Non-Markov property of certain eigenvalue processes analogous to Dyson's model

TL;DR

This paper demonstrates that altering the coefficient in Dyson's eigenvalue process for a 2x2 matrix causes the eigenvalue dynamics to lose the Markov property, highlighting the delicate nature of Markovianity in such models.

Contribution

It shows that changing the off-diagonal coefficient in Dyson's 2x2 matrix eigenvalue process breaks the Markov property, revealing a non-Markovian behavior.

Findings

Eigenvalue process becomes non-Markov when coefficient is changed.

Markov property is sensitive to the off-diagonal coefficient.

Original Dyson process with coefficient 1/√2 is Markovian.

Abstract

It is proven that the eigenvalue process of Dyson's random matrix process of size two becomes non-Markov if the common coefficient $1/\sqrt{2}$ in the non-diagonal entries is replaced by a different positive number.

Equations

$$\Xi (t)=\left(\begin{array}{cccc}{B}_{1,1}(t)& \frac{1}{\sqrt{2}}{B}_{1,2}(t)& \cdots & \frac{1}{\sqrt{2}}{B}_{1,N}(t)\\ \frac{1}{\sqrt{2}}\overline{B_{1,2}(t)}& B_{2,2}(t)& \cdots & \frac{1}{\sqrt{2}}{B}_{2,N}(t)\\ ⋮& ⋮& \ddots & ⋮\\ \frac{1}{\sqrt{2}}\overline{B_{1,N}(t)}& \frac{1}{\sqrt{2}}\overline{B_{2,N}(t)}& \cdots & B_{N,N}(t)\end{array}\right)$$

$$d{\lambda }_i(t)=d{\beta }_i(t)+\frac{\beta }{2}\sum _{j\ne i}\frac{1}{{\lambda }_i(t)-{\lambda }_j(t)}dt$$

$${\Xi }^{c,\delta }(t)=\left(\begin{array}{cc}{B}_1(t)& \sqrt{c/2}{\xi }^{\delta }(t)\\ \sqrt{c/2}{\xi }^{\delta }(t)& B_2(t)\end{array}\right)$$

$${\tilde{\Xi }}^{c,\delta }(t)=\left(\begin{array}{cc}{B}_1(t)& \sqrt{c/2}{B}_3(t)\\ \sqrt{c/2}\overline{B_3(t)}& B_2(t)\end{array}\right)$$

$$d{\lambda }_1(t)=$$

$$d{\lambda }_2(t)=$$

$${\lambda }_1(t)=$$

$${\lambda }_2(t)=$$

$${\lambda }_1(t)=$$

$${\lambda }_2(t)=$$

$${\lambda }_1+{\lambda }_2=$$

$${\lambda }_1-{\lambda }_2=$$

$$d\tilde{\lambda }(t)=d{B}_4(t)+\frac{\delta }{2}\frac{1}{\tilde{\lambda }(t)}dt$$

$$\displaystyle\Bigl{(}\Xi^{c,\delta}(t)\Bigr{)}_{t\geq 0}\stackrel{{\scriptstyle\rm law}}{{=}}\left(U_{\delta}(t)\widetilde{\Xi}^{c,\delta}(t)U_{\delta}^{*}(t)\right)_{t\geq 0}.$$

$$U_{\delta }(t)=\left(\begin{array}{cc}1& 0\\ 0& \frac{B_3(t)}{|B_3(t)|}\end{array}\right)1_{B_3(t)\ne 0}+\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)1_{B_3(t)=0}$$

$$U_{\delta }(t){\tilde{\Xi }}^{c,\delta }(t)U_{\delta }^{\ast }(t)=\left(\begin{array}{cc}{B}_1(t)& \sqrt{c/2}|B_3(t)|\\ \sqrt{c/2}|B_3(t)|& B_2(t)\end{array}\right),$$

$$p_t^{\delta }(x,y)=\frac{1}{2t}{\left(\frac{y}{x}\right)}^{(\delta -2)/4}\exp \left(-\frac{x+y}{2t}\right)I_{(\delta -2)/2}\left(\frac{\sqrt{xy}}{t}\right)$$

$$\displaystyle I_{\nu}(x)\sim\frac{1}{\Gamma(\nu+1)}\Bigl{(}\frac{x}{2}\Bigr{)}^{\nu}\quad\text{as $x\downarrow 0$},$$

$$I_{\nu }(x)\sim \frac{e^x}{\sqrt{2\pi x}}\text{as x↑∞}.$$

$$p_t^{\delta }(0+,y)=\frac{y^{(\delta /2)-1}}{(2t{)}^{\delta /2}\Gamma (\delta /2)}\exp \left(-\frac{y}{2t}\right)$$

