Leadership Exponent in the Pursuit Problem for 1-D Random Particles

Abstract

For n + 1 particles moving independently on a straight line, we study the question of how long the leading position of one of them can last. Our focus is the asymptotics of the probability p_T,n that the leader time will exceed T when n and T are large. It is assumed that the dynamics of particles are described by independent, either stationary or self-similar, Gaussian processes, not necessarily identically distributed. Roughly, the result for particles with stationary dynamics of unit variance is as follows: $L: = - \ln p_{T,n}/(T\ln n) = 1/d_{0} + o(1),$where d₀/(2π) is the power of the zero frequency in the spectrum of the leading particle, and this value is the largest in the spectrum. Previously, in some particular models, the asymptotics of L was understood as a sequential limit first over T and then over n. For processes that do not necessarily have non-negative covariances, the limit over T may not exist. To overcome this difficulty, the growing parameters T and n are considered in the domain $c\;{\text{ln}}\;T < n \le CT$ where c > 1. The Lamperti transform allows us to transfer the described result to self-similar processes by changing the $\ln p_{T,n}$ normalization to the value $\ln T\ln n$.

Appendix

1.1 Proof of Lemma 4

Following [3], we use the following representation

$$ g_{N} (t) = \sum\limits_{{\left| {n < N/2} \right|}} {\frac{\sin \pi (t - n)}{\pi (t - n)} = 1 + \frac{\sin \pi t}{2\pi i}\oint_{K} {\frac{dz}{(t - z)\sin \pi z}}} $$

where K is the boundary of the square of size N, centered at the origin. On the sides of K that are parallel to the xaxis one has $\left| {t - {\text{z}}} \right| \ge N/2,\left| {\sin \pi z} \right|^{2} = \cosh \pi N.$

The analogous estimates for the sides of $K$ that are parallel to the $y$ axis are

$$ \left| {t - {\text{z}}} \right| \ge (N - T_{\delta})/2,\left| {\sin \pi z} \right|^{2} = \cosh (2\pi \text{Im} z).$$

Substituting these estimates into the integral, we obtain

$$ \left| {g_{N} (t) - 1} \right| < \upsilon _{0} /(N - T_{\delta } ) + e^{{ - \pi N/2}} ,\left| t \right| \le T_{\delta } /2 < N/2. $$

1.2 Proof of Lemma 6

Let $X(t)$ be a stationary process with covariance function $r_{0} (t)$ and continuous spectral density $f_{0} (\lambda)$. Then the sequence $\{X(k\theta)\}$ has spectral function

$$ f_{\theta} (\lambda) = \theta^{- 1} f_{0} (\lambda/\theta) + 2\sum\nolimits_{k \ge 1} {\theta^{- 1} f_{0} ((\lambda + 2\pi k})/\theta),\left| \lambda \right| \le \pi. $$

(4.1)

By the assumptions, $f_{0} (\lambda/\theta) \le f_{0} (0)$ and $f_{0} (\lambda)$ is majorized by a monotone function $\tilde{f}_{0} (\lambda) \in L_{1}$.Therefore

$$ \begin{aligned} & \sum\nolimits_{k \ge 1} {f_{0} ((\lambda + 2\pi k})/\theta) \le \tilde{f}_{0} (\pi/\theta) + \sum\nolimits_{k \ge 1} {\tilde{f}_{0} ((2k + 1)\pi/\theta)} \\ & \le 2\pi^{- 1} \theta \int_{\pi/(2\theta)}^{\pi/\theta} {\tilde{f}_{0} (\lambda)d\lambda + (2\pi)^{- 1}} \theta \int_{\pi/\theta} {\tilde{f}_{0} (\lambda)d\lambda} \\ & \le 2\pi^{- 1} \theta \int_{\pi/(2\theta)} {\tilde{f}_{0} (\lambda)d\lambda} \\ \end{aligned} $$

Hence, putting $\sigma_{\theta}^{2} = \sup_{0 \le \lambda \le \pi} 2\pi f_{\theta} (\lambda)$, we get

$$ \theta \sigma_{\theta}^{2} = \sup_{0 \le \lambda \le \pi} [2\pi f_{0} (\lambda/\theta) + o(\theta)] = 2\pi f_{0} (0) + o(1),\theta \to 0. $$

