Suppose you have a stock, its initial price is $$1$$. After a period of time $$\Delta t$$, its price will either get multiplied by $$1+\sqrt{\Delta t}$$ (goes up), or get multiplied by $$1-\sqrt{\Delta t}$$ (goes down). Both case has $$1/2$$ probability. Now let $$n\Delta t=1$$, What is the probability of the stock price being $$x$$ at $$t=1$$ for large $$n$$?

Here is my idea:

Well, the probability $$P(x)$$ is only non-zero for certain $$x$$. The stock price can go up $$a$$ times with $$0 \leq a \leq n$$, and $$a$$ being an integer. If the stock price goes up $$a$$ times, then its price would be $$(1+\frac{1}{\sqrt{n}})^a(1-\frac{1}{\sqrt{n}})^{n-a}$$, so the set of possilbe prices would be $$X=\{ (1+\frac{1}{\sqrt{n}})^a(1-\frac{1}{\sqrt{n}})^{n-a}\}^n_{a=0}$$, each with probability $${\frac{n \choose a}{2^n}}$$.

When $$n$$ gets larger and later, $$X$$ will get denser and denser on $$\Bbb R$$, eventually every real number will be a possible price up to a small error $$\epsilon$$ (can be made arbitrarily small by choosing some $$n$$).

Observe the identity $$\lim_{n\to\infty} (1+\frac{1}{\sqrt{n}})^{\frac{n}{2}+\sqrt{n}r} (1-\frac{1}{\sqrt{n}})^{\frac{n}{2}-\sqrt{n}r}=e^{2r-\frac{1}{2}}$$, this means to achieve a stock price $$e^{2r-\frac{1}{2}}$$, the stock needs to go up $${\frac{n}{2}+\sqrt{n}r}$$ times. In other words, let $$x=e^{2r-\frac{1}{2}}$$, the stock price will be $$x$$ if it goes up $$a={\frac{n}{2}+\frac{\sqrt{n}}{2}(\ln x+\frac{1}{2})}$$ times (choose appropriate $$n$$ and $$x$$ so that $$a$$ is an integer).

Consider $$P(X|x\leq X \leq x+\Delta x)$$, for small $$\Delta x$$. It should be $$P(X=x)$$ times $$M$$=number of possible prices between $$x$$ and $$x+\Delta x$$. Where $$M=\Delta a=\frac{\sqrt{n}}{2x}\Delta x$$

Putting everything together, $$P(X|x\leq X \leq x+\Delta x)$$ should be $$\lim_{n\to\infty}\frac{\sqrt{n} {n \choose {\frac{n}{2}+\frac{\sqrt{n}}{2}(\ln x+\frac{1}{2})}}\Delta x}{2x2^n}=\frac{e^{-\frac{(\ln x-\frac{1}{2})^2}{2}}\Delta x}{x^2\sqrt{2\pi}}$$

Which means the PDF is:

$$\rho(x)=\frac{e^{-\frac{(\ln x-\frac{1}{2})^2}{2}}}{x^2\sqrt{2\pi}},x>0$$.

This is just a log-normal distribution.

Then $$Y\sim log(X)$$ has a normal distribution, and therefore $$X\sim \text{lognormal}$$
• sorry quick question. On going over your working again, I found $\lim_{n\to\infty}\frac{\sqrt{n} {n \choose {\frac{n}{2}+\frac{\sqrt{n}}{2}(\ln x+\frac{1}{2})}}\Delta x}{2x2^n}=\frac{e^{-\frac{(\ln x-\frac{1}{2})^2}{2}}\Delta x}{x^2\sqrt{2\pi}}$ Why does the $x$ on the LHS simplify to $x^2$ in the denominator on the RHS? Note the pdf of the lognormal: en.wikipedia.org/wiki/Log-normal_distribution corresponds to $x$ in the denominator – Rahul Madhavan Apr 17 at 18:01
• as regards your question in the comments, we are summing over, $\frac{1}{2}ln(1+x)+\frac{1}{2}ln(1-x)$, which cancel out in the first term, and as you point out the second terms add up (we ignore the 4th and higher order terms), so yes – Rahul Madhavan Apr 17 at 18:12