Extreme Statistics of Superdiffusive Lévy Flights and Every Other Lévy Subordinate Brownian Motion

Abstract

The search for hidden targets is a fundamental problem in many areas of science, engineering, and other fields. Studies of search processes often adopt a probabilistic framework, in which a searcher randomly explores a spatial domain for a randomly located target. There has been significant interest and controversy regarding optimal search strategies, especially for superdiffusive processes. The optimal search strategy is typically defined as the strategy that minimizes the time it takes a given single searcher to find a target, which is called a first hitting time (FHT). However, many systems involve multiple searchers, and the important timescale is the time it takes the fastest searcher to find a target, which is called an extreme FHT. In this paper, we study extreme FHTs for any stochastic process that is a random time change of Brownian motion by a Lévy subordinator. This class of stochastic processes includes superdiffusive Lévy flights in any space dimension, which are processes described by a Fokker–Planck equation with a fractional Laplacian. We find the short-time distribution of a single FHT for any Lévy subordinate Brownian motion and use this to find the full distribution and moments of extreme FHTs as the number of searchers grows. We illustrate these rigorous results in several examples and numerical simulations.

Appendix

In this appendix, we prove the results in the main text.

Lemma 4

Assume $S=\{S(t)\}_{t\ge 0}$ is a compound Poisson process plus a drift, meaning its Laplace exponent is in (13) with $b\ge 0$ and $\int _{0}^{\infty }\,\nu (d z)\in (0,\infty )$. If $F:[0,\infty )\rightarrow [0,1]$ is continuous and satisfies (24), then (25) holds.

Proof of Lemma 4

By assumption, we have that $S(t)=bt+\sum _{m=1}^{M(t)}Z_{m}$, where $M=\{M(t)\}_{t\ge 0}$ is a Poisson process with rate $\lambda =\int _{0}^{\infty }\,\nu (d z)\in (0,\infty )$ and $\{Z_{m}\}_{m\ge 1}$ are iid nonnegative random variables independent of M. In this case, the probability measure of $Z_{m}$ is $\nu (d z)/\lambda$. Decomposing the mean based on the value of M(t) yields

$$\begin{aligned} {\mathbb {E}}[F(S(t))] ={\mathbb {E}}[F(S(t))1_{M(t)=0}]+{\mathbb {E}}[F(S(t))1_{M(t)=1}]+{\mathbb {E}}[F(S(t))1_{M(t)\ge 2}], \end{aligned}$$

where $1_{A}$ denotes the indicator function on an event A. Since M(t) is a Poisson random variable with mean $\lambda t$ and F is bounded, we have that ${\mathbb {E}}[F(S(t))1_{M(t)\ge 2}]=o(t)$ as $t\rightarrow 0+$. Furthermore, since M and $Z_{1}$ are independent, we have that

$$\begin{aligned} {\mathbb {E}}[F(S(t))1_{M(t)=1}]&={\mathbb {P}}(M(t)=1){\mathbb {E}}[F(bt+Z_{1})] =\lambda t e^{-\lambda t}{\mathbb {E}}[F(bt+Z_{1})],\\ {\mathbb {E}}[F(S(t))1_{M(t)=0}]&={\mathbb {P}}(M(t)=0){\mathbb {E}}[F(bt)] =e^{-\lambda t}F(bt). \end{aligned}$$

Since F is bounded, F is continuous, and $\int _{0}^{\infty }\,\nu (d s)<\infty$, we complete the proof by applying the Lebesgue dominated convergence to conclude

$$\begin{aligned} {\mathbb {E}}[F(bt+Z_{1})] =\frac{1}{\lambda }\int _{0}^{\infty }F(bt+s)\,\nu (d s) \rightarrow \frac{1}{\lambda }\int _{0}^{\infty }F(s)\,\nu (d s)\quad \text {as }t\rightarrow 0+. \end{aligned}$$

$\square$

Proof of Proposition 1

The boundedness of F and (24) ensure that the integral in (25) is finite. Let $\varepsilon =2^{-j}>0$ for some $j\in \{0,1,2,\dots \}$ and define

$$\begin{aligned} \begin{aligned} S_{[\varepsilon ,\infty )}(t)&:=bt+\iint _{z\in [\varepsilon ,\infty ),\,t'\in [0,t]} z\,{\textbf{N}}(d t',d z),\\ S_{(0,\varepsilon )}(t)&:=\iint _{z\in (0,\varepsilon ),\,t'\in [0,t]} z\,{\textbf{N}}(d t',d z), \end{aligned} \end{aligned}$$