$$\begin{array}{rl}\underset{y\to 0+}{\lim }{y}^{1-\delta /2}{p}_t^{\delta }(x,y)& =x^{1-\delta /2}{p}_t^{\delta }(0+,x)\\ & =\frac{1}{(2t{)}^{\delta /2}\Gamma (\delta /2)}\exp \left(-\frac{x}{2t}\right)\end{array}$$

$$p_t^{\delta }(x,y)\sim \frac{1}{2t\sqrt{2\pi }}\frac{y^{(\delta -3)/4}}{x^{(\delta -1)/4}}\exp \left(-\frac{x+y-2\sqrt{xy}}{2t}\right)$$

$$P\left(Z^c(2)\in d{z}_3\mid Z^c(\epsilon )=z_1,Z^c(1)=z_2\right)\text{for }0<\epsilon <1$$

$$P\left(Z^c(2)\in d{z}_3\mid Z^c(\epsilon )=z_1,Z^c(1)=z_2\right)=\frac{q(z_2,z_3;\epsilon ,z_1)}{q(z_2;\epsilon ,z_1)}d{z}_3,$$

$$q(z_2,z_3;\epsilon ,z_1)$$

$$q(z_2;\epsilon ,z_1)$$

$$A_{1,1}=$$

$$A_{1,2}=$$

$$A_{1,3}=$$

$${\int }_0^1{e}^{-\lambda \varphi (x)}f(\lambda ,x)x^{\nu -1}dx\sim f(\infty ,0)\frac{\Gamma (\nu )}{b^{\nu }}{\lambda }^{-\nu }{e}^{-a\lambda }\text{as λ→∞}.$$

$${\lambda }^{-\nu }{e}^{-a\lambda }{\int }_0^{\lambda }{e}^{-\lambda \{\varphi (u/\lambda )-a\}}f(\lambda ,u/\lambda )du.$$

$$\underset{\lambda \to \infty }{\lim }{\int }_0^{\lambda }{e}^{-\lambda \{\varphi (u/\lambda )-a\}}f(\lambda ,u/\lambda )du=f(\infty ,0){\int }_0^{\infty }{e}^{-bu}{u}^{\nu -1}du$$

$$\underset{\epsilon \to 0+}{\lim }\frac{q(z_2,z_3;\epsilon ,z_1)}{q(z_2;\epsilon ,z_1)}=\frac{q(z_2,z_3;z_1)}{q(z_2;z_1)}$$

$$q(z_2,z_3;z_1)={\int }_0^{z_2}d{x}_2{\int }_0^{z_3}d{x}_3{A}_{2,1}{A}_{2,2},q(z_2;z_1)={\int }_0^{z_2}d{x}_2{A}_{2,1}$$

Full text

The research of the third author was supported by KAKENHI (20740060)

1. Introduction

Dyson [3] has introduced the matrix-valued stochastic process

[TABLE]

to model the dynamics of particles with the Coulomb type interactions, where $B_{i,i}$’s are real Brownian motions and $B_{i,j}$’s for $i<j$ are complex Brownian motions all of which are mutually independent. He proved that the eigenvalue processes $\lambda_{1},\ldots,\lambda_{N}$ satisfy the (system of) stochastic differential equations

[TABLE]

with $\beta=2$. It has been proven later that if the complex Brownian motions are replaced by real or quaternion Brownian motions, the eigenvalue processes satisfy similar stochastic differential equations with $\beta=1$ or $4$, respectively. (See [1, 4] for discussions based on the stochastic analysis.) These processes are now called Dyson’s Brownian motion models for GOE, GUE, and GSE when $\beta=1,2$, and $4$, respectively. In any case, it is remarkable that the process $\Lambda=(\lambda_{1},\ldots,\lambda_{N})$ is Markov.

We may ask the following question: “Does the process $\Lambda$ remain Markov if we replace the common coefficient $1/\sqrt{2}$ by a different positive number?” In this paper, we give the negative answer to this question when the matrix size $N=2$.

