We know that, for a 1D symmetrical random walk, ${\displaystyle p = 1/2, q = 1/2}$, with equal walk step length, after n steps, its probability distribution will be proportional to binomial coefficient:

${\displaystyle f(k) = {n \choose k}}$

or in the continuous limit, its probability distribution will be Gaussian-normal distribution.

My question is, which type of random walk can have "scaled" binominal coefficient distribution ? that is, its distribution is:

${\displaystyle g(k) = f(ak), a > 0}$

I tried different walk step length for "walk to the left" and "walk to the right", (asymmetrical random walk), I could not get above distribution.

Can anyone help ?

Thank you in advance.

  • $\begingroup$ So you mean $g(k)=\large{{\binom{n}{a\cdot k}}}$? $\endgroup$ Aug 6, 2018 at 18:31
  • $\begingroup$ That is what I mean $\endgroup$
    – david
    Aug 6, 2018 at 21:46

1 Answer 1


First of all, the random walk probabilities are proportional to $$\binom{n}{(n+k)/2},$$ not $\binom{n}k$. The walk is symmetric, so the formula needs to be symmetric when the sign of $k$ changes.

If you want to scale the distribution, just scale the step size. The simple random walk whose step sizes are $\pm 1/a$ will have distribution proportional to $$ \binom{n}{(n+xa)/2} $$ where $x$ is the position you are trying to find the probability of.

  • $\begingroup$ Thank you for the post. $\endgroup$
    – david
    Aug 6, 2018 at 22:35
  • $\begingroup$ Thank you for the post. Maybe my question is not clear. Here is what I am looking for. For example, original symmetric random walk, after walk 100 steps, the possible end positions can be [0, 1, 2, ...., 100] with probability f(k) where k = 0, 1, 2, ..., 100. I am looking for another random walk, after 100 steps, the possible end position can be [0, 1, ..., 10] with distribution g(k) = f(10k). What kind of random walk can satisfy above requirements ? Maybe I shall not call this "scaled" binomial coefficient, maybe i shall call it "squeezed" binomial coefficient. $\endgroup$
    – david
    Aug 6, 2018 at 22:47
  • 1
    $\begingroup$ I now understand your question and have no idea how to answer it. $\endgroup$ Aug 6, 2018 at 23:05

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