Geometric Brownian motion/stock price

[Awatarn]

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

I read another book. They told me that if $\mu > \sigma^2/2$ then
$Prob\{X(t \rightarrow \infty )=\infty \} = 1$,
and if $\mu < \sigma^2/2$
$Prob\{X(t \rightarrow \infty ) = 0 \} = 1$,
where $X(t)$ 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?

 

[Valkarie]

The two different answers that you mentioned come from the fact that the expectation value of a stochastic differential equation is not necessarily equal to the probability of a given outcome. The expectation value is an average of all possible outcomes, while the probability of a given outcome is the likelihood of that outcome occurring. For example, if \mu > \sigma^2/2, then the expectation value of X(t) will increase with time, but there is still a small probability that X(t) will go to zero instead of increasing with time. Similarly, if \mu < \sigma^2/2, then the expectation value of X(t) will decrease with time, but there is still a small probability that X(t) will go to infinity instead of decreasing with time. In summary, the expectation value and the probability of a given outcome in a stochastic differential equation can be different.