(4.2)

Consider the $m \times m$ matrix $R_{m} = [r_{0} (i\theta - j\theta)]_{i,j = 1 - m}$. Because $\sigma_{\theta}^{2} = \sup_{0 \le \lambda \le \pi} 2\pi f_{\theta} (\lambda)$, we have the following relation for the quadratic forms $({\mathbf{x}} ,R_{m} {\mathbf{x}}) = \int_{- \pi}^{\pi} {\left| {\sum\nolimits_{1}^{m} {x_{k} e^{ik\lambda}}} \right|}^{2} f_{\theta} (\lambda)d\lambda \le \int_{- \pi}^{\pi} {\left| {\sum\nolimits_{1}^{m} {x_{k} e^{ik\lambda}}} \right|}^{2} \sigma_{\theta}^{2}/(2\pi)d\lambda = ({\mathbf{x}},{\mathbf{x}})\sigma_{\theta}^{2}.$

We will see later that $R_{m}$ is non-degenerate. Assuming the existence $R_{m}^{- 1}$, we have

$$ {\text{exp[}} - ({\mathbf{x}} ,R_{m}^{- 1} {\mathbf{x}})/2] \le \exp [- ({\mathbf{x}},{\mathbf{x}})/2\sigma_{\theta}^{2}]. $$

The last one means that for any $\{u_{i} \}$ we have

$$ P\{X(i\theta) > u_{i},i = 1, \ldots,m\} \le P(\sigma_{\theta} \eta_{i} > u_{i},i = 1, \ldots,m)Q_{m} $$

where $\{\eta_{i},i = 1, \ldots,m\}$ are i..i.d. standard Gaussian variables, $Q_{m} = \sqrt {\sigma_{\theta}^{2m}/D_{m}}$ and $D_{m} = \det R_{m}$.

According to the theory of the Toeplitz forms [11], $\delta_{m}^{2} = D_{m}/D_{m - 1}$ is the mean-square error of the $X(0)$ prediction based on $\{X(i\theta),i = 1, \ldots,m - 1\}$ data. Moreover, $\delta_{m}^{2} = D_{m}/D_{m - 1}$ decreases and converges to the value $\delta_{\infty}^{2} = \exp \{(2\pi)^{- 1} \int_{- \pi}^{\pi} {\ln f_{\theta}} (\lambda)d\lambda \}$.

Since $f_{\theta} (\lambda) \ge \theta^{- 1} f_{0} (\lambda/\theta)$ we have$D_{m} = \delta_{m}^{2} \cdot \delta_{m - 1}^{2} \cdot \cdots \cdot \delta_{1}^{2} \ge (\delta_{\infty}^{2})^{m} \ge \exp \{m(\theta/\pi)\int_{0}^{\pi/\theta} {\ln f_{0}} (\lambda)d\lambda \} \cdot \theta^{- m}$.Therefore

$$ Q_{m} = \sqrt {\sigma_{\theta}^{2m}/D_{m}} \le (\theta \sigma_{\theta}^{2})^{m/2} \exp \{- m/2 \cdot (\theta/\pi)\int_{0}^{\pi/\theta} {\ln f_{0}} (\lambda)d\lambda \} \cdot $$

This estimate also proves the non-degeneracy of the matrix $R_{m}$. According to (4.2), $\theta \sigma_{\theta}^{2} = 2\pi f_{0} (0) + o(1),\theta \to 0$.

Hence for small $\theta$ we have $Q_{m} \le \exp (\tilde{A}_{\theta} m/2)$ with

$$ \tilde{A}_{\theta} = \exp \{(\theta/\pi)\int_{0}^{\pi/\theta} {\ln (2\pi f_{0} (0)/f_{0}} (\lambda))d\lambda + 1 \cdot $$

Obviously, $\tilde{A}_{\theta}$ can be replaced by a non-decreasing function of $\theta^{- 1}$, namely

$$ A_{\theta} = \sup_{0 \le \varLambda \le \pi/\theta} < \ln [2\pi f_{0} (0)/f_{0} (\lambda)] >_{\varLambda} + 1,\theta \le \theta_{0} $$

where $< \psi (\lambda) >_{\varLambda}$ is the mean of $\psi (\lambda)$ in the interval $(0,\varLambda)$.□