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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 $\lim_{n\to\infty}\frac{\sqrt{n} {n \choose {\frac{n}{2}+\frac{\sqrt{n}}{2}(\ln x+\frac{1}{2})}}\Delta x}{2x2^n}=\frac{e^{-\frac{(\ln x-\frac{1}{2})^2}{2}}\Delta x}{x^2\sqrt{2\pi}}$ enter image description here

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.

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You can also apply the CLT.

The sum of a large number of random variables tends towards a normal distribution. Here you are taking the product of a large number of random variables. Take log on both sides and apply the central limit theorem.

Then $Y\sim log(X)$ has a normal distribution, and therefore $X\sim \text{lognormal}$

PS: Your working looks correct to me.

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  • $\begingroup$ I see. X~lognormal(-1/2,1), the -1/2 is actually coming from the second term of the taylor expansion of ln(1+x). $\endgroup$ – Mango Apr 17 at 17:47
  • $\begingroup$ 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 $\endgroup$ – Rahul Madhavan Apr 17 at 18:01
  • $\begingroup$ 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 $\endgroup$ – Rahul Madhavan Apr 17 at 18:12
  • $\begingroup$ I think RHS simplifies to lognormal(-1/2,1) $\endgroup$ – Mango Apr 17 at 19:00

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