Comparison Techniques for Competing Brownian Particles

Abstract

Consider a finite system of Brownian particles on the real line. Each particle has drift and diffusion coefficients depending on its current rank relative to other particles, as in Karatzas et al. (Ann I H Poincare-PR 52(1):323–354, 2016). We prove some comparison results for these systems. As an example, we show that if we remove a few particles from the top, then the gaps between adjacent particles become stochastically larger, the local times of collision between adjacent particles become stochastically smaller, and the remaining particles shift upward, in the sense of stochastic ordering.

Appendix

Lemma 5.1

Take a $d\times d$-reflection nonsingular $\mathcal M$-matrix R and fix a nonempty subset $J \subseteq \{1, \ldots , d\}$. Then

$$\begin{aligned} 0 \le [R]_J^{-1} \le [R^{-1}]_J. \end{aligned}$$

Proof

Since $R = I_d - Q$, where $Q \ge 0$ is a $d\times d$-matrix with spectral radius strictly less than one, we can apply the Neumann series:

$$\begin{aligned} R^{-1} = I_d + Q + Q^2 + \cdots \end{aligned}$$

(43)

By Lemma 5.6, $[R]_J = I_{|J|} - [Q]_J$ is also a reflection nonsingular $\mathcal M$-matrix, so we have:

$$\begin{aligned}{}[R]_J^{-1} = I_{|J|} + [Q]_J + [Q]_J^2 + \cdots \end{aligned}$$

But from (43) we get:

$$\begin{aligned}{}[R^{-1}]_J = I_{|J|} + [Q]_J + [Q^2]_J + \cdots \end{aligned}$$

That $[Q^k]_J \ge [Q]_J^k$ for $k = 1, 2, 3, \ldots$ can be proved by induction using Lemma 5.2. The rest is trivial. $\square$

Lemma 5.2

Take nonnegative matrices A ($m\times d$) and B ($d\times n$), and let $I \subseteq \{1, \ldots , m\}$, $J \subseteq \{1, \ldots , d\}$, $K \subseteq \{1, \ldots , n\}$ be nonempty subsets. Then

$$\begin{aligned}{}[A]_{IJ}[B]_{JK} \le [AB]_{IK}. \end{aligned}$$

Proof

Let $A = (a_{ij})$ and $B = (b_{ij})$. Then for $i \in I$ and $k \in K$,

$$\begin{aligned} \left( [A]_{IJ}[B]_{JK}\right) _{ik} = \sum \limits _{j \in J}a_{ij}b_{jk} \le \sum \limits _{i=1}^da_{ij}b_{jk} = (AB)_{ik} = \left( [AB]_{IK}\right) _{ik}. \end{aligned}$$

$\square$

Lemma 5.3

Take a $d\times n$-matrix A and a vector $a \in \mathbb {R}^n$. Let $I \subseteq \{1, \ldots , d\}$ be a nonempty subset. Then $[Aa]_I = [A]_{I\times \{1, \ldots , n\}}a$.

The proof is trivial.

Lemma 5.4

Take a $d\times d$-nonnegative matrix A and a nonnegative vector $a \in \mathbb {R}^d$. Let $J \subseteq \{1, \ldots , d\}$ be a nonempty subset. Then $[Aa]_J \ge [A]_{J}[a]_J$.

The proof is trivial.

Lemma 5.5

Let $R \le \overline{R}$ be two $d\times d$-reflection nonsingular $\mathcal M$-matrices. Then $R^{-1} \ge \overline{R}^{-1} \ge 0$.

Proof

Apply Neumann series again: If

$$\begin{aligned} R = I_d - Q,\ \ \overline{R} = I_d - \overline{Q}, \end{aligned}$$

then $Q \ge \overline{Q} \ge 0$, and so $Q^k \ge \overline{Q}^k \ge 0$, $k = 1, 2, \ldots$. Thus,

$$\begin{aligned} R^{-1} = I_d + Q + Q^2 + \cdots \ge I_d + \overline{Q} + \overline{Q}^2 + \cdots = \overline{R}^{-1}. \end{aligned}$$

$\square$

Lemma 5.6

If R is a $d\times d$-reflection nonsingular $\mathcal M$-matrix and $I \subseteq \{1, \ldots , d\}$ is a nonempty subset, then $[R]_I$ is also a reflection nonsingular $\mathcal M$-matrix.

Proof

Use [32, Lemma 2.1]. A $d\times d$-matrix $R = (r_{ij})$ is a reflection nonsingular $\mathcal M$-matrix if and only if

$$\begin{aligned} r_{ii} = 1,\ i = 1, \ldots , d;\ \ r_{ij} \le 0,\ i \ne j, \end{aligned}$$

and, in addition, R is completely-$\mathcal S$, which means that for every principal submatrix $[R]_J$ of R there exists a vector $u > 0$ such that $[R]_Ju > 0$. Now, switch from R to $[R]_I$. The same conditions hold:

$$\begin{aligned} r_{ii} = 1,\ i \in I;\ \ r_{ij} \le 0,\ i \ne j,\ i, j \in I, \end{aligned}$$

and, in addition, for every principal submatrix $[[R]_I]_J = [R]_J$ of $[R]_I$, where $J \subseteq I$, there exists a vector $u > 0$ such that $[R]_Ju > 0$. This means that $[R]_I$ is also a reflection nonsingular $\mathcal M$-matrix. $\square$

Lemma 5.7

If $A \ge B \ge 0$ and $C \ge D \ge 0$ are matrices such that the matrix products AC and BD are well defined, then $AC \ge BD \ge 0$.

The proof is trivial.