# Random Walk of N Correlated Assets

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:

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

Where:

• 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?

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 $\mathbf{W}$.
• @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. May 13 '18 at 10:15