4
$\begingroup$

Consider a one-dimensional random walk whose steps are $+2$ and $-1$ with probabilities $p$ and $1-p$ respectively, starting from $0$ and in the interval {$-n$, $n$}. The walk ends at $-n$ or $n$ or $n+1$. Let $m$ be the number of integers "jumped" during the walk.

Is there a limit for the ratio $\frac{m}{2n+1}$ for $n \rightarrow \infty$?

Three examples to clarify:

1) n=15 p= 1/2 Steps = {-1, 2, -1, 2, -1, 2, -1, 2, -1, -1, 2, 2, -1, -1, 2, 2, 2, 2, -1, \ -1, 2, -1, -1, 2, -1, 2, -1, -1, 2, 2}

Positions = {0, -1, 1, 0, 2, 1, 3, 2, 4, 3, 2, 4, 6, 5, 4, 6, 8, 10, 12, 11, 10, \ 12, 11, 10, 12, 11, 13, 12, 11, 13, 15}

Missed (jumped) positions = {-15, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, 7, 9, 14}

m = 17; r = m/(2 n +1) = 0.548387

2) n= 15 p=1/2 Steps = {2, 2, 2, 2, -1, -1, 2, 2, 2, -1, -1, -1, -1, 2, 2, -1, -1, -1, 2, \ -1, 2, -1, -1, 2, 2, 2}

Positions = {0, 2, 4, 6, 8, 7, 6, 8, 10, 12, 11, 10, 9, 8, 10, 12, 11, 10, 9, 11, \ 10, 12, 11, 10, 12, 14, 16}

Missed positions = {-15, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 1, 3, 5, 13, 15}

m =20; r = m/(2 n +1) = 0.645161

3) n=20 p=1/2 Steps = {-1, 2, -1, 2, 2, 2, -1, -1, -1, -1, -1, -1, 2, -1, 2, 2, 2, -1, 2, \ -1, 2, -1, -1, -1, -1, -1, 2, -1, -1, -1, 2, -1, -1, -1, 2, -1, 2, \ -1, 2, 2, -1, -1, -1, -1, -1, -1, -1, 2, 2, -1, -1, 2, 2, -1, 2, -1, \ 2, 2, 2, -1, 2, 2, 2, 2}

Positions = {0, -1, 1, 0, 2, 4, 6, 5, 4, 3, 2, 1, 0, 2, 1, 3, 5, 7, 6, 8, 7, 9, \ 8, 7, 6, 5, 4, 6, 5, 4, 3, 5, 4, 3, 2, 4, 3, 5, 4, 6, 8, 7, 6, 5, 4, \ 3, 2, 1, 3, 5, 4, 3, 5, 7, 6, 8, 7, 9, 11, 13, 12, 14, 16, 18, 20}

Missed positions = {-20, -19, -18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, \ -6, -5, -4, -3, -2, 10, 15, 17, 19}

m =20; r = m/(2 n +1) = 0.560976

A simulation with the range {-2000,2000}, iterated 1000 times provides r as 0.572958.

The question is: Is there a limit for n -> Infinity based on: (n, p, base steps {-1,2}) ?

  • 0
    'number of integers "jumped" ' I'm not sure of what that means2012-08-06
  • 0
    maybe "missed" is meant...?2012-08-06
  • 0
    When you say "limit", do you mean a bound? Or the limit of the ratio's expectation value? Or one of various notions of convergence for the random variable $m/(2n+1)$?2012-08-06

2 Answers 2