I am trying to value an option on N assets, say $S^1, S^2,..., S^N$ that expires in $\Delta T$ years using Monte Carlo simulation. I have read many sources that state I should use the following formula for each asset:

$S_T^i = S_0^i exp( (\mu_i - \sigma_i^2/2)\Delta T + \alpha_i\sigma_i\sqrt{\Delta T})$


  • The $i$'s are used to differentiate the different assets.
  • $S_t^i$ denotes the price of asset $S^i$ at time t.
  • $(\alpha_1,...,\alpha_N)$ are derived by taking the Cholesky decomposition $LL^*$ of the "correlation matrix" and then applying it to N iid standard normal random variables $(\epsilon_i,...,\epsilon_N)$.

My questions are:

  1. Does the "correlation matrix" represent the correlations between the Assets or of the Asset returns?
  2. Does the Cholesky method simply accomplish drawing from multivariate normal distribution with mean $(0,...,0)$ and variance-covariance matrix of the answer to my first question?

Thank you in advance.

  • $\begingroup$ why is this question titled "random walk", but the answer is all about "Brownian motion"? $\endgroup$
    – develarist
    Oct 14, 2020 at 18:28

1 Answer 1


The correlation matrix refers to the correlations between the asset returns. In fact, it can be seen as follows. Each asset follows a geometric Brownian motion, i.e., $$ \frac{{\rm d}S_t^i}{S_t^i}=\mu_i{\rm d}t+\sigma_i{\rm d}W_t^i, $$ where the correlation between $W_t^i$ and $W_t^j$ is supposed to be $$ \text{Corr}\left(W_t^i,W_t^j\right)=\rho_{ij}. $$ Therefore, the correlation matrix referes to the correlations between the asset returns.

The Cholesky decomposition helps to transform $N$ independent normal random variables into $N$ correlated normal random variables, with the correlation matrix $\rho_{ij}$ as above. This can be seen as follows. Solve the SDE for each asset, and $$ S_t^i=S_0^i\exp\left[\left(\mu_i-\frac{1}{2}\sigma_i^2\right)t+\sigma_iW_t^i\right]. $$ As we are only interested in the sampling of $S_T^i$, the above formula yields $$ S_T^i=S_0^i\exp\left[\left(\mu_i-\frac{1}{2}\sigma_i^2\right)T+\sigma_iW_T^i\right]. $$ Here each $W_T^i\sim\mathcal{N}(0,T)$. By contrast, when all $W_T^i$'s are viewed as a whole, i.e., a vector of $N$ random variables, we have $$ \mathbf{W}\sim\mathcal{N}(\mathbf{0},T\Sigma), $$ where the $\left(i,j\right)$-th entry of the square matrix $\Sigma$ reads $$ \Sigma_{ij}=\rho_{ij}, $$ because the components of $\mathbf{W}$ are correlated. Now, suppose we have another vector of $N$ random variables, denoted by $\mathbb{Z}$, that follows $$ \mathbf{Z}\sim\mathcal{N}(\mathbf{0},I_N), $$ meaning that the components of $\mathbf{Z}$ are independent and identically distributed as standard normal. This is what we could generate numerically. Our target is to make use of this $\mathbf{Z}$ to get $\mathbf{W}$. This could be implemented by $$ \sqrt{T}L\mathbf{Z}, $$ where $L$ satisfies $$ \Sigma=LL^{\top}. $$ Since $$ \sqrt{T}L\mathbf{Z}\sim\mathcal{N}(\mathbf{0},T\Sigma), $$ it may provide sampling of $\mathbf{W}$.

  • $\begingroup$ Thank you very much. As a follow up... you wrote that W ~ N(0, TSigma) where Sigma(i,j) = rho(i,j)*T. Is that supposed to read: TSigma(i,j) = rho(i,j)*T? Further, I thought that when we write N(0, Sigma), Sigma is supposed to denote the variance-covariance matrix, whereas here, it denotes the correlation matrix. $\endgroup$
    – quantnoob
    May 12, 2018 at 17:08
  • $\begingroup$ @quantnoob: Thank you for your comment! That should be a typo there (I seemed to have successfully confused myself while typing...). $\Sigma$ should be the correlation coefficient matrix, i.e., $\Sigma_{ij}=\rho_{ij}$. As $\Sigma$ is the correlation coefficient matrix, $T\Sigma$ is exactly the variance-covariance matrix. Hence $\mathcal{N}(0,T\Sigma)$ is not ambiguous any more. $\endgroup$
    – hypernova
    May 13, 2018 at 10:15
  • $\begingroup$ @hypernova why is this question titled "random walk", but the answer is all about "Brownian motion"? $\endgroup$
    – develarist
    Oct 14, 2020 at 18:29

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