I wrote this to simulate m random walks of n steps

Li[n_] := 2*RandomVariate[BinomialDistribution[n, 1/2], n] - n;
Tb[n_, m_] := Table[Li[n], {i, 1, m}];
y = table[10, 10]

The walker has to start at (0) I don't know how to adjust the function to get that. And I have to write a function that finds the average position after n steps and the average of the square of the distance between the walker after n steps and the origin without using Mean or StandardDeviation. We can take n=10 and m=10

For the average position I wrote that but I don't get the same result as when I use Mean ( and I think I have to do that in one function)

 averagepos[n_] := Total[y]/n;

For the average of the square of the distance walker-origin I don't get what I should calculate.

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    – Michael E2
    Oct 9, 2015 at 0:56
  • $\begingroup$ Shouldn't a 1-dimensional random walk start at 0, not {0, 0}? $\endgroup$
    – m_goldberg
    Oct 9, 2015 at 1:57
  • $\begingroup$ You can use Prepend to get the right starting position, like I did here. $\endgroup$
    – Karsten7
    Oct 9, 2015 at 2:03

1 Answer 1


To make sure all the walks start at {0}, you could use

Li[n_] := 2*RandomVariate[BinomialDistribution[n, 1/2], n] - n; 
Tb[n_, m_] := Table[Join[{0}, Li[n]], {i, 1, m}]; 
ListPlot[Tb[10, 5], Joined -> True]

enter image description here

Then the average distance for 1 run is calc using

    Ave[n_] := Abs[Total[Flatten[Tb[n, 1], 1]]]/n
(* 0.891*)

You can repeat the process for m runs as well. The output is highly sensitive to the parameter (1/2 in this case) which appears in the BionomialDistribution

Regarding the remark about the mean, I don't see it going to zero. Try

Li[n_, p_] := 2*RandomVariate[BinomialDistribution[n, p], n] - n; 
Tb[n_, p_, m_] := Table[Join[{0}, Li[n, p]], {i, 1, m}]; 
Ave[n_, p_] := Abs[Total[Flatten[Tb[n, p, 1], 1]]]/n
Aver[n_, p_, m_] := N[Sum[Ave[n, p]/m, {i, 1, m}]]
 Aver[n, p, m], {n, {100, 500, 1000}}, {p, 0.1, 
  1}, {m, {100, 200, 300}}]

enter image description here

For p=1/2, you get average < 1, for p values that deviate from 1/2, the average explodes. Unless of course, you have used some other relation for the mean.

  • $\begingroup$ When I do that I get that the mean is always 0 , but I have to show that it tends to 0 when m is big $\endgroup$
    – ferrou
    Oct 9, 2015 at 6:06
  • $\begingroup$ I found my error $\endgroup$
    – ferrou
    Oct 9, 2015 at 6:19
  • $\begingroup$ For the mean it is correct I was just extracting the wrong element from the tab to calculate it ! thanks for the answer :) $\endgroup$
    – ferrou
    Oct 9, 2015 at 6:57

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