Why Discrete GBM Simulations Can Produce Negative Values
Summary
A geometric Brownian motion process has a continuous-time solution that remains positive when its starting value is positive. The note addresses why a numerical simulation of the stochastic differential equation may nevertheless produce negative values. Its explanation is that a discrete update can apply a sufficiently large negative shock in one step and jump across zero, even though the continuous process approaches zero with changes that shrink as the current value shrinks.
The discussion also identifies a spreadsheet formula error in the example and says that correcting it does not eliminate negative simulated values when volatility is large relative to the time step. Reducing the step size makes such artifacts less likely. This is a discretization issue, not evidence that the continuous GBM solution becomes negative. The note gives an intuition rather than a convergence analysis, and its displayed closed-form exponent is inaccurate: the standard drift correction uses one half of the variance term.
Key ideas
- A continuous GBM started above zero remains positive under its standard closed-form solution.
- A discrete numerical update can jump from a positive value to a negative one when a shock is large relative to the step size.
- The likelihood of such simulation artifacts depends on volatility and time-step size.
- The example contains a spreadsheet update error and an inaccurate displayed drift correction.
Tags
Full text
# negative values in geometric brownian motion
# negative values in geometric brownian motion
A GBM (Geometric Brownian Motion)
$ \frac{dx}{x} = \mu dt + \sigma dW $
solves to
$x_t = x_o e^{(\mu - \sigma^2)t + \sigma W_t}$
From the solution, it is clear that $x_t$ cannot become negative. However, it is not so clear from the SDE. In fact, if I do simulations using the SDE, x very frequently becomes negative for certain parameter combinations.
Is the non-negativity of $x_t$ valid only in case $dt->0$
Here is one simulation:
## Answer by polarbear (score 6, accepted)
https://quant.stackexchange.com/a/43758
I agree with wrong formula in simulation, but think i understand the question. Here's my take on it:
The reason SDE may seem to allow a negative value of x is because dW can be a large negative number, and thus can move x from positive to negative values. If that's the confusion, that is justified only in discrete case.
Do a thought experiment. Imagine the transition point of x from positive to negative in discrete case. Now, to move closer to the continuous case, you can decrease dt and imagine x as going from positive to zero first, and then to negative. But your SDE states that dx = x multiplied by something. So when x->0, dx also ->0. This should help you understand that the negative values are only coming from discretization of an otherwise continuous process.
Now, if you fix your formulas in the simulation and make it E3= E2+ D2*E2 (instead of D2+ D2*E2 which you used), you can still get negative values if sigma is really large and dt is not small enough. But try changing dt relative to sigma and you will find it harder to get negative values. Moreover, excel might have a limit of how small you can go.
Hope it helps.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.