarks on the formula for $d(xI - A)$:

  • It is a polynomial of degree $n-1$ in $x$ with leading term $x^{n-1}$.
  • The coefficient of $x^{n-2}$ in $d(xI-A)$ is the sum of all variables $a_{ij}$, $i \neq j$.
  • The coefficient of $x^{n-3}$ in $d(xI-A)$ is the sum of all products of pairs of variables $a_{ij} a_{kl}$, $i \neq j$, $k \neq l$, where we additionally require that $a_{ij}$ and $a_{kl}$ not occupy positions that are reflections of one another across the diagonal, or in the same row - or more suggestively for the general case, where the directed edges from $i$ to $j$ and from $k$ to $l$ do not together form a loop (preventing the monomial from representing part of a tree) or point in two "directions" from the same point (preventing the monomial from representing part of a rooted tree).
  • In general, the coefficient of $x^{n-(d+1)}$ in $d(xI-A)$ is a sum of degree $d$ monomials $a_{i_1 j_1} a_{i_2 j_2} \cdots a_{i_d j_d}$ in the variables $a_{ij}$, where a given monomial appears if and only if the directed graph with vertices $\{1,\dots,n\}$ and edges $\{(i_k, j_k): 1 \leq k \leq d\}$ is a rooted forest. When $d = n-1$ this is the requirement that $\{(i_k, j_k): 1 \leq k \leq d\}$ define a rooted tree. There are famously $n^{n-1}$ such trees (and hence such monomials) by Cayley's formula (which tells us there are $n^{n-2}$ labeled trees on $\{1,2,\dots,n\}$, each of which can be rooted in $n$ different ways).
  • The coefficient of $x^{k-1}$ in $d(xI - A)$ is in general a sum of $|\mathcal{F}_k|$ monomials in the variables $a_{ij}$ (each monomial having degree $n-k$). As just observed, it is well known that $|\mathcal{F}_1| = n^{n-1}$; I do not know of a simple citation for where the numbers $|\mathcal{F}_k|$ (equivalently, the number of rooted forests on $\{1,2,\dots,n\}$ involving $n-k$ edges) have been computed in print, but this MO post provides express (if somewhat ugly) formulas for the number of labeled forests with $k$ components on $\{1,\dots,n\}$ (that is, counting the objects in $\mathcal{F}_k$ up to differences in choices of roots).
  • One can use computer software to expressly compute and check all of the formulas above, although they involve large numbers of terms even when $n$ is as small as $3$ or $4$, and the large amount of symmetry in the problem makes it difficult (at least for me) to visually process, let alone independently evaluate, the resulting output.
  • 2
    $\begingroup$ Like a commenter mentioned, the $x$ can be dropped. Does this make your solution easier? $\endgroup$ Mar 8 at 17:35
  • $\begingroup$ @BenjaminWang I suppose it depends on what one means by "easier"; $x$ does not appear in a particularly complicated way in these formulas, and depending on how you look at it, is arguably helpful in keeping track of where monomials of different degree in the $a_{ij}$ appear in the relevant determinants. The main formulas do simplify (there is certainly no need to parameterize sums over rooted forests by the number "$k$" of components), but this simplification is primarily visual. For purposes of computing anything nontrivial, IMVHO it is not any easier to take $x=1$ $\endgroup$ Mar 8 at 17:45
  • $\begingroup$ The above also shows directly e.g. that if $x$ and the $a_{ij}$ lie in some subset of your field (or commutative ring) that is closed under $+$ and $*$, then so do the ingredients in the relevant formulas (of course division is necessary to get the matrix entries themselves). The "rescale by $x^{-1}$" reduction makes one think for a second about determinants and how this plays out in a field of quotients before seeing this, while the above has this on the surface. (For purposes of e.g. integer computation on computers, it may also be helpful not to "just" rescale by $x^{-1}$ at the outset) $\endgroup$ Mar 8 at 18:14

The missing part of this post is a proof that $B$ is a stochastic matrix. Such a proof can quickly lead to a proof that OP's claim is true. I suggest using Inverse of strictly diagonally dominant matrix for reference.

Mimicking the link, we have $A' := I-A$ is a strictly diagonally dominant matrix with positive diagonal and negative off diagonal and $\mathbf 1 = A'\mathbf 1$. Now $\delta A'$ has all diagonal entries $\lt \frac{1}{n}$ for some $\delta \gt 0$ small enough. Write $Q = I- \delta A'$ which is a positive matrix and has Perron vector $\mathbf 1$ that satisfies $Q\mathbf 1 = (1-\delta) \mathbf 1$.

$\delta^{-1}\big(I-A\big)^{-1}=\big(\delta A'\big)^{-1}= \big(I-Q\big)^{-1}=I + Q + Q^2 + Q^3+\dots $
where the spectral radius of $Q$ is $1-\delta$ hence the series converges

(i.) This proves $\big(I-A\big)^{-1}$ is a positive matrix and $\big(I-A\big)^{-1}\mathbf 1 = \mathbf 1$ hence it is a stochastic matrix.
(ii.) $Q$, being strictly substochastic in each row, is a transition matrix for a transient chain hence it also proves OP's claim. i.e. $N=\big(I-Q\big)^{-1}= I+QN$ is the fundamental matrix for some absorbing state markov chain and $n_{k,j}$ is the expected number of times the process is in transient state $k$ given a start in transient state $j$.

For $j\neq k$, we may look at $N = I + QN$ to get $n_{k,j} = \sum_{r=1}^n q_{k,r}n_{r,j}$ i.e. it is a sub-convex combination of elements in $\mathbf n_j$ . Compare against $n_{k,k} = 1+\sum_{r=1}^n q_{k,r}n_{r,k}$ which is not a sub-convex combination. [This is equivalent to "first step analysis" for expected visits, though the probability interpretation is not needed.] There are finitely many values in each column of $N$, all of which are positive, hence a column max exists but said maximum cannot be a sub-convex combination of other values in that column $\implies n_{k,k}$ is maximal for each column.

$\implies \delta \cdot N =\big(I-A\big)^{-1} = B$ has diagonal elements strictly larger than other elements in their respective column $\implies B-B^2$ has a positive diagonal since the $k$th diagonal of $B^2=BB$ is a non-trivial average of elements in $\mathbf b_k$ hence is strictly less than the column max, $b_{k,k}$.

If $A$ were symmetric, then we can get to the result via spectral theory since $\big(I-A\big)$ has all eigenvalues in $\mathbb R_{\geq 1}$. So all eigenvalues of $\big(I-A\big)^{-1}$ are $\in (0,1]$ and $\lt 1$ except for the Perron root which implies all eigenvalues of $B-B^2$ are $\lambda - \lambda^2 \geq 0$ with equality only for the Perron Root. Since the diagonal of a real symmetric matrix is in the convex hull of the eigenvalues of said matrix [majorization], the result follows. [With minor modification this gives the result for normal $A$ as well.] Corollary: If the chain is reversible then $B$ is diagonally similar to a symmetric matrix and this does not change eigenvalues or the diagonals of $B-B^2$ so the same spectral argument gives the result.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .