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If C is the correlation matrix, then we can do the cholesky decomposition: Then I can easily generate correlated random variables: where X are uncorrelated values and Y are correlated values. If I want two correlated random variables then L is:
When to use a covariance matrix in statistics?
When the population contains higher dimensions or more random variables, a matrix is used to describe the relationship between different dimensions. In a more easy-to-understand way, covariance matrix is to define the relationship in the entire dimensions as the relationships between every two random variables.
How to generate a random vector with a covariance matrix?
Hence, To generate a random vector with a given covariance matrix Q, look at the Cholesky decomposition of Q i.e. Q = L L T. Note that it is possible to obtain a Cholesky decomposition of Q since by definition the co-variance matrix Q is symmetric and positive definite. Now look at the random vector Z = L X. We have
When to use a covariance m easure matrix?
Covariance m easures how much two random variables vary together in a population. When the population contains higher dimensions or more random variables, a matrix is used to describe the relationship between different dimensions.
Is there an alternative to the Cholesky decomposition?
As you can see, the post-hoc estimated correlation matrix drastically differs from the prior one. Is there a bug in my code, or is there some alternative to using the Cholesky decomposition? I beg your pardon for this mess.
How is the Cholesky decomposition used in Python?
I use Cholesky decomposition to simulate correlated random variables given a correlation matrix. The thing is, the result never reproduces the correlation structure as it is given. Here is a small example in Python to illustrate the situation.
Is there anything wrong with the Cholesky factorization?
There’s nothing wrong with the Cholesky factorization. There is an error in your code. See edit below. Here is MATLAB code and results, first for n_obs = 10000 as you have, then for n_obs = 1e8. For simplicity, since it doesn’t affect the results, I don’t bother with means, i.e.,