# Geometric Brownian Motion

1. Jan 29, 2014

### motherh

Hi, I am trying to answer the following question:

Consider a geometric Brownian motion S(t) with S(0) = S_0 and parameters μ and σ^2. Write down an approximation of S(t) in terms of a product of random variables. By taking the limit of the expectation of these compute the expectation of S(t).

Let Z_i = σ√Δ (with probability 1/2(1+ $\frac{μ}{σ}$√Δ)) or -σ√Δ (with probability 1/2(1- $\frac{μ}{σ}$√Δ)).

Then logS(t) ≈ Z_1 + ... + Z_($\frac{t}{Δ}$) (as logS(t) is Brownian motion)

And this approximation gets better as $\frac{t}{Δ}$ tends to infinity

S(t) ≈ exp(Z_1) * ... * exp(Z_($\frac{t}{Δ}$))

So now exp(Z_i) = exp(σ√Δ) (with probability 1/2(1+ $\frac{μ}{σ}$√Δ)) or exp(-σ√Δ) (with probability 1/2(1- $\frac{μ}{σ}$√Δ)).

E[S(t)] ≈ (E[exp(Z_1)])^$\frac{t}{Δ}$

I know that the answer I am looking for is S_0*exp(tμ+$\frac{tσ^2}{2}$). Also

E[exp(Z_1)] =1/2exp(σ√Δ)*(1+ $\frac{μ}{σ}$√Δ) -1/2exp(-σ√Δ)*(1- $\frac{μ}{σ}$√Δ).

So I believe that

E[S(t)] = lim [ 1/2exp(σ√Δ)*(1+ $\frac{μ}{σ}$√Δ) -1/2exp(-σ√Δ)*(1- $\frac{μ}{σ}$√Δ) ]^$\frac{t}{Δ}$ (as $\frac{t}{Δ}$ tends to infinity)

Where do I go from here? I want to make use of lim (1+x/t)^t = exp(x) but how do I go from here? Help is MUCH appreciated!

2. Jan 30, 2014

### Ray Vickson

Use standard probability results to figure out what is the distribution of $\sum_i Z_i$ in the limit as $n \to \infty, \Delta \to 0$ with $t = n \Delta$ fixed.

Last edited: Jan 30, 2014