1
$\begingroup$

I would like to find a smart way to generate a $N\times N$ random matrix $M$ with arbitrary correlation: \begin{equation} \boxed{\langle M_{ij}M_{kl}\rangle=\tau_{ijkl}} \end{equation} Where the mean and variance of the elements are given by: \begin{align} \langle M_{ij}\rangle&=0 \\ \langle M_{ij}^2\rangle&=\sigma^2 \end{align}


The case I am interested in is actually a sub-problem of this. I would like to generate a matrix whose elements follow a normal distribution of mean $0$ and variance $1/N$, and whose elements are correlated the following way: \begin{equation} \langle M_{ij}M_{ki}\rangle=\tau_{ijk} \end{equation} When $\tau_{ijk}=\delta_{jk}N^{-1}$ I recover a symmetric matrix.

$\endgroup$
1
  • $\begingroup$ Use Cholesky factorisation $\endgroup$
    – mikado
    Apr 12 at 5:23
2
$\begingroup$

Simplify $ij\to u$ and $kl\to v$: find $M_u$ such that $\langle M_u\rangle=0$ and $\langle M_u M_v\rangle=\tau_{u v}$. This notation makes the analysis a bit simpler.

We can achieve this effect by using the matrix-square-root of the matrix $T$ of elements $\tau_{u v}$. Example:

n = 3;
T = {{2, -1, 0.3}, {-1, 4, 1.3}, {0.3, 1.3, 2}};
SymmetricMatrixQ[T] && PositiveSemidefiniteMatrixQ[T]
(*    True    *)

Using the matrix square root MatrixPower[T, 1/2] to generate $10^5$ lists of random numbers and computing their covariance matrix:

V = RandomVariate[NormalDistribution[], {10^5, n}] . MatrixPower[T, 1/2];
Covariance[V]
(*    {{1.99821, -1.01062, 0.29987},
       {-1.01062, 4.03928, 1.28961},
       {0.29987, 1.28961, 1.99284}}    *)

We see that this covariance matrix matches the desired T.

$\endgroup$
2
  • $\begingroup$ How can I construct back my matrix $M$? It seems that for $n=3$ in your code my matrix $M$ does not exist. Sorry if I misunderstood something. $\endgroup$
    – Matt
    Apr 12 at 6:07
  • $\begingroup$ If at the end you want $4\times4$-matrices $M$ correlated by a $4\times4\times4\times4$-tensor of elements $\tau_{ijkl}$, then set $n=16$, flatten the correlation tensor out into a $16\times16$-matrix T, and generate random 16-vectors. Then, take each 16-vector and reshape it into a $4\times4$ matrix. $\endgroup$
    – Roman
    Apr 12 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.