(49)

where ${\textbf{N}}$ is a Poisson point process on the first quadrant with intensity measure $d t'\,\nu (d z)$. The process S can then be written as $S(t)=S_{[\varepsilon ,\infty )}(t)+S_{(0,\varepsilon )}(t)$. Since F is Lipschitz, there exists a constant $\kappa >0$ so that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}[F(S_{[\varepsilon ,\infty )}(t))] -\kappa {\mathbb {E}}[S_{(0,\varepsilon )}(t))] \le {\mathbb {E}}[F(S(t))]\\&\qquad \qquad \qquad \le {\mathbb {E}}[F(S_{[\varepsilon ,\infty )}(t))] +\kappa {\mathbb {E}}[S_{(0,\varepsilon )}(t))]\quad \text {for all }t>0. \end{aligned} \end{aligned}$$

(50)

Since $S_{[\varepsilon ,\infty )}$ is a compound Poisson process plus a drift, Lemma 4 implies that

$$\begin{aligned} \lim _{t\rightarrow 0+}t^{-1}{\mathbb {E}}[F(S_{[\varepsilon ,\infty )}(t))] =\rho _{\varepsilon } :=bF'(0)+\int _{\varepsilon }^{\infty }F(s)\,\nu (d s) <\infty . \end{aligned}$$

(51)

To handle the terms in (50) involving $S_{(0,\varepsilon )}(t)$, recall that $\varepsilon =2^{-j}$ and observe that a dyadic partitioning of the interval $(0,\varepsilon )$ yields

$$\begin{aligned} S_{(0,\varepsilon )}(t)&:=\iint _{z\in (0,\varepsilon ),\,t'\in [0,t]} z\,{\textbf{N}}(d t',d z) \le \sum _{k=j}^{\infty }2^{-k}{\textbf{N}}([0,t]\times [2^{-k-1},2^{-k}]). \end{aligned}$$

Since ${\textbf{N}}$ is a Poisson point process, we have that

$$\begin{aligned} 2^{-k}{\mathbb {E}}\big [{\textbf{N}}([0,t]\times [2^{-k-1},2^{-k}])\big ] =2t\int _{2^{-k-1}}^{2^{-k}}2^{-k-1}\,\nu (d z) \le 2t\int _{2^{-k-1}}^{2^{-k}}z\,\nu (d z). \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathbb {E}}[S_{(0,\varepsilon )}(t)] \le 2t\int _{0}^{\varepsilon }z\,\nu (d z). \end{aligned}$$

(52)

Combining (50) with (51) and (52) yields

$$\begin{aligned}&\rho _{\varepsilon } -2\kappa \int _{0}^{\varepsilon }z\,\nu (d z) \le \liminf _{t\rightarrow 0+}\frac{{\mathbb {E}}[F(S(t))]}{t} \le \limsup _{t\rightarrow 0+}\frac{{\mathbb {E}}[F(S(t))]}{t} \\&\quad \le \rho _{\varepsilon } +2\kappa \int _{0}^{\varepsilon }z\,\nu (d z). \end{aligned}$$

Since these bounds converge to $\rho$ as $\varepsilon \rightarrow 0+$, the proof is complete. $\square$

Lemma 5

Let $H:[0,\infty )\rightarrow [0,1]$ be nondecreasing and satisfy (24). Then, $\limsup _{t\rightarrow 0+}{\mathbb {E}}[H(S(t))]/t<\infty$.

Proof of Lemma 5

Using the definitions in (49), we have that

$$\begin{aligned} H(S(t)) =H(S_{[\varepsilon ,\infty )}(t)+S_{(0,\varepsilon )}(t))&\le H(2 S_{[\varepsilon ,\infty )}(t)) +H(2S_{(0,\varepsilon )}(t)). \end{aligned}$$

Since $2 S_{[\varepsilon ,\infty )}(t)$ is a compound Poisson process plus a drift, Lemma 4 ensures that $\lim _{t\rightarrow 0+}{\mathbb {E}}[H(2 S_{[\varepsilon ,\infty )}(t))]/t <\infty$. Since H satisfies (24), there exists an $s_{0}\in (0,1]$ and a $\theta \ge 1$ so that $H(s)\le \theta s$ for all $s\in (0,s_{0}]$. Therefore, $H(s)\le \theta s/s_{0}$ for all $s\ge 0$. The proof is complete since (52) implies

$$\begin{aligned} {\mathbb {E}}[H(2S_{(0,\varepsilon )}(t))] \le \frac{2\theta }{s_{0}}{\mathbb {E}}[S_{(0,\varepsilon )}(t)] \le \frac{4\theta t}{s_{0}}\int _{0}^{\varepsilon }z\,\nu (d z). \end{aligned}$$

