Random Walk on $\mathbb{R}$ with Uniformly Distributed Steps and "Reflective" Boundary at Origin

Question

A particle lies on the real number line at the origin. For each step taken, the particle moves from its current position a distance (and direction) chosen equi-probably from range $[-1,r]$. However, if the particle would otherwise move left of the origin, the particle is set back at the origin before the next step.

Given $r$, can a closed expression be derived for the probability $P_{n,r}$ of the particle being at the origin after step $n$? Can an expression be derived for the centered variance $E_{n,r}(X^2)$?

This is a variation of a problem featured here, where instead of the particle "dying" upon crossing the origin it is simply set at the origin and allowed to continue. Also, I am asking about the probability of it simply being at the origin at the $n$th step, not whether it has at some point been there.

Being only a graduate, all I've managed to do is conjecture what the expression might be for particular cases. Inspired by the excellent answers to the original problem, I directly simulated the problem to find the probabilities and presumed they were given by a rational expression $\frac{N_r(n)}{(1+r)^n n!}$. Surprisingly, my results indicated that the numerator was likely an integer in these cases. My conjectures as to the numerator for those particular cases are:

• $r=0$: $N_0(n)=n!$ (Known precisely, since $P_{n,0} = 1$ for all $n$)
• $r=1$: $N_1(n) = (2n-1)!!$
• $r\rightarrow\infty$: $N_r(n)\rightarrow (n+1)^{n-1}$

Edit: The conjecture for the case when $r$ gets arbitrarily large has been changed. There was a typo. (I was considering its similarity to Cayley's Formula and typed that out instead!)

At this point I'm lost. Your consideration is appreciated.

Update: I've managed to derive a stronger conjecture for the expression for the "re-center" probability:

$$P_{n,r}=\left(\frac{a}{c}\right)^n\frac{\Gamma(1+\frac{d}{a}) \Gamma(1+\frac{b}{a}+n)}{\Gamma(1+\frac{b}{a}) \Gamma(1+\frac{d}{c}+n)}$$

I have made no progress in determining the dependence of the four variables $a$, $b$, $c$, $d$ on the parameter $r$, except in the obvious cases when $r=0$ and $r=1$. If this expression suggests any accessible combinatorial argument I'd be happy to hear it.

Second Update: I have a closed form conjecture for the "re-center" probability when $r \geq 1$:

$$P_{n,r}=\frac{\left(\frac{3r}{1+r}\right)_{n-1}}{(1+r)\left(2r\right)_{n-1}}, r\geq 1$$

This expression uses the rising Pochhammer symbol. It appears to fail for $r<1$ for whatever reason.

Answer 1

Let $X_1,X_2,\dots$ be independent identically distributed random variables representing the successive jumps of the particle. Let $M_n$ be the position of the particle at time $n\in\{0,1,\dots\}$, so that $M_0=0$ and $M_n=\max(0,M_{n-1}+X_n)$ for $n\in\{1,2,\dots\}$. Let now $F_n(x):=\mathbb{P}(M_n\le x)$ for real $x$. Then $F_0(x)=\mathrm{I}\{x\ge0\}$ and $F_n(x)=\frac{\mathrm{I}\{x\ge0\}}{1+r}\,\int_{-r}^1 F_{n-1}(x-y)\,dy$ for all real $x$ and $n\in\{1,2,\dots\}$. In particular, the probability in question is $P_{n,r}=\mathbb{P}(M_n=0)=F_n(0)$.

The work in Mathematica notebook at n=1,2,3 (the corresponding pdf file at n=1,2,3 - pdf) shows that the expression $$\frac{\left(\frac{3r}{1+r}\right)_{n-1}}{(1+r)\left(2r\right)_{n-1}},\quad r\geq 1,$$ which you most recently proposed for $P_{n,r}$, is incorrect already for $n=3$ (and $r\ne1$), even though it seems pretty close to the correct one.

I doubt that there is such a simple closed form expression for $P_{n,r}$. However, using Spitzer's identity and work done in c.f. of X_+, one can obtain an integral expression for the generating function of $P_{n,r}$: $$ (*)\qquad \sum_{n=0}^\infty P_{n,r}z^n= \frac1{\sqrt{1-z}}\exp \Big(\frac1{2 \pi i}\,\int_0^{\infty } \ln\frac{(r+1) u-z (e^{i r u}-e^{-i u})}{(r+1) u-z (e^{i u}-e^{-i r u})} \, \frac{du}{u}\Big) $$ for complex $z$ with $|z|<1$. To obtain (*), one can start with formula (23) in c.f. of X_+, analytically extend it to complex $s$ with $\tau:=\Im s\ge0$, and then let $\tau\to\infty$. Alternatively, one can start with the second displayed formula on page 159 (with $\lambda=0$) in Spitzer 1960 and then reason as in c.f. of X_+; here one should also have in mind statement (2.4) on page 155 in Spitzer 1960.

In particular, it easily follows from (*) that your expression for $P_{n,r}$ is correct in the symmetric case, when $r=1$, but apparently only in this case.

Addendum: Even though there does not seem to be a simple closed form expression for $P_{n,r}$, an explicit but rather complicated expression for $P_{n,r}$ can be obtained from (*). Indeed, expand the logarithm back in powers of $z$. Then use the Cauchy integral theorem to see that for natural $n$ $$\frac1{\pi i}\int_0^\infty\frac{f(u)^n-f(-u)^n}u\,du =1-2a_n,$$ where $$f(u):=\mathbb{E}e^{iuX_1}=\frac{e^{iru}-e^{-iu}}{i(r+1)u}$$ for $u\ne0$ and $$a_n:=a_{n,r}:=\frac1{2n}\Big[1+\frac1{n!}\sum_{j=0}^n(-1)^j \binom nj \Big(\frac n{r+1}-j\Big)^n\,\text{sign}\Big(\frac n{r+1}-j\Big)\Big].$$

It follows from (*) that for complex $z$ with $|z|<1$ $$\sum_{n=0}^\infty P_{n,r}z^n=\exp\sum_{k=1}^\infty a_k z^k =\prod_{k=1}^\infty\exp(a_k z^k) =\prod_{k=1}^\infty\sum_{q=0}^\infty \frac{a_k^q z^{kq}}{q!},$$ whence $$P_{n,r}=\sum\prod_{k=1}^n\frac{a_k^{q_k}}{q_k!},$$ where the sum is taken over all $n$-tuples $(q_1,\dots,q_n)$ of nonnegative integers such that $1q_1+2q_2+\dots+nq_n=n$. In particular, for $n=0$ the set of all such $n$-tuples is the singleton set $\{\emptyset\}$, and, as usual, $\prod_{k=1}^0\ldots:=1$, so that $P_{0,r}=1$. Also, $$P_{1,r}=a_1=a_{1,r},\quad P_{2,r}=a_2+a_1^2/2!, \quad P_{3,r}=a_3+a_1a_2+a_1^3/3!. $$ Substituting here the expressions for the $a_k$'s, one sees that the above results for $n=1,2,3$ agree with the ones previously found in the Mathematica notebook at n=1,2,3 (the corresponding pdf file at n=1,2,3 - pdf) by iterative integration.

Answer 2

Not an answer, but I was exploring a similar random walk, so I thought I would include an image of a simulation. In my walk, each step is of a random length drawn from a normal distribution with mean $\mu=0$ and $\sigma=1$. So $x_{i+1} = x_i + \cal{N}$$(0,1)$ if that is nonnegative, and otherwise $x_{i+1}=0$.

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Vertical axis is $x_i$; horizontal $i$, the number of steps. The walk wanders rather far from zero.