We know that binominal random walk will approach to the normal distribution in the limit.
For a random walk, each step $X_i$ has the step size of 1 or -1 with probability $p = 1/2$ and $q = 1 – p = 1/2$.
$$X_i = \left\{
\begin{array}{ll}
1 & \mbox p = 1/2 \\
-1 & \mbox q = 1 – p = 1/2
\end{array}
\right.$$
Then the sampled average
$$\bar{X}_n = \frac{X_1 + X_2 + \dots + X_n}{n}$$
in the limit, will approach to the normal distribution:
$$\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$
Question: What kind of (additive) random walk can produce a distribution that is proportional to the following in the limit ?
If we let:
$$ u = e^x$$
then above distribution becomes:
$$e^{-u^2}$$
So it seems that we only need to take a logarithm transform from normal distribution to get the distribution we want. The question is, how to realize such transform for the each step of random walk ?
Because of central limit theorem, if such random walks exist, each step can NOT be independent and identical, what do we need to make change to each $X_i$ to get the distribution we want in the limit ? For example, each step has different probability and step size for left and right movement ? and steps are NOT independent ?
Best Answer
Fix $n$ and let $\{\xi_m:m\in\mathbb{N}\}$ be i.i.d. and uniform on $\{1,\ldots,n\}$. We regard the $\xi_m$'s as independent draws from a collection of $n$ coupon-types. Let $\tau_0=0$ and $$ \tau_k^n = \inf\{m: |\{\xi_1, \ldots, \xi_m\}| = k\} $$ for $k\ge1$, so that $\tau_k$ is the number of draws needed to collect $k$ distinct coupons. Let $X_k=\tau_k^n-\tau_{k-1}^n$ and $T_n=\tau_n^n$, so that $$ T_n = X_1 + X_2 + \cdots + X_n. $$ Here, $T_n$ is the number of draws to obtain a complete collection, and, after collecting $k$ types, $X_k$ is the number of additional draws needed to obtain a new distinct coupon. In this case, the $X_k$'s are independent, but not identically distributed.
In Probability: Theory and Examples, by Rick Durrett, it is shown that $$ \frac{T_n - n\log n}n \Rightarrow X $$ as $n\to\infty$, where $X$ has distribution function $F(x)=\exp(-e^{-x})$ for all real $x$. (Here, $\Rightarrow$ means convergence in distribution and the logarithm is a natural logarithm.)
Version 5 of Durrett's book is freely available on his website, and the proof of the above convergence result in on page 173. (You will also need to read Theorem 3.6.10 and Example 2.2.7.) Here is a link: https://services.math.duke.edu/~rtd/PTE/PTE5_011119.pdf#page=181.