Suppose that I have $N$ time series $x_{1t},x_{2t},\dots,x_{Nt},$, that are correlated with each other. A $N\times N$ correlation matrix is $R=\rho_{ij}$. It can be represented with eigen value dcomposition: $$R=\sum_{j=1}^N\lambda_j\xi_j\xi_j'$$ Assuming the eigenvalues are sorted in descending order $\lambda_1$ is the largest eigenvalue. It represents the overall correlation of series. My question: How to estimate $\lambda_1$ when you have only one observation at time $t$ of each series $x_{it}$? We have only $N$ values $x_{it}$ where $i\in[1,N]$.

If it helps, I suspect that the series might have $K$ common factors, so that $x_{it}=\sum_{j=1}^KB_{ij}f_{jt}+\varepsilon_i,$ where the noise is independent, i.e. $corr[\varepsilon_i,\varepsilon_j]=\delta_{ij}$ and $corr[\varepsilon_i,f_j]=0$.

Me thinks that maybe when $K=1$, we can somehow estimate $\hat\lambda_1$. However, I'd like a more general solution if possible.


Eigen value $\lambda_1$ represents overall correlation. Let's write a covariance of a sum of series: $$cov(\sum_i x_i)=\sum_{ij}\sigma_i\sigma_j\rho_{ij}=\sum_i\sigma^2+\sum_{i\ne j}\sigma_i\sigma_j\rho_{ij}$$

Assuming $\rho_{ij}=\rho$ we get the following metric: $$r=\frac{(\sum_ix_i)^2-\sum_ix_i^2}{\sum_{i\ne j}x_ix_j}$$

Suppose that one day I have two series observations (1,3), then another day they are (5,-1). Here's how we calculate the metrics: $$r_1=\frac{4^2-(1^2+3^2)}{2\times1\times 3}=\frac{6}{6}=1$$ $$r_2=\frac{4^2-(5^2+1^2)}{2\times5\times (-1)}=\frac{-10}{-10}=-1$$

In the first day both series moved in the same direction, and although the moves were smaller than in the second day, they still produced the same average move 2 because they were more correlated. Hence, this metric seems to be related to $\lambda_1$.

The idea's inspired by Eqs.6 and 7 in Bouchaud's paper.

  • $\begingroup$ Would you be wanting to deconvolve the different temporal frequencies across the N series (wavelet or Fourier transform?), then rank them by variance? $\endgroup$ – ReneBt Feb 24 at 20:36

I suggest you a very simple online update solution:

At each time $t$ you provide me a point of a time series $x$ of size $N$, then

(1) $\hat\mu=\hat\mu+\frac{1}{1+t}(x-\hat\mu)$

(2) $\hat\Sigma=\frac{t}{1+t}\hat\Sigma+\frac{t}{(1+t)^2}(x-\hat\mu)^\prime (x-\hat\mu)$

(3) $v^{new}=\hat\Sigma v$

(4) $\hat\lambda=\frac{1}{N}\sum_{i=1}^{N} v_{i}^{new}/v_{i}$

(5) $v=v^{new}$

(6) $v=v/||v||$


  1. Eq. (1) and (2) are basic online update formula for the first and second moment. See for instance this paper.

  2. For understanding Eq. (3), note que $Av=\lambda v$. Thus, the right side is original eigenvector multiplied by $\lambda$. If $A$ is a constant matrix, one can prove that it converges to the eigenvector associated with the largest eigenvalue.

  3. It is just an estimation of lambda. I average the division of each element of $\lambda v$ by each element of $v$, which is an estimation of $\lambda$.
  4. Eq. (6) is just to avoid numerical instability.
  5. Steps (3), (4), (5) and (6) are based on the basic algorithm to evaluate the numerical value of the largest eigenvalue of a matrix. See for instance Scientific Computing: An Introductory Survey - Michael T. Heath.

See the code below to note that it converges for the desired value. In this code I create a time series that come from a multinormal distribution with known mean and known covariance matrix with known eigenvalues and eigenvectors.

import numpy as np
from scipy.linalg import norm
def rotationMatrix(theta):
    return np.array([[np.cos(theta), -np.sin(theta)],[np.sin(theta), np.cos(theta)]])

def generateCovarianceMatrix(theta,lambda1,lambda2):
    eigMatrix=np.array([[lambda1, 0],[0, lambda2]])
    return Sigma

def generateData(mu,sigma):
    return np.random.multivariate_normal(mu, sigma, 1)    

if __name__=="__main__":
    mu = [1, 2]
    hatSigma=np.array([[1, 0],[0, 1]])
    hatLargeVector=[1,1] # eigenvector associated with the largest eigenvalue
    for i in range(T):
            hatSigma=np.array([[1, 0],[0, 1]])

        hatLargestLambda=np.sum([(hatLargeVector[i]+epsilon)/(oldHatLargeVector[i]+epsilon) for i in range(len(hatLargeVector))])/len(hatLargeVector)
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  • $\begingroup$ @Aksakal does this solution not work for you? It really works. $\endgroup$ – DanielTheRocketMan Mar 3 at 0:28

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