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I'm trying to work through the process talked about in the article "Fixing a broken correlation matrix" by Phil Joubert and Stephen Langdell (found here https://www.nag.co.uk/IndustryArticles/fixing-a-broken-correlation-matrix.pdf), but I'm having trouble replicating the results in the 3x3 example they give. I'm attempting the "quick and dirty" method of patching up the broken matrix $C$ given by

\begin{bmatrix}1&0.95&0\\0.95&1&0.95\\0&0.95&1\end{bmatrix}

It says $C$ can be decomposed as $C = Q\Lambda Q^T$ where $Q$ is the matrix whose columns are eigenvectors of $C$ and $\Lambda$ is a matrix with the corresponding eigenvalues on the diagonal and zeros elsewhere. The article claims that we can get the "patched up" matrix $C'= Q\Lambda' Q^T$ where $\Lambda'$ is $\Lambda$, but with negative entries replaced by zero. However, I do not get the $C'$ they get given by

\begin{bmatrix}1&0.73454&0.07908\\0.73454&1&0.73454\\0.07908&0.73454&1\end{bmatrix}

The eigenvalues I get for $C$ are $\Lambda = \{-\frac{19\sqrt{2}-20}{20}, 1, \frac{19\sqrt{2}+20}{20}\}$ (these match the paper) and the corresponding eigenvectors (in matrix form, i.e. $Q$) are

\begin{bmatrix}1&1&1\\-\sqrt{2}&0&\sqrt{2}\\1&-1&1\end{bmatrix}

From what it sounds like in the article, I now need to replace the first eigenvalue with 0, and come up with $C'$ as described earlier where $\Lambda'$ is given by

\begin{bmatrix}0&0&0\\0&1&0\\0&0&\frac{19\sqrt{2}+20}{20}\end{bmatrix}

However, $C'= Q\Lambda' Q^T$ with these matrices does not give the $C'$ found in the article. Did I calculate my eigenvalues and eigenvectors incorrectly? At first I thought I needed to calculate a different eigenvector with $\Lambda_1 = 0$, but using that eigenvector did not give me their $C'$ either. Any help would be greatly appreciated!

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I do not yet see exactly, what happened here, but I'm able to reproduce the given matrix $C'$ First, I get the same eigenvalues by similarity rotations on C. But by rotations, the matrix $Q$ in my setting is a scaling of your $Q$. If you take $Q \cdot Q^t $ with your matrices, you don't get the identity matrix - but which is expected if $Q$ is a rotation-matrix/is orthonormal. So my matrix $R = \operatorname{norm_columns}(Q)$ which has the sum-of-squares over a column equalling $1$ does the job better. (my earlier remark $R=Q/2$ was wrong and based only on a too short view on the two matrices)

The next step is not yet sufficient: doing $S = R^t \cdot \Lambda' \cdot R$ gives now a covariance-matrix: $$ \text{ S =} \left[ \begin{array} {rrr} 1.09& 0.83& 0.09\\ 0.83& 1.17& 0.83\\ 0.09& 0.83& 1.09 \end{array} \right]$$ The final step could be the (standard) norming of a covariance to a corrleation-matrix: you put the diagonal elements $s_{k,k}$ of $S$ into the diagonal matrix $D = \operatorname{diag}(\{1/\sqrt{s_{k,k}} \}) $ and multiply $C' = D \cdot S \cdot D $ getting $$ \text{ C' =} \left[ \begin{array} {rrr} 1.000& 0.735& 0.079\\ 0.735& 1.000& 0.735\\ 0.079& 0.735& 1.000 \end{array} \right] $$ which looks trustworthy. So the final step in the described method was possibly forgotten to be written down...
[update] Well, I just looked into the article. There they write (on page 3)

(...) but with negative entries replaced by zero and scale the result to give a matrix C' (...)

Well, they mentioned it - but somehow cursory...

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  • $\begingroup$ Gottfried, it sounds like they leave a lot to the imagination in the article! Do you have any insight on how to come up with the scaling factor (2 in this case) other than guessing? $\endgroup$ – mjswartz Sep 7 '16 at 16:29
  • $\begingroup$ @mjswartz : (my earlier comment and my answer had a mistake) Your matrix Q does not fit the requirement of a rotation-matrix $R$ that $ R \cdot R^t = I$ But it can be reached by norming the columns such that the sum-of-squares along a column should equal 1. If you do that norming you get the correct result (notice, that eigenvectors at different eigenvalues are unique only up to scaling of the columns by some arbitrary scalar factