Correlation between brownian motions and Cholesky decomposition [closed]

I know it is a pretty basic question (I'm new at Quantitative Finance), but what's the logic behind the Brownian Motions correlation?

The expression is: Where is this formula coming from?

On the other hand, when there are more than two motions, the process is to apply Cholesky decomposition to the covariance matrix. Why is this necessary?

Many thanks!

• If you want to simulate correlated normal variables, you need to draw from a multivariate normal distribution with a given correlation matrix. In particular, if you have a vector of independent standard normals (or in general, uncorrelated random variables) $u= (X_1, \cdots , X_n)^T$, then it is a result that if your covariance matrix is $\Sigma = AA^*$ then the covariance matrix of the random vector $Au$ is $\Sigma$. May 17 '21 at 15:52

I assume, the first equation is about creating 2 correlated standard normal random variables. Then $$X_1 = Z_1$$ and $$X_2 = \rho Z_1 + \sqrt{1- \rho^2}Z_2$$ are correlated with correlation $$\rho$$. One can prove this by calculateing the covariance. $$\text{Cov}(X_1, X_2) = \mathbb{E}(X_1X_2) - \mathbb{E}(X_1) \mathbb{E}(X_2) = \rho \mathbb{E}(Z_1^2) + 0 = \rho$$ $$\text{Corr}(X_1, X_2) = \frac{\rho}{\sigma_{X_1}\sigma_{X_2}} = \rho$$
The case is a little more complicated if you want more $$X_i$$ to correlate. Then you will end up with a covariance/correlation matrix. Let's consider the covariance matrix ($$\Sigma$$). We want the following property to hold: $$\text{Cov}(X, X) = \Sigma$$ Then if you have $$\{U_i\}$$ i.i.d. standard normal variables, and the Cholesky factorization of your covariance matrix ($$\Sigma = J J^T$$), you can create the wanted correlated X variables as follows: $$X = JU \text{ , then }$$ $$\text{Cov}(X, X) = \mathbb{E}(X X^T) - \mathbb{E}(X) \mathbb{E}(X^T) = \mathbb{E}(JUU^TJ^T) - 0 = J \mathbb{E}(UU^T)J^T = J I J^T = \Sigma$$
You can use this approach for the 2 variable case as well, in this case the covariance matrix looks like this $$\begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}$$. But since the only variable in this matrix is $$\rho$$, we like to simplify this case. For more variables it's not doable as the number of free parameters is $$n (n-1)/2$$, which grows quadratically, so there you have to work with the matrix solution.
Note: It's not necessary to use the Cholesky-decomposition. You can use any $$A$$ matrix which satisfies $$\Sigma = A A^T$$, but it's the easiest choice most of the time.