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.
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.