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Simulating Correlated Geometric Brownian Motion for Multiple Assets

Article Quant Q&A · Author: quantnoob

Summary

The document explains how to simulate terminal prices for several assets whose returns are correlated, in order to value a multi-asset option by Monte Carlo. In a geometric Brownian motion model, dependence is specified through correlations among Brownian motions, which correspond to correlations among the assets’ returns, rather than correlations among their price levels.

To generate joint shocks, draw independent standard normal variables and multiply the vector by a Cholesky factor of the return correlation matrix. Scaling by the square root of the time horizon gives Brownian increments with the desired covariance; multiplying each asset’s increment by its volatility then produces the correlated terms in the terminal-price formula. The explanation derives this from the multivariate normal distribution. It assumes the GBM model and a valid correlation matrix, and does not discuss estimating correlations, time-varying dependence, or other model risks.

Key ideas

  • The correlation matrix describes dependence between asset returns, represented by correlated Brownian motions.
  • Cholesky factorization transforms independent standard normal draws into correlated normal shocks.
  • Scaling the correlated shocks by the square root of the horizon gives Brownian increments with the required covariance.
  • Each asset’s volatility scales its Brownian increment in the GBM terminal-price equation.

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Full text
# Random Walk of N Correlated Assets


# 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:

- Does the "correlation matrix" represent the correlations between the Assets or of the Asset returns?

- 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.

## Answer by hypernova (score 3, accepted)

https://quant.stackexchange.com/a/39744

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 $\mathbf{W}$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.