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I am currently trying to simulate correlated GBM paths and I found the Cholesky Composition for it. From my understanding, the Cholesky Decomposition can be used to create correlated random variables from uncorrelated random variables. However, it does not take into account the drift, which is exactly where I am struggling to understand.

Lets say I have two processes $P1$ and $P2$, both follow a GBM. The correlation $\rho$ between the two is $0.8$. The drift of $P1$ is $-0.03$ and of $P2$ it is $0.02$. When using the Cholesky Decomposition to generate the correlated random variables, the negative drift of $P1$ is not considered for $P2$, meaning that $P1$ tends to go down while $P2$ still tends to go up, since the correlated random variables do not take into account the drift. Am I missing something or did I misunderstand something?

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I see a bit of confusion. Let's say you have a model of $n$ stochastic processes $X_i$. Let us assume that they follow a GBM. This means their dynamics are: $$dX_t^i = \mu^i_t X_t^i dt + \sigma^i_t X_t^i dW_t^i,$$ where $W_t^i, \ i \in \{1, ..., n\}$ are $n$ correlated standard Brownian motion.

In this setting let us assume (without loss of generality) that $\mu$ and $\sigma$ do not depend on time. We thus have a vector $$\mu = \{\mu^1, \mu^2, ..., \mu^n\}$$ and the variance covariance matrix $\Sigma$ (please spare me to write it in TeX).

Now, let us move to the simulation. Since (I believe) you have to simulate this in a computer, the setting has to be discretized. Let us fix a time horizon $[0, T]$ and let us assume to have a tenor structure $$\tau = \{t_0 = 0, \ t_1, \ t_2, ..., t_k = T \}.$$

The trick here is to observe that we can generate each process with a (composed) Euler scheme: $$X_{t_{j+1}}^i = X_{t_{j}}^i \cdot e^{(\mu^i - \frac{1}{2}\sigma^2_i)(t_{j+1} - t_{j}) + \sigma_i \Delta W^i(t_{j+1})}.$$ To clarify notation $$ \Delta W^i(t_{j+1}) = W^i_{t_{j+1}} - W^i_{t_{j}} \ \sim N(0, \ t_{j+1} - t_j)$$ is our increment for the $i$-th brownian.

As you can see, the correlation between assets is encapsulated in the brownian increments. Therefore, the problem of generating correlated assets (processes) boils down to a problem of generating correlated normal random variables.

Here it is where Cholesky decomposition is used. Indeed, call $L$ the Cholesky decomposition of $\Sigma$. It will be a $n$ square matrix. The steps to construct correlated (with $\Sigma$) brownian increments are:

  1. For each time step, simulate $n$ random variables $Z^I$.
  2. Multiply them by $\sqrt{t_{j+1} - t_j}$ to obtain the wanted variance
  3. You get $$\Delta W^i(t_{j+1}) = \sum_{q = 1}^nL_{i,q}Z_q.$$

As you can see, the correlation is on the stochastic driver and NOT on the drift. The drift is a deterministic movement.

Hope this clarifies.

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  • $\begingroup$ This explanation helps to clarify this, but I am still wondering, doesn't that yield to unrealistic results in my example where one process $X1$ has a negative drift and the other $X2$ a positive drift? Without correlation, $X1$ tends to go down and $X2$ tends to go up, but with reasonable correlation (e.g. $0.8$) I would expect $X2$ instead to also have downward movement over time, which is not the case. The result doesn't look like there is a correlation in that case. $\endgroup$
    – Merwin
    Sep 24 '21 at 7:49
  • $\begingroup$ No. Try to think about this way: the drift is the direction your process is following, kind of a straight line. The dW part is how it is sway around this line. If two assets have opposite drift, they will have the same trend. If their stochastic driver is correlated, then the sway around their direction similarly. $\endgroup$ Sep 24 '21 at 8:02
  • $\begingroup$ I understand that now, but I am still missing something. I am basing my whole idea on historical data, e.g. I have two stocks and compute the correlation of the two ($0.8$) and the parameters for the GBM (negative drift for stock 1, positive for stock 2). A correlation of $0.8$ means that they tend to have the same movements, which is why I would also expect them to have it in the future. However, I think my mistake is that two stocks, one with negative drift and one with positive drift would probably never have a correlation of $0.8$ in the first place. $\endgroup$
    – Merwin
    Sep 24 '21 at 8:46

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