Geometric Brownian Motion

In summary, the question discusses a geometric Brownian motion and its approximation in terms of a product of random variables. The expectation of S(t) is computed by taking the limit and using the standard probability result. The distribution of the sum of Z_i is also discussed in the limit.
  • #1
motherh
27
0
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+ [itex]\frac{μ}{σ}[/itex]√Δ)) or -σ√Δ (with probability 1/2(1- [itex]\frac{μ}{σ}[/itex]√Δ)).

Then logS(t) ≈ Z_1 + ... + Z_([itex]\frac{t}{Δ}[/itex]) (as logS(t) is Brownian motion)

And this approximation gets better as [itex]\frac{t}{Δ}[/itex] tends to infinity

S(t) ≈ exp(Z_1) * ... * exp(Z_([itex]\frac{t}{Δ}[/itex]))

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

E[S(t)] ≈ (E[exp(Z_1)])^[itex]\frac{t}{Δ}[/itex]

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

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

So I believe that

E[S(t)] = lim [ 1/2exp(σ√Δ)*(1+ [itex]\frac{μ}{σ}[/itex]√Δ) -1/2exp(-σ√Δ)*(1- [itex]\frac{μ}{σ}[/itex]√Δ) ]^[itex]\frac{t}{Δ}[/itex] (as [itex]\frac{t}{Δ}[/itex] 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!
 
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  • #2
motherh said:
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+ [itex]\frac{μ}{σ}[/itex]√Δ)) or -σ√Δ (with probability 1/2(1- [itex]\frac{μ}{σ}[/itex]√Δ)).

Then logS(t) ≈ Z_1 + ... + Z_([itex]\frac{t}{Δ}[/itex]) (as logS(t) is Brownian motion)

And this approximation gets better as [itex]\frac{t}{Δ}[/itex] tends to infinity

S(t) ≈ exp(Z_1) * ... * exp(Z_([itex]\frac{t}{Δ}[/itex]))

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

E[S(t)] ≈ (E[exp(Z_1)])^[itex]\frac{t}{Δ}[/itex]

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

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

So I believe that

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


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:

1. What is Geometric Brownian Motion?

Geometric Brownian Motion is a mathematical model used to describe the random movement of a variable over time. It is commonly used in finance to model stock prices and other financial data.

2. How is Geometric Brownian Motion different from regular Brownian Motion?

Geometric Brownian Motion differs from regular Brownian Motion in that it includes a drift term, which accounts for the overall trend or growth of the variable being modeled. This drift term makes the model more realistic for financial data, which often exhibits an overall upward trend.

3. What are the assumptions of Geometric Brownian Motion?

The main assumptions of Geometric Brownian Motion are that the variable being modeled is continuous, has a constant volatility, and follows a log-normal distribution. Additionally, the model assumes that the increments of the variable are independent and identically distributed.

4. How is Geometric Brownian Motion used in finance?

Geometric Brownian Motion is commonly used in finance to model the behavior of stock prices, interest rates, and other financial data. It is also used in the Black-Scholes model for pricing options. By simulating different scenarios of Geometric Brownian Motion, analysts can make predictions about the future behavior of these variables.

5. What are the limitations of Geometric Brownian Motion?

Although Geometric Brownian Motion is a useful model for financial data, it has several limitations. It assumes that the volatility of the variable being modeled remains constant over time, which may not always be the case. Additionally, the model does not account for extreme events or market shocks, which can lead to inaccurate predictions.

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