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I am way more physicist than mathematician and this question arises from experimental physics/engineering. The motivation is dealing with small amount of random discrete shifts between measured periods of digital signal:

  1. Suppose there is a $1D$ random walk with a possibility of remaining at the place, i.e. steps $\left(~-k,-k + 1,\ldots,\pm 0,\ldots,+k~\right)$ for $k \in \mathbb{N}$ in each step.
  2. The probability is symmetric and equal for all $2k + 1$ possibilities $\left(~\mbox{i.e.}\ p=0.2\ \mbox{for}\ k = 2~\right)$.
  3. I need a formula for probability of reaching a particular discrete distance of $x$ after $n$ steps.

It must be a solved problem, I just didn't get lucky googling.

Felix Marin
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  • Do you mean the probability of reaching that distance at exactly $n$ steps, or just at some number of steps $\leq n$? – Ben Grossmann Jul 13 '16 at 21:23
  • I need $p$ as a function $p \equiv p\left( x(n), n\right)$. So exactly. – Victor Pira Jul 13 '16 at 21:34
  • Being there at step $n$, or being there for the first time at step $n$? – Anon Jul 13 '16 at 21:36
  • Being there at the step $n$, no matter for how many times. – Victor Pira Jul 13 '16 at 21:38
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    Would it be OK for you if I close this as a duplicate of http://math.stackexchange.com/questions/1805754? I'll be happy to explain how to map your problem onto that one if it's not obvious. – joriki Jul 13 '16 at 21:44
  • Thank you, I would be grateful for an elaboration. I see the connection, but I don't see the result. It's up to you whether you want post that as a comment or an answer, I would accept it certainly. – Victor Pira Jul 13 '16 at 22:25

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Your problem is equivalent to toss a $2k+1$ facets die at each step, subtract $k+1$ and get the
result as the $\Delta x$ to move.
Equivalently you can toss a die, with $2k+1$ facets, numbered $0,\; \ldots ,\,r=2k$, and subtract $k$.
Let's take this model (it simplifies the treatment).
So, you are asking what is the probability that after $m$ tosses you'd get $s=X+mk$.
The total number of equi-probable events will be $(2k+1)^m$. That is the number of integer points in a $m$-dimensional cube with side $0,\; \ldots ,\,r$.
The number of events leading to $s$ will be $$ N_{\,b} (s,r,m) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered} 0 \leqslant \text{integer }x_{\,j} \leqslant r \hfill \\ x_{\,1} + x_{\,2} + \cdots + x_{\,m} = s \hfill \\ \end{gathered} \right. $$ which geometrically is the number of integer points on the diagonal plane summing to $s$, delimited to be within the cube, and which is given by $$ N_{\,b} (s,r,m)\quad \left| {\;0 \le {\rm integers}\;s,r,m} \right.\quad = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,{s \over r}\, \le \,m} \right)} {\left( { - 1} \right)^{\,j} \left( \matrix{ m \cr j \cr} \right)\left( \matrix{ s + m - 1 - j\left( {r + 1} \right) \cr s - j\left( {r + 1} \right) \cr} \right)} $$ Refer to Rolling dice problem for further considerations.

G Cab
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    Thanks a lot! I just need a little clarification for a dumb physicists. In your notation $X$ is a target location and the sum is for $k$, i.e. for $k=2$ the indices should be 0,1,2? – Victor Pira Jul 14 '16 at 11:27
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    @VictorPira Sorry, there was a confusing superposition of notation: I amended that, should be clear now (sorry, it is coming..from a dumb engineer) – G Cab Jul 14 '16 at 14:45