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I want to simulate stock price paths with different stochastic processes. I started with the famous geometric brownian motion. I simulated the values with the following formula:

$$R_i=\frac{S_{i+1}-S_i}{S_i}=\mu \Delta t + \sigma \varphi \sqrt{\Delta t}$$


$\mu= $ sample mean

$\sigma= $ sample volatility

$\Delta t = $ 1 (1 day)

$\varphi=$ normally distributed random number

I used a short way of simulating: Simulate normally distributed random numbers with sample mean and sample standard deviation.

Multiplicate this with the stock price, this gives the price increment.

Calculate Sum of price increment and stock price and this gives the simulated stock price value. (This methodology can be found here)

So I thought I understood this, but now I found the following formula, which is also the geometric brownian motion:

$$ S_t = a \exp\left[\left(\mu - \frac{\sigma^2}{2}\right) t + \sigma W_t \right] $$

I do not understand the difference? What does the second formula says in comparison to the first? Should I have taken the second one? How should I simulate with the second formula?

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This method is really close to be off-topic, but it can be interesting for later users so I'll still answer it. –  SRKX Nov 22 '12 at 8:37
@SRKX, by the way, why would this question be close to be off-topic? I find it more on target than 30%-40% of all other questions recently asked. You will be surprised how many market practitioners cannot answer this seemingly simple question, even those on the derivatives and exotics side. –  Matt Wolf Nov 22 '12 at 9:12
Please do not hesitate to register in order to help the site grow and make it out of beta! –  SRKX Nov 22 '12 at 9:24
@Freddy we try our best to close the one we find off-topic, maybe some of them went through. About registration, I was talking to the user who wrote the question, and it helps in the Area51 stats for make it out of beta. –  SRKX Nov 22 '12 at 9:30
see if you can get your hands on a copy of Glasserman's book Monte Carlo Methods in Financial Engineering, truly a masterpiece in many regards. –  user25064 Jan 27 at 17:54

2 Answers 2

up vote 12 down vote accepted

The way you do it in the first place is a discretization of the Geometric Brownian Motion (GBM) process. This method is most useful when you want to compute the path between $S_0$ and $S_t$, i.e. you want to know all the intermediary points $S_i$ for $0 \leq i \leq t$.

The second equation is a closed form solution for the GBM given $S_0$. A simple mathematical proof showed that, if you know the initial point $S_0$ (which is $a$ in your equation), then the value of the process at time $t$ is given by your equation (which contains $W_t$, so $S_t$ is still random). However, this method will not tell you anything about the path.

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nice concise explanation. Upvoted –  Matt Wolf Nov 22 '12 at 9:03
ok, thanks @SRKX –  user1690846 Nov 22 '12 at 9:06
There is no reason at all that paths cannot be simulated using the second method. Solving the SDE over a single interval will still allow a conditional formula such as $S_t = S_{t-1} \exp \{ (\mu - \sigma^2/2)\Delta t + \sigma (W_t - W_{t-1})\}$ with the standard method of simulation for the sample path of the brownian motion. –  user25064 Jan 27 at 17:57
@user25064 this is not what I meant, you can indeed do these multiple steps with the closed form. I meant that if you use it to compute $S_T$ directly, then you don't know what happened until then. There was no judgement here. –  SRKX Dec 1 at 13:25

To complement @SRKX comment ,i'll try to explain the "simple mathematical proof" beetween both formula : I assume you know the geometric or arithmetic brownian motion :

Geometric: \begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*} Arithmetic : \begin{equation*} dS = \mu dt + \sigma dz \end{equation*}

Then another important stochastic tool you need to know is the so called Ito Lemma : Loosely speaking, if a random variable $x$ follows an Ito process : (drift = $ a(x,t) $ et variance = $ b(x,t)^{2} $):

\begin{equation*} dx = a(x,t) dt + b(x,t) dz \end{equation*} Then another function $G$ which depends of $x$ and $t$ will respect also (ito lemma) the following process : \begin{equation*} dG = (\frac{\partial G}{\partial x}a + \frac{\partial G}{\partial t}+ \frac{1}{2}\frac{\partial^{2} G}{\partial x^{2}} b^{2}) dt + \frac{\partial G}{\partial x} bdz \end{equation*}

If we replace $x$ by the stock price and take its logarithm: $ G = ln(S)$. We also know : \begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*} then $ a = \mu S $ et $b = \sigma S $ and \begin{equation*} \frac{\partial G}{\partial S} =\frac{1}{S}, \frac{\partial^{2} G}{\partial S^{2}} = - \frac{1}{S^{2}},\frac{\partial G}{\partial t} =0 \end{equation*} using Ito lemma : \begin{equation*} dG = (\mu - \frac{\sigma^{2}}{2})dt + \sigma dz \end{equation*} Thus if we investigate the variation of $ln(S)$ (=G) between date zero and date $T$ : \begin{equation*} ln(S_{T})-ln(S_{0}) \sim \phi[(\mu - \frac{\sigma^{2}}{2})T, \sigma \sqrt{T}] \end{equation*} \begin{equation*} ln(S_{T}) \sim \phi[ln(S_{0})+(\mu - \frac{\sigma^{2}}{2})T, \sigma \sqrt{T}] \end{equation*} If we integrate : \begin{equation*} S(t) = S(0) \exp{(\mu - \frac{\sigma^{2}}{2})t + \sigma (z(t)-z(0))} \end{equation*} or \begin{equation*} S(t) = S(0) \exp{(\mu - \frac{\sigma^{2}}{2})t + B_{t}} \end{equation*} where $ B_{t} $ is a brownian motion.

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I might be wrong but it seems like your arithmetic and geometric BM are the same. You might want to drop S for the arithmetic. –  user7227 Feb 13 at 6:36
you are right, sorry, i corrected it. –  Malick Feb 13 at 18:51

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