Let's take a process $S$ that satisfies: \begin{equation} dS = \mu S dt + \sigma S dz \end{equation} with $dz$ a Wiener process, $\sigma$ the volatility of $S$, $\mu$ the expected return of $S$.

From Ito's lemma, we have that the process verified by $ln(S)$ is: \begin{equation} d(ln (S)) = (\mu - \sigma^2/2)dt + \sigma dz \end{equation}

Why is it more accurate to use the second equation to simulate a path for S rather than the first one?

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    $\begingroup$ Hi: In the first one, S can go below zero which is unrealistic. $\endgroup$ – mark leeds Feb 25 at 18:52
  • $\begingroup$ @Victor, Your first equation is arithmetic Brownian Motion, whereas your second equation is geometric Brownian Motion.If we want to compare the current stock price,X(0) and future stock price X(t) which is unknown, the best way to compare them is the ratio of stock price following lognormal distribution. $\endgroup$ – Dhamnekar Winod Mar 28 at 14:22
  • $\begingroup$ If i recall, the euler discretization in log spot space is equivalent to the milstein scheme (ie higher order) in log space, while in spot space they are not equivalent, so when you use an euler discretization, log spot is more accurate than spot. $\endgroup$ – will Mar 29 at 23:45

The specification of $ln(S)$ is based on the explicit assumption security prices and interest rates cannot go below zero.

And for the behaviour of securities, it has been well-established via empirical research that the security absolute price grows at an exponential rate rather than absolute rate.... i.e. after $T$ years, security price tends to be $S(0)e^{r_fT}$, instead of $S(0)\cdot(1 + r_tT)$.

Since the financial crisis, this assumption for funding interest rates have proven to be false, and there are multiple models where the $S$ is modeled instead of $ln(S)$.

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    $\begingroup$ Imho the question was about numerical accuracy, i.e. do you step forward by calculating $dS$ from one step to the next or $d(ln S)$ $\endgroup$ – ZRH Feb 25 at 20:37
  • $\begingroup$ @ZRH: I see your point but, in that case, won't his path using $dS$ still have the possibility of S going below zero ? It depends on the parameters of course, but still possible. $\endgroup$ – mark leeds Feb 26 at 18:02
  • $\begingroup$ As $S\rightarrow 0$, the factor $\sigma S$ that precedes $dz$ tends to zero fast enough to prevent that from happening. $\endgroup$ – ZRH Feb 26 at 18:53
  • $\begingroup$ @ZRH, with regards to numerical accuracy (prob I did misunderstand his query), is that my understanding : taking difference due to ln(S) generates continuous (compounding) which is a basic assumption of black-scholes (i.e. assume no discrete moves). Hence, if the time-step was insanely large, d(lnS) is still accurate for large time-steps with long periods, whereas taking dS, it will start to diverge. $\endgroup$ – Kiann Mar 29 at 12:23

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