What kind of random walk can lead to a lognormal distribution in the limit?

Question

For the binominal random walk with step sizes: $S_i = \begin{cases} &+1 &\text{probability} &p, \\ &-1 &\text{probability} &q=1-p \end{cases}$

Let $T_n = \sum_{i=1}^mS_i$ be the displacement after a fixed, not random, number of steps $n$.

then the probability distribution: $P(T_n=t)$ will approach a normal distribution when $n$ is large.

My question is, what kind of random walk (maybe with different step sizes, different probability) can generate a lognormal distribution in the limit ?

Answer 1

$T_0 = 1, T_{n+1} = \begin{cases} &2 * T_n &\text{probability} &p, \\ &1/2 * T_n &\text{probability} &q=1-p \end{cases}$

additive version: $S_0 = 1, S_{n+1} = \begin{cases} &T_n &\text{probability} &p, \\ &-1/2 * T_n &\text{probability} &q=1-p \end{cases}$

main challenge is that you have to adjust step size $S_n$ to place where you are now $T_n$, here I used factor $2$ but you can tweek this parameter to get desired result

this version should produce skewed displacement $T_0 = 1, S_{n} = \begin{cases} &2/n &\text{probability} &p, \\ &-1/n &\text{probability} &q=1-p \end{cases}$

Answer 2

I believe there is a fairly well-known solution to this problem which is called the Metropolis algorithm. This is one of the most common Markov Chain Monte Carlo methods. The idea is to make the random walk $(T_n)$ a Markov Chain such that the desired log-normal distribution is the unique stationary distribution for $(T_n)$. Then $T_n$ will be approximately log-normal distributed for large $n$.

Let $f$ denote the density of the log-normal distribution with some fixed parameters $\mu\in\mathbb{R}$ and $\sigma^2>0$. That is, $$ f(x)= \begin{cases} \frac{1}{x\sigma\sqrt{2\pi}}\exp\left(-\frac{(\log x-\mu)^2}{2\sigma^2}\right)&\text{if }x>0, \\ 0&\text{if }x\leq0. \end{cases} $$ Moreover, let $(U_n)$ and $(Z_n)$ be two independent sequences of i.i.d. random variables such that $U_1\sim U(0,1)$ and $Z_1\sim N(0,1)$. Now, put $T_0:=1$ and define for $n\geq1$: $$ T_n:= \begin{cases} T_{n-1}+Z_n&\text{if }U_n\leq\frac{f(T_{n-1}+Z_n)}{f(T_{n-1})}, \\ T_{n-1}&\text{if }U_n>\frac{f(T_{n-1}+Z_n)}{f(T_{n-1})}. \end{cases} $$ To get closer to the formulation in the question we can also write: $$ T_n:= \begin{cases} T_{n-1}+Z_n&\text{with probability }1\wedge\frac{f(T_{n-1}+Z_n)}{f(T_{n-1})}, \\ T_{n-1}&\text{with probability }1-\left(1\wedge\frac{f(T_{n-1}+Z_n)}{f(T_{n-1})}\right). \end{cases} $$ So the steps in this random walk are either $0$ or a realization of the $N(0,1)$-distribution. The probabilities which determine which of these steps to take depend on the current position, i.e. $T_n$, and the position we would go to if we were to take the $N(0,1)$ step, i.e. $T_{n-1}+Z_n$.