Let $c\geq 0$ and $\delta>0$. Consider the $2\times 2$-matrix-valued process

[TABLE]

where $B_{1}$ and $B_{2}$ are two independent standard Brownian motions and $\xi^{\delta}$ is a Bessel process of dimension $\delta$ starting from 0 which is independent of $B_{1}$ and $B_{2}$. We see in Lemma 2.2 that $\Xi^{c,\delta}$ with $\delta=1,2$, or 4 is unitarily equivalent in law to

[TABLE]

with $B_{3}$ a real, complex, or quaternion Brownian motion independent of $B_{1}$ and $B_{2}$, respectively. Let $\lambda_{1}(t)$ and $\lambda_{2}(t)$ for $t\geq 0$ denote the eigenvalues of the Hermitian matrix $\Xi^{c,\delta}(t)$ such that $\lambda_{1}(t)\geq\lambda_{2}(t)$. Define the two-dimensional process $\Lambda^{c,\delta}=(\lambda_{1},\lambda_{2})$.

When $c=0$, $\lambda_{1}(t)$ and $\lambda_{2}(t)$ are nothing but the order statistics of $B_{1}(t)$ and $B_{2}(t)$, that is, $\lambda_{1}(t)=\max\left\{B_{1}(t),B_{2}(t)\right\}$ and $\lambda_{2}(t)=\min\left\{B_{1}(t),B_{2}(t)\right\}$. Hence it is obvious that the process $\Lambda^{0,\delta}$ is Markov.

When $c=1$, the process (1.1) is a time-dependent version of Dumitriu-Edelman’s matrix model for beta-ensembles (cf. [2]) and we see in Lemma 2.1 that the processes $\lambda_{1}(t)$ and $\lambda_{2}(t)$ satisfy Dyson’s stochastic differential equations with index $\beta=\delta$ given by

[TABLE]

for two independent Brownian motions $\beta_{1}(t)$ and $\beta_{2}(t)$. In particular, the process $\Lambda^{1,\delta}(t)$ is Markov.

Theorem 1.1**.**

The process $\Lambda^{c,\delta}$ is Markov if and only if $c\in\{0,1\}$.

We prove this theorem by reducing it to the following.

Theorem 1.2**.**

Let $\delta_{1},\delta_{2}>0$. Let $X^{\delta_{1}}$ and $Y^{\delta_{2}}$ be two independent squared Bessel processes starting from 0 of dimension $\delta_{1}$ and $\delta_{2}$, respectively. Then the process $Z^{c}(t)=cX^{\delta_{1}}(t)+Y^{\delta_{2}}(t)$ for $c\geq 0$ is Markov if and only if $c\in\{0,1\}$.

Theorems 1.1 and 1.2 seem similar to Matsumoto-Ogura’s $cM-X$ theorem [6]. Let $X$ be a Brownian motion and set $M(t)=\sup_{0\leq s\leq t}X(s)$. When $c=0,1,2$, the process $cM-X$ is Markov; indeed, $-X$ is a Brownian motion, $M-X$ is a reflecting Brownian motion by Lévy’s theorem (see, e.g., [7, Thm.VI.2.3]), and $2M-X$ is a three-dimensional Bessel process by Pitman’s theorem (see, e.g., [7, Thm.VI.3.5]).

Theorem 1.3** ([6]).**

The process $cM-X$ is Markov if and only if $c\in\{0,1,2\}$.

2. Non-Markov property of the eigenvalue processes

Proof of Theorem 1.1 provided Theorem 1.2 is justified.

An elementary calculation shows that $\lambda_{1}$ and $\lambda_{2}$ are given by

[TABLE]

Set $B_{3}(t)=\{B_{1}(t)+B_{2}(t)\}/\sqrt{2}$, $X^{1}(t)=\{B_{1}(t)-B_{2}(t)\}^{2}/2$ and $Y^{\delta}(t)=\xi^{\delta}(t)^{2}$. Then $B_{3}$ is a real Brownian motion, $X^{1}$ and $Y^{\delta}$ are squared Bessel processes of dimension 1 and $\delta$, respectively. Moreover, $B_{3}$, $X^{1}$, and $Y^{\delta}$ are mutually independent. It follows that

[TABLE]

It is obvious that the two dimensional process $\Lambda^{c,\delta}=(\lambda_{1},\lambda_{2})$ is Markov if and only if so is the process $(\lambda_{1}+\lambda_{2},\lambda_{1}-\lambda_{2})$. Since

[TABLE]

and they are independent, for the process $\Lambda^{c,\delta}$ to be Markov it is necessary and sufficient that the process $X^{1}+cY^{\delta}$ is Markov. This is equivalent to $c=0$ or $1$ by Theorem 1.2. ∎

Lemma 2.1**.**

For $c=1$ and $\delta>0$, consider the $2\times 2$-matrix-valued process $\Xi^{1,\delta}$ defined by (1.1). Then the corresponding eigenvalue processes satisfy the stochastic differential equations (1.3)–(1.4).

