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Geometric Brownian motion/stock price

  1. Jun 28, 2012 #1
    I'm tying to solve a stochastic differential equation of stock price. The equation is
    [itex]dX = X(\mu dt + \sigma dW)[/itex]
    where [itex]\mu, \sigma[/itex] are constants and greater than zero.
    It is easy to show analytically that the expectation value to the solution is
    [itex]E[X(t)] = E[X(0)] e^{\mu t}[/itex]
    Then I solved this equation numerically by standard Euler method to check my analysis.
    I found that if [itex]\sigma[/itex] is less than some particular value, my numerical solution is
    consistent with the analytical solution which is increasing with time. The problem is
    when [itex]\sigma[/itex] is greater some number, the numerical solution tends to go to zero
    instead of increasing with time. Note that [itex]\mu[/itex] is the same.
    What happens?

    I read another book. They told me that if [itex]\mu > \sigma^2/2[/itex] then
    [itex]Prob\{X(t \rightarrow \infty )=\infty \} = 1[/itex],
    and if [itex]\mu < \sigma^2/2[/itex]
    [itex]Prob\{X(t \rightarrow \infty ) = 0 \} = 1[/itex],
    where [itex]X(t)[/itex] is the Ito solution.
    For the second case, the probabilistic value is not consistent the expectation value, isn't it?
    Can you help me about these two different answers?
     
  2. jcsd
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