1

Suppose I have a small particle and put it on the center of an interval on a 1-D axis. If the particle undergoes a motion that satisfies:

  1. It chooses its direction freely and randomly
  2. It chooses the span of each step of equal probability in range (0,r]
  3. The total number of the steps is unlimited

So I'd like to ask that what's the probability that the particle finally jump out of the interval specified as a function of r.

YiFei
  • 316
  • It sounds like you're doing a random walk on an interval. In which case, every point is visited infinitely often. – Michael Burr Aug 17 '15 at 14:52
  • @MichaelBurr Doesn't that the random walk is something discrete so can I simply use the results of random walk in one dimension – YiFei Aug 17 '15 at 14:55
  • "2.it chooses the span of each step freely and of equal probability". Isn't this the same as defining a uniform probabiltiy distribution on the positive real numbers? That isn't possible. – Santeri Aug 17 '15 at 15:03
  • @Santeri Basically, yes. But why that's impossible? – YiFei Aug 17 '15 at 15:21
  • There's a good discussion here: http://math.stackexchange.com/questions/14777/why-isnt-there-a-uniform-probability-distribution-over-the-positive-real-number – Santeri Aug 17 '15 at 15:22
  • @Santeri Thanks. I've edited my question. – YiFei Aug 17 '15 at 15:31
  • The answer is 1. –  Aug 17 '15 at 19:26

2 Answers2

1

The position at time $t$ is $x_t = \sum_{i=1}^t e_i$ where $e_i$ are iid, uniform in $[-r,r]$, $E(e_i)=0$, $Var(e_i)=r^2/3$. Hence $x_t$ tends to a gaussian with zero mean and variance $ t \, r^2/3$. Hence, when $t\to \infty$ the probability that it's inside some finite interval goes to zero.

leonbloy
  • 63,430
  • As $t \rightarrow \infty$ you have $Var(x_t) \rightarrow \infty$. – BruceET Aug 17 '15 at 21:17
  • Does your $Var(x_t)=\frac{tr^2}{3}$ comes from the formula of the variance of the summation of two random variables? – YiFei Aug 18 '15 at 00:49
  • @BruceTrumbo That the variance tends to infinity is not enough to prove that the probability to fall inside an interval tends to zero. – leonbloy Aug 18 '15 at 14:05
  • @leonbloy: Agreed. I did not say, or mean to imply that. It does however preclude a legitimate long-run distribution. Perhaps should have written a more detailed comment. – BruceET Aug 18 '15 at 16:45
0

This is a continuous-valued symmetrical random walk on the real line. At each step it moves (is 'displaced') by a number $D \sim Unif(-r, r).$ It is a Markov Chain because the movement of the particle from step $i-1$ (now) to step $i$ (next) depends only on the distribution of $D$ and the position of the particle now..

[Condition 1 says displacement can be positive or negative with equal probabilities, and Condition 2 says the absolute displacement is random in $Unif(0, r)$. That implies $D \sim Unif(-r,r).$]

The resulting long-run behavior, with respect to occupancy of the interval $(0,r]$, is not dependent on $r> 0$, which is essentially a scale parameter. Over the long run this process exhibits strange behaviors I won't discuss here. The comment by @MichaelBurr gives the relevant answer. The process will continue to go in and out of $(0, r]$ forever. In particular, it will typically leave this interval for the first time after only a few steps.

Note: In order to get a related process that settles to a long run distribution, one can contrive the displacements to be biased towards the origin. (The farther from the origin, the more likely movement is toward the origin.)

Below is a simulation (in R software) of one run of 5000 steps of the process you specified, in which I chose $r = 1.$ From the graph you can see it is outside $(0, 1)$ after only a few steps. (Index denotes the step $i$.) Sometimes plots such as this one are called 'history plots'.

 r = 1
 m = 5000;  x = numeric(m);  x[1] = 0
for(i in 2:m) {
   d = runif(1, -r, r)
   x[i] = x[i-1] + d  }
plot(x, type = "l")

enter image description here

BruceET
  • 51,500
  • But why's that the position of the particle affects its future directions in a long-run in your 'note' part? – YiFei Aug 18 '15 at 00:20
  • Attraction toward the origin' keeps the variance from going to 0 with increasing numbers of steps. In your model, there is no limiting distribution. The informal jargon is that the 'probability escapes to infinity'. If you are going to examine long-run behavior for symmetrical random walks, it is easier to do for a random walk that goes L or R by one integer at each step. Even there, lots of counter-intuitive facts. (In applications, this is moot; no such thing as a PERFECTLY fair coin in the real world.) Maybe Google 'random walk coin toss Feller". – BruceET Aug 18 '15 at 00:43