Proof.

Set $\widetilde{\lambda}=(\lambda_{1}-\lambda_{2})/\sqrt{2}$. Then, by (2.2) for $c=1$ and by Shiga-Watanabe’s theorem (see, e.g., [7, Thm.XI.1.2]), we see that the process $\widetilde{\lambda}$ is a Bessel process of dimension $1+\delta$. Hence we have

[TABLE]

where $B_{4}$ is a real Brownian motion independent of $B_{3}$. If we set $\beta_{1}=(B_{3}+B_{4})/\sqrt{2}$ and $\beta_{2}=(B_{3}-B_{4})/\sqrt{2}$, then $\beta_{1}$ and $\beta_{2}$ are two independent real Brownian motions. Therefore, combining (2.3) with (2.1), we conclude that (1.3)–(1.4) hold. ∎

Lemma 2.2**.**

Let $c>0$, $\delta=1,2$, or $4$, and $\Xi^{c,\delta}$ and $\widetilde{\Xi}^{c,\delta}$ be the matrix-valued processes defined by (1.1) and (1.2), respectively. Then, there exists a unitary matrix-valued process $U_{\delta}(t)$ such that

[TABLE]

In particular, eigenvalue processes associated with $\Xi^{c,\delta}$ and $\widetilde{\Xi}^{c,\delta}$ have the same law.

Proof.

We define

[TABLE]

by using $B_{3}$ in (1.2). Then we have

[TABLE]

which shows the desired result since $|B_{3}|\stackrel{{\scriptstyle\rm law}}{{=}}\xi^{\delta}$. ∎

3. Transition probability density of squared Bessel processes

In this section, we recall some basic asymptotic estimates on the transition probability density $p_{t}^{\delta}(x,y)$ of squared Bessel processes of dimension $\delta$ which we shall use later. We first note that it has an expression

[TABLE]

for $x,y>0$, where $I_{\nu}$ stands for the modified Bessel function of index $\nu$ (see, e.g., [7, Cor.XI.1.4]). Now let us recall following two asymptotic estimates on the modified Bessel function (see, e.g., Sect. 5.16.4 of [5]):

[TABLE]

Here, $f(x)\sim g(x)$ means $f(x)/g(x)\to 1$ in the subsequently indicated limit.

Using (3.2) in (3.1), we can derive

[TABLE]

for $t,y>0$ and

[TABLE]

for $t,x>0$. On the other hand (3.3) together with (3.1) yields

[TABLE]

as $\sqrt{xy}\to\infty$.

4. Non-Markov property of weighted sums of two independent squared Bessel processes

For the proof of Theorem 1.2, we may restrict ourselves to $0<c<1$; otherwise consider $Z^{c}/c$ instead. We prove that $Z^{c}$ is non-Markov by checking that the conditional law

[TABLE]

does depend on $(\varepsilon,z_{1})$. This conditional law has the density

[TABLE]

where $q(z_{2},z_{3};\varepsilon,z_{1})$ and $q(z_{2};\varepsilon,z_{1})$ are the densities of the joint laws of $(Z^{c}(\varepsilon),Z^{c}(1),Z^{c}(2))$ and $(Z^{c}(\varepsilon),Z^{c}(1))$, respectively. Thus it suffices to prove that the fraction ${q(z_{2},z_{3};\varepsilon,z_{1})}/{q(z_{2};\varepsilon,z_{1})}$ depends on $(\varepsilon,z_{1})$.

To this end, we shall use the integral expression

[TABLE]

where

[TABLE]

We divide the proof into several steps. First of all, we prove

Lemma 4.1**.**

Let $f(\lambda,\cdot)$ for $\lambda>0$ be a bounded measurable function on $(0,1)$. Suppose that $f(\lambda,x/\lambda)$ converges to a constant $f(\infty,0)$ for any $x\in(0,1)$ as $\lambda\to\infty$. Let $\phi\in C^{1}((0,1))$ and suppose that $\phi(0+)=a\in\mathbb{R}$, $\phi^{\prime}(0+)=b>0$ and $\phi^{\prime}(x)>0$ for $x\in(0,1)$. Let $\nu>0$. Then

[TABLE]

Proof.

Changing variables to $u=\lambda x$, we find that the left hand side of (4.2) equals

[TABLE]

Note that $\lambda\{\phi(u/\lambda)-a\}\geq Ku$ for $u\in(0,\lambda)$ and $\lambda>0$ where $K=\inf_{x\in(0,1)}\{\phi(x)-\phi(0+)\}/x>0$. Hence we see that

[TABLE]

by the dominated convergence theorem. ∎

Second, we take the limit as $\varepsilon\to 0$.

