Reinforced Brownian Motion: A Prototype

Abstract

We analyze the Brownian Motion limit of a prototypical unit step reinforced random-walk on the half-line. A reinfoced random walk is one which changes the weight of any edge (or vertex) visited to increase the frequency of return visits. The generating function for the discrete case is first derived for the joint probability distribution of $S_N$ (the location of the walker at the $N^{th}$ step) and $A_N$, the maximum location the walker achieved in $N$ steps. Then the bulk of the analysis concerns the statistics of the limiting Brownian walker, and of its “environment”, both parametrized by the amplitude $\delta$ of the reinforcement. The walker marginal distribution can be interpreted as that of free diffusion with a source serving as a diffusing soft confinement, details depending very much on the value of $-1< \delta < \infty$.

Appendices

Appendix A. Marginal Distribution of the Maximum, Large $b^2/t$

It is sufficient to work directly with the Laplace Transform in time and then extract the time dependence. From (15) we have

$$\begin{aligned} \hat{Q}_\delta (s,b)&\equiv \int ^b_0 Q(s,y,b) \,dy = \frac{2(1+\delta )}{\sqrt{2s}} \frac{\sinh b\sqrt{2s}}{\left(\cosh b\sqrt{2s}\right)^{2+\delta }}\\&= -\frac{2}{b} \frac{\partial }{\partial s} \left(\cosh b \sqrt{2s}\right)^{ -(1+\delta )} \ . \nonumber \end{aligned}$$

(43)

Therefore

$$\begin{aligned} Q_\delta (t,b)&= \frac{2t}{b}\; \mathcal {L}^{-1}_{t,s} \left(\cosh b\sqrt{2s}\right)^{-(1+ \delta )} \ . \end{aligned}$$

It is “only” the technical issue of carrying out the inverse Laplace Transform $\mathcal {L}^{-1}_{t,s}$ that remains. Even for $\delta =0$, this requires introduction of the Jacobi Theta function, not the best way to visualize the resulting situation. There are alternatives. One alternative is to make use of the rapid convergence of the series expansion

$$\begin{aligned} \left(\cosh b\sqrt{2s}\right)^{-(1+\delta )}&= \left[\frac{1}{2}\left(e^{b\sqrt{2s}} + e^{-b\sqrt{2s}}\right) \right]^{-(1+\delta )}\nonumber \\&= 2^{1+\delta } \, e^{-(1+\delta )\,b\sqrt{2s}} \left(1+ e^{-2b \sqrt{2s}}\right)^{-(1+\delta )}\nonumber \\&= 2^{1+\delta } \sum ^\infty _{j=0} (-1)^j \left( \begin{matrix} j+\delta \\ j \end{matrix} \right) e^{-\left(2j+1 + \delta \right) b\sqrt{2s}}. \end{aligned}$$

(44)

Using $\mathcal {L}^{-1}_{t,s} \,e^{-c\sqrt{s}} = ce^{-c^2/4t}/2\sqrt{\pi }\,t^{3/2}$, we therefore have

$$\begin{aligned} Q_\delta \,(t,b)=2^{2+\delta }\Big /\sqrt{2\pi t}\; \sum ^\infty _{j=0} (-1)^j \begin{pmatrix} j+\delta \\ j \end{pmatrix} \left(2j+1+\delta \right) e^{-\frac{1}{2}\left(\frac{b^2}{t}\right)(2j+1+\delta )^2}\ . \end{aligned}$$

(45)

Eq. (45) is an alternating series $\sum ^\infty _0 (-1)^j\, a_j$, with

$$\begin{aligned} a_{j+1}/a_j = \frac{j+\delta }{j} \; \frac{2j+1+\delta }{2j-1+\delta } \; e^{-2\,(2+\delta )\,b^2/t} \le (3+\delta )\, e^{-2\,(2+\delta )\,b^2/t} \end{aligned}$$

and so is certainly dominated by the first (Gaussian) term if

$$\begin{aligned} \frac{b^2}{t} \ge \ell n \left(3+\delta \right)\!\bigl /\! 2\,(2+ \delta ) \end{aligned}$$

(46)

Appendix B. Marginal Distribution of the Maximum, Small $b^2/t$

The analysis simplifies materially if $\delta$ is an integer, which we henceforth assume. Then, in (45), we make the replacement $j\rightarrow -1-j-\delta$, so that

$$\begin{aligned} Q_\delta \,(t,b) =2^{2+\delta }\!\Bigl /\!\sqrt{2\pi t} \;\sum ^{-(\delta +1)}_{j=-\infty } (-1)^j \begin{pmatrix} j+ \delta \\ j \end{pmatrix} \left(2j+1+\delta \right) e^{-\frac{1}{2}\left(\frac{b^2}{t}\right)(2j+1+\delta )^2}. \end{aligned}$$

