If we assume a risk-neutral random walk for an underlying [$] S [$] to be [$]dS = rS dt + \sigma S dX[$] then an obvious discretised monte-carlo algorithm for simulating this is:[$] S(t+\Delta t) = r S(t) \Delta t + \sigma S(t) \sqrt{\Delta t} \phi [$] where [$]\phi[$] is drawn from a normal distribution. However, every google search has revealed the following algorithm for the time-stepping:[$] S(t+\Delta t) = S(t) e^{((r-\frac{1}{2}\sigma^2) \Delta t + \sigma \sqrt{\Delta t} \phi)}[$]I understand how to get the second equation, but what I wonder is what is wrong with the first? Why does nobody ever simulate a risk-neutral random walk using the first?

The first discretization generates normally distributed [$]S_t[$] (i.e. it can become negative). The second is equivalent to discretizing the SDE for [$]\ln \left( S_t \right)[$] which follows a constant coefficient arithmetic Brownian motion and thus there is no discretization error.

no, the first generates a lognormal S(t) as well, it is the plain Euler discretization of S, though it has a stochastic drift requiring small time steps (and S might become negative for numerical reasons).

It seems you lost term - S ( t ) on the left hand side of the first equation. There is a difference between using discretization of S or discretization of the SDE which specifies S. First approximates solution of the SDE , second approximates SDE. Both of them lead to stepwise approximation of the S. And the difference is probably o ( delta t )

Would it be an idea to write the SDE as a stochastic integral equation SIE and move it from there? QuoteIt seems you lost term - S ( t ) on the left hand side of the first equation. Indeed Having a SIE we can write it as a combination of Riemann and Ito integrals let's say on an interval [t, t + d] using numerical integration which is more intuitive and robuster than the SDE form?

QuoteOriginally posted by: list1It seems you lost term - S ( t ) on the left hand side of the first equation. There is a difference between using discretization of S or discretization of the SDE which specifies S. First approximates solution of the SDE , second approximates SDE. Both of them lead to stepwise approximation of the S. And the difference is probably o ( delta t )A SDE is only a symbolic representation of a stochastic integral equation, consisting of Riemann integral + Ito integral as mentioned. An SDE is just symbols.And most classical FD schemes (e.g. Euler) are awful numerical approximations.QuoteAnd the difference is probably o ( delta t)This is not true. The small o is wrong. And "difference" is ill-defined concept.Example: (norm) Convergence can be weak or strong, each one leading to a different O(dt^p) (BIG O). p can take on values like 1 or 1/2.

When mathematicians talk about SDE it always implies it in integral form tough discrete time form makes sense.The difference between integral SDE and its finite difference approximation calculated straightforward by making its difference. The simple way is consider mean squared difference ( Sorry I missed that ) and we arrive at o( [$]\Delta t[$] )

QuoteOriginally posted by: list1The simple way is consider mean squared difference ( Sorry I missed that ) and we arrive at o( [$]\Delta t[$] )What do you mean o(dt) difference? (really)

The Euler-Maruyama can diverge for MLMC A bit hit and miss..

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1The simple way is consider mean squared difference ( Sorry I missed that ) and we arrive at o( [$]\Delta t[$] )What do you mean o(dt) difference? (really)Cuchulainn, sorry for delay. I did not have enough time to answer. I also did not find books or papers to refer. I tried to make a sketch of derivation to present order of convergence solution of Euler scheme to correspondent solution of the continuous SDE. I should say that squared mean convergence is O ( dt ), ie [$]E | S ( t ) - S_\delta ( t ) | ^ 2 < C \Delta t [$]. If you interested to look on this sketch I could send it to you.

As a general aside, does anyone have any numbers on how long it takes in their pet language to generate e.g. 1 million such walks? In my python code it takes around 1 minute for a million.

QuoteOriginally posted by: barnyAs a general aside, does anyone have any numbers on how long it takes in their pet language to generate e.g. 1 million such walks? In my python code it takes around 1 minute for a million.It depends on the language and whether serial or parallel. Python is probably slow, but I don't know. Example: MLMC has a speedup on GPU in the range [10,190] compared to CPU. OpenMP and PPL are similar because problem is embarrassingly parallel.