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Can anyone provide a source that formulates how to generate multivariate geometric Brownian motion returns using the Cholesky method with target correlation matrix, instead of correlated GBM prices?

If instead, correlated GBM prices are started, and then transformed to returns, I found that the correlation matrix of the prices following this route does not carry over to the returns, and is lost. so it would be better to model correlated GBM returns immediately, not prices. But where is a source that formulates the cholesky method with multivariate GBM (not just randn)?

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  • $\begingroup$ I didn't read carefully but this link looks relevant. quant.stackexchange.com/questions/25219/… $\endgroup$ – mark leeds Oct 7 '20 at 17:21
  • $\begingroup$ saw that but the Goddard link only shows a formula of univariate GBM price, not returns, not multivariate, not correlated. it is also impossible to extrapolate a formula from any codes because they simply tend to just loop univariate GBMs to construct the multivariate $\endgroup$ – develarist Oct 7 '20 at 17:30
  • $\begingroup$ Just to be clear about what you are asking, the process for univariate GBM is $\frac{dS}{S} = \mu \, dt + \sigma\, dZ$ and can be simulated over discrete steps by $\frac{S_{t+\delta t} - S_t}{S_t} = \mu \delta t + \sigma \sqrt{\delta _t} \xi$ where $\xi$ is a draw from a standard normal distribution. Somewhat better is $\log S_{t + \delta t} = \log S_t + \left(\mu - \frac{1}{2} \sigma^2 \right) \delta t + \sigma \sqrt{\delta t} \xi$. Do you call that generating prices or generating returns? $\endgroup$ – RRL Oct 7 '20 at 18:35
  • $\begingroup$ everything before "somewhat better" looks like returns, whereas after that point looks like price development and actually the definition of univariate GBM. how do you move (derive) from the second half mentioned to the first half $\endgroup$ – develarist Oct 7 '20 at 18:53
  • $\begingroup$ The first is the Euler method for the approximate numerical solution of the stochastic differential equation. By Ito's lemma GBM also follows $d \log S = \left(\mu - \frac{1}{2} \sigma^2 \right ) \, dt + \sigma \, dZ$ and the second difference equation happens to solve it exactly. They are time-stepping methods for equivalent representations of GBM. A path followed by GBM sampled at discrete steps is generated by drawing from the distribution of returns and aggregating. $\endgroup$ – RRL Oct 7 '20 at 22:15

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