(47)

Since $\left(\begin{array}{l} j+ \delta \\ j \end{array}\right)=0$ when $-\delta <j<0$, the upper bound in (47) can be replaced by $-1$. Doing so, and adding to (47), we get

$$\begin{aligned} Q_\delta \,(t,b) = \frac{2^{1+\delta }}{\sqrt{2\pi t}} \;\sum ^\infty _{j=-\infty } (-1)^j \begin{pmatrix} j+ \delta \\ j \end{pmatrix} \left(2j+1+\delta \right) e^{-\frac{1}{2}\left(\frac{b^2}{t}\right) (2j+1+\delta )^2}. \end{aligned}$$

(48)

We can then make use of the extended Poisson transformation, which takes the form (see [14]),

$$\begin{aligned}&\sum ^\infty _{j=-\infty } (-1)^j \, f\!\left(j+\frac{1}{2}\right) = \frac{1}{i}\, \sum ^\infty _{-\infty } (-1)^j \, \bar{f}\!\left(j+\frac{1}{2}\right) \nonumber \\&\qquad \qquad \text {where}\qquad \bar{f}(j) \equiv \int ^\infty _{-\infty } f(w)\, e^{2\pi i\,jw}\, dw. \end{aligned}$$

(49)

For this purpose, we first assume that $\delta$ is an even integer, and let $j\rightarrow j-\frac{\delta }{2}$ in (48):

$$\begin{aligned} Q_\delta \,(t,b) = \frac{2^{2+\delta }}{\sqrt{2\pi t}} \;\sum ^\infty _{j=-\infty } (-1)^j \, (-1)^{\delta /2} \begin{pmatrix} j+ \frac{\delta }{2} \\ j- \frac{\delta }{2} \end{pmatrix}\!\left(j+\frac{1}{2}\right) e^{-\left(2b^2/t\right)\left(j+\frac{1}{2}\right)^2}. \end{aligned}$$

(50)

But $(-1)^{\delta /2} \left( \begin{array}{l} j+ \delta /2 \\ j-\delta /2 \end{array} \right) = \frac{1}{\delta !} \prod ^{\delta /2}_{k=1} \left( \left(k-\frac{1}{2}\right)^2 - \left(j +\frac{1}{2}\right)^2\right), \quad \text {and so}$

$$\begin{aligned} Q_\delta \,(t,b)&= \frac{2^\delta }{\delta !} \prod ^{\delta /2}_{k=1} \left( \left(k-\frac{1}{2}\right)^2 + \frac{1}{2}\;\frac{\partial }{\partial b^2/t}\right) Q_0 \,(t,b)\nonumber \\&\quad \text {where}\quad Q_0\,(t,b) = \frac{4}{\sqrt{2\pi t}} \;\sum ^\infty _{-\infty } (-1)^j \left(j+\frac{1}{2}\right) e^{-\left(2b^2/t\right) \left(j+\frac{1}{2}\right)^2}. \end{aligned}$$

(51)

With minor changes of notation, this expression for $Q_0\,(t,b)$ was obtained in Ref. [14], and (49) was used to transform it to

$$\begin{aligned} Q_0\,(t,b) = \frac{\pi }{b}\;\frac{t}{2b^2} \,\sum ^\infty _{-\infty } (-1)^j \left(j+\frac{1}{2}\right) e^{-\frac{\pi }{2}\;\frac{t}{b^2} \left(j+\frac{1}{2}\right)^2}. \end{aligned}$$

(52)

The first example of the use of (51) is for $\delta =2$:

$$\begin{aligned} Q_2\,(t,b)&= \frac{2\pi }{b}\,\sum ^\infty _{-\infty } (-1)^j \left[ \frac{1}{4}\;\frac{t}{2b^2} \left(j+\frac{1}{2}\right) - \left(\frac{t}{2b^2}\right)^2 \left(j+\frac{1}{2}\right) \right. \nonumber \\&\qquad \qquad \qquad \left. + \pi ^2 \left(\frac{t}{2b^2}\right)^3 \left(j+\frac{1}{2}\right)^3 \right] e^{-\frac{\pi ^2}{2}\;\frac{t}{b^2} \left(j+\frac{1}{2}\right)^2}, \end{aligned}$$

(53)

and $Q_\delta$ for higher values of $\delta$ is even more involved.