Lemma 4.2**.**

[TABLE]

with

[TABLE]

where $A_{2,2}=A_{1,3}$ and

[TABLE]

Proof.

We know that

[TABLE]

from (3.4). Now we can rewrite $q(z_{2},z_{3};\varepsilon,z_{1})/q(z_{2};\varepsilon,z_{1})$ as $F_{1}/G_{1}$ with

[TABLE]

where $\widetilde{c}=(1-c)/2$ and

[TABLE]

Using Lemma 4.1 in the integrals (4.3) and (4.4), we have

[TABLE]

as $\varepsilon\to 0+$. Here we have used the fact that $A_{1,4}(\varepsilon,x_{1})$ and $A_{1,5}(\varepsilon,x_{1})$ are continuous in $\varepsilon\in[0,\infty)$ and $x_{1}\in[0,z_{1}]$. Therefore, $F_{1}/G_{1}$ approaches to $A_{1,4}(0,0)/A_{1,5}(0,0)=q(z_{2},z_{3};z_{1})/q(z_{2};z_{1})$. ∎

Third, we study the asymptotic behavior of the numerator $q(z_{2},z_{3};z_{1})$ as $z_{3}\to 0+$.

Lemma 4.3**.**

[TABLE]

with

[TABLE]

where $A_{3,1}=A_{2,1}$ and

[TABLE]

Proof.

Recall that

[TABLE]

where

[TABLE]

Here we note that $A_{3,1}$ does not depend on $z_{3}$ nor $x_{3}$. If we take $x_{3}=z_{3}u$ for $0<u<1$, we have

[TABLE]

Using (3.5), we have, as $z_{3}\to 0+$,

[TABLE]

Changing variables to $u=x_{3}/z_{3}$ in the integral (4.5), we obtain

[TABLE]

which converges to $C_{1}\widetilde{q}(z_{2};z_{1})$ as $z_{3}\to 0+$. ∎

Fourth, we study the asymptotic behaviors of $\widetilde{q}(z_{2};z_{1})$ and $q(z_{2};z_{1})$ as $z_{2}\to\infty$. Recall that

[TABLE]

and that

[TABLE]

Lemma 4.4**.**

Let $r>0$. Then

[TABLE]

where $C_{2}$ is some positive constant depending only on $\delta_{1}$ and $\delta_{2}$ and

[TABLE]

Proof.

If we express $\widetilde{q}(z_{2};z_{2}r)$ as

[TABLE]

using

[TABLE]

with $b_{1}=1-c$ and $a_{1}=(\sqrt{r}-1)^{2}/2+1/2$, then $f_{1}(z_{2},\cdot\,)$ turns out to be a bounded continuous function such that $f_{1}(z_{2},u/z_{2})$ converges to a constant depending only on $\delta_{1}$ and $\delta_{2}$ as $z_{2}\to\infty$, by (3.6). Since $\phi_{1}$ and $f_{1}$ satisfies the assumptions, we can use Lemma 4.1 and hence we obtain

[TABLE]

with some constant $C_{2,1}$ depending only on $\delta_{1}$ and $\delta_{2}$.

We also have a similar expression

[TABLE]

for $q(z_{2};z_{2}r)$ using

[TABLE]

with $b_{2}=(1-c)/2$ and $a_{2}=(\sqrt{r}-1)^{2}/2$ and a function $f_{2}(z_{2},\cdot\,)$ as before. Thus the same argument yields

[TABLE]

with some constant $C_{2,2}$ depending only on $\delta_{1}$ and $\delta_{2}$.

Using (4.7) and (4.8) together with $\phi_{1}^{\prime}(0+)=b_{1}+\sqrt{r}c/2$ and $\phi_{2}^{\prime}(0+)=b_{2}+\sqrt{r}c/2$, we obtain (4.6). ∎

Now we are in a position to prove Theorem 1.2.

Proof of Theorem 1.2.

Let $0<c<1$. We combine Lemmas 4.2, 4.3 and 4.4 to obtain

[TABLE]

for some constant $C_{3}$ which depends only on $\delta_{1}$, $\delta_{2}$ and $c$. Therefore we conclude that the conditional probability (4.1) does depend on $(\varepsilon,z_{1})$, which proves that $Z^{c}$ is non-Markov. ∎

Acknowledgements.

The authors thank Professors Hideki Tanemura and Makoto Katori for helpful discussions. They also thank Professor Tomoyuki Shirai for drawing their attention to [2].