Let $\Sigma \in \mathbb{R}^{n \times n}$ be a circulant, symmetric, positiv definite matrix. To generate correlated random variables $Y$ with the covariance matrix $\Sigma$, one has: $$ Y = C X $$ where $X$ are uncorrelated gaussian random variables and $C$ is the square root of $\Sigma$, i.e. $C C^T = \Sigma$. The Eigenvalues $\Lambda$ and Eigenvectors $V$ are derived simply with the discrete Fourier Transform (DFT), but $V \Lambda^{1/2}$ is not necessarily real.

Is there a way to compute (numerically) a real $Y$ without creating $C$ explicitly?

An example for $n=3$ is $$ \Sigma = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$$ then $$ C = \sqrt{2} \begin{bmatrix} 1 & 0.5 & 0.5 \\ 0 & 0.86 & 0.28 \\ 0 & 0 & 0.81\end{bmatrix}$$ The eigenvalues are $\Lambda = diag([4,1,1])$. The Eigenvectors $V$ of circulant matrices are given by the DFT matrix, which are though complex. Given a matrix $X \in \mathbb{R}^{n \times k}$. Is it possible to calculate $Y$ by means of a vector operation or DFT?


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  • $\begingroup$ Could you give an example of the kind of covariance matrix you are interested in, say for $n=3$? $\endgroup$ – Dilip Sarwate Jan 20 '16 at 11:32
  • $\begingroup$ @DilipSarwate I added an example. Hope it helps. $\endgroup$ – Sebastian Schlecht Jan 20 '16 at 12:22
  • $\begingroup$ I should have said: covariance matrix in which all the off-diagonal terms are not the same. If the off-diagonal terms have the same value (and all the on-diagonal terms have the same but larger value as in your $C$ above), then this answer of whuber's gives an explicit construction of a multivariate normal distribution with that covariance matrix. $\endgroup$ – Dilip Sarwate Jan 20 '16 at 21:19

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