For a binomial random walk, each step $S_i$, can move up or down with equal step size with probability $1/2$, then we know that the generating function for the distribution of the summation:
$$S_1 + S_2 + … + S_n$$
is:
$$\frac{1}{2^n}(1 + x)^n$$
Question 1: What is the generating function for the distribution of the this binomial random walk:
$$(S_1 + S_2 + … + S_n)^2$$
Q2: As we know the distribution for the summation of regular binomial random walk will approach normal distribution in the continuous limit. What will be distribution of the above random walk in the continuous limit ?
Best Answer
The answer for Q1 is $2^{-n} \sum_{k=0}^n {n\choose k} x^{k^2}$. There is no closed form for this sum.
Concerning Q2, denote $X_n = S_1 + \dots + S_n$ and write $$ \frac{X_n^2 - n^2/4}{n\sqrt{n/4}} = \frac{X_n - n/2}{\sqrt{n/4}}\cdot \left(\frac{X_n}n + \frac12\right). $$ The first term converges to the standard normal distribution thanks to CLT, the second one converges to 1 almost surely thanks to SLLN. So by Slutsky's theorem, the whole expression converges to the standard normal distribution. Put differently, $X_n^2\approx N(n^2/4, n^3/4$).