$\square$

Proof of Theorem 3

Define $F(s):={\mathbb {P}}(B(s)+X(0)\in U)\in [0,1]$ for $s\ge 0$. Using the independence of B and X(0), we have

$$\begin{aligned} F(s) =\frac{1}{(4\pi s)^{d/2}}\iint _{U\times U_{0}}\exp \Big (\frac{-\Vert x-x_{0}\Vert ^{2}}{4s}\Big )\,\mu _{0}(d x_{0})\,d x,\quad \text {if }s>0, \end{aligned}$$

(53)

where $\mu _{0}$ is the probability measure of X(0) with support $U_{0}\subset {\mathbb {R}}^{d}$. Using standard results for interchanging differentiation with integration (for example, see Theorem A.5.3 in Durrett (2019)), F(s) is infinitely differentiable and each derivative is bounded. Furthermore, (33) ensures that $F(0)=F'(0)=0$, and thus Proposition 1 implies

$$\begin{aligned} \lim _{t\rightarrow 0+}\frac{{\mathbb {P}}({{X}}(t)\in U)}{t} =\lim _{t\rightarrow 0+}\frac{{\mathbb {E}}[F(s)]}{t} =\rho :=\int _{0}^{\infty }{\mathbb {P}}({{B}}(s)+X(0)\in U)\,\nu (d s). \end{aligned}$$

(54)

In the first equality in (54), we have used the independence of B, S, and X(0). Note that $\rho \in (0,\infty )$. Indeed, Proposition 1 implies $\rho <\infty$. Further, $\rho >0$ by (i) the assumption in (31), (ii) the fact that ${{B}}(s)\in {\mathbb {R}}^{d}$ is a Gaussian random variable with variance proportional to $s>0$, and (iii) U has strictly positive Lebesgue measure (since U is nonempty and the closure of its interior).

To complete the proof, we therefore need to show that

$$\begin{aligned} \lim _{t\rightarrow 0+}t^{-1}{\mathbb {P}}(\tau \le t) =\lim _{t\rightarrow 0+}t^{-1}{\mathbb {P}}({{X}}(t)\in U). \end{aligned}$$

(55)

For $t>0$, define the enlarged target $U^{\delta (t)}:=\{x\in {\mathbb {R}}^{d}:\inf _{y\in U}\Vert x-y\Vert \le \delta (t)\}$, where we set $\delta (t):=t^{1/4}>0$ in order to satisfy

$$\begin{aligned} \lim _{t\rightarrow 0+}\delta (t)=0 \quad \text {and}\quad \lim _{t\rightarrow 0+}\delta (t)t^{-1/2}=\infty . \end{aligned}$$

(56)

Decomposing the event $\tau \le t$ based on the position of X(t) yields

$$\begin{aligned} {\mathbb {P}}(\tau \le t)&={\mathbb {P}}({{X}}(t)\in U) +{\mathbb {P}}(\tau \le t,{{X}}(t)\in U^{\delta (t)}\backslash U) +{\mathbb {P}}(\tau \le t,{{X}}(t)\notin U^{\delta (t)}). \end{aligned}$$

Therefore, showing (55) amounts to showing that

$$\begin{aligned} \lim _{t\rightarrow 0+}t^{-1}{\mathbb {P}}(\tau \le t,{{X}}(t)\in U^{\delta (t)}\backslash U)&=0=\lim _{t\rightarrow 0+}t^{-1}{\mathbb {P}}(\tau \le t,{{X}}(t)\notin U^{\delta (t)}). \end{aligned}$$

(57)

We first prove the first equality in (57). Since ${{X}}(t)={{B}}(S(t))+X(0)$ and B, X(0), and S are independent, integrating over the possible values of S(t) yields

$$\begin{aligned} {\mathbb {P}}(\tau \le t,{{X}}(t)\in U^{\delta (t)}\backslash U)&\le {\mathbb {P}}({{X}}(t)\in U^{\delta (t)}\backslash U) ={\mathbb {E}}[F_{0}(S(t);t)], \end{aligned}$$

where $F_{0}(s;t):={\mathbb {P}}({{B}}(s)+X(0)\in U^{\delta (t)}\backslash U)$. By the assumption in (33), we may take $t_{0}$ sufficiently small so that $U^{\delta (t_{0})}\cap U_{0}=\varnothing$. Therefore, if $t\in (0,t_{0}]$, then $F_{0}(s;t)$ satisfies the assumptions of Proposition 1 [by the same argument used for F(s) in (53)]. Therefore, Proposition 1 implies that we may take t sufficiently small so that,

$$\begin{aligned} t^{-1}{\mathbb {P}}(\tau \le t,{{X}}(t)\in U^{\delta (t)}\backslash U) \le 2\int _{0}^{\infty }{\mathbb {P}}({{B}}(s)+X(0)\in U^{\delta (t_{0})}\backslash U)\,\nu (d s)<\infty . \end{aligned}$$

Now, it is immediate that ${\mathbb {P}}({{B}}(s)+X(0)\in U^{\delta (t_{0})}\backslash U)\rightarrow 0$ as $t_{0}\rightarrow 0$ for each $s\ge 0$. Hence, the Lebesgue dominated convergence theorem implies

$$\begin{aligned} \lim _{t_{0}\rightarrow 0+}\int _{0}^{\infty }{\mathbb {P}}({{B}}(s)+X(0)\in U^{\delta (t_{0})}\backslash U)\,\nu (d s)=0, \end{aligned}$$

and thus the first equality in (57) holds. Turning to the second equality in (57), conditioning that $\tau \le t$ implies

$$\begin{aligned} {\mathbb {P}}(\tau \le t,{{X}}(t)\notin U^{\delta (t)}) ={\mathbb {P}}({{X}}(t)\notin U^{\delta (t)}\,|\,\tau \le t){\mathbb {P}}(\tau \le t), \end{aligned}$$

and the fact that ${\widetilde{\tau }}\le \tau$ almost surely and Lemma 5 imply

$$\begin{aligned} \limsup _{t\rightarrow 0+}t^{-1}{\mathbb {P}}(\tau \le t) \le \limsup _{t\rightarrow 0+}t^{-1}{\mathbb {P}}({\widetilde{\tau }}\le t) <\infty , \end{aligned}$$

since ${\mathbb {P}}({\widetilde{\tau }}\le t)={\mathbb {E}}[H(S(t)]$ where $H(s):={\mathbb {P}}(\sigma \le s)$ is nondecreasing. Next, it follows from the strong Markov property (Bertoin 1996) that

$$\begin{aligned} {\mathbb {P}}({{X}}(t)\notin U^{\delta (t)}\,|\,\tau \le t) \le \sup _{r\in (0,t]}{\mathbb {P}}_{0}(\Vert {{X}}(r)\Vert \ge \delta (t)), \end{aligned}$$

where ${\mathbb {P}}_{0}$ denotes the probability measure conditioned that ${{X}}(0)=0$. Again using that B and S are independent, we have that

$$\begin{aligned} \begin{aligned} {\mathbb {P}}_{0}(\Vert {{X}}(r)\Vert \ge \delta (t)) ={\mathbb {E}}[F_{1}(S(r);t)] \le {\mathbb {E}}[F_{1}(S(t);t)],\quad \text {if }r\in [0,t], \end{aligned} \end{aligned}$$

(58)

since $F_{1}(s;t):={\mathbb {P}}(\Vert B(s)\Vert \ge \delta (t))$ is an increasing function of s and S is almost surely nondecreasing. Define

$$\begin{aligned} \delta _{1}(t) :=(1+b)t>0, \end{aligned}$$

(59)

and observe that (58) implies that for $r\in (0,t]$,

$$\begin{aligned} \begin{aligned} {\mathbb {P}}_{0}(\Vert {{X}}(r)\Vert \ge \delta (t))&\le {\mathbb {E}}[F_{1}(S(t);t)1_{S(t)<\delta _{1}(t)}] +{\mathbb {E}}[F_{1}(S(t);t)1_{S(t)\ge \delta _{1}(t)}]. \end{aligned} \end{aligned}$$

Since S(t)/t converges in probability to $b\ge 0$ as $t\rightarrow 0+$ (Bertoin 1996), we have that

$$\begin{aligned} {\mathbb {E}}[F_{1}(S(t);t)1_{S(t)\ge \delta _{1}(t)}] \le {\mathbb {P}}(S(t)\ge \delta _{1}(t)) ={\mathbb {P}}(S(t)\ge (1+b)t)\rightarrow 0\quad \text {as }t\rightarrow 0+. \end{aligned}$$

Next, since $F_{1}(s;t)$ is an increasing function of s, we have that

$$\begin{aligned} {\mathbb {E}}[F_{1}(S(t);t)1_{S(t)<\delta _{1}(t)}] \le F_{1}(\delta _{1}(t);t) ={\mathbb {P}}(\Vert B(\delta _{1}(t))\Vert \ge \delta (t)). \end{aligned}$$

The Brownian scaling in (16) and the choices of $\delta (t)$ in (56) and $\delta _{1}(t)$ in (59) imply

$$\begin{aligned} {\mathbb {P}}(\Vert B(\delta _{1}(t))\Vert \ge \delta (t)) ={\mathbb {P}}(\Vert B(1)\Vert \ge \delta (t)(\delta _{1}(t))^{-1/2}) \rightarrow 0\quad \text {as }t\rightarrow 0+. \end{aligned}$$

Hence, the second equality in (57) holds and the proof is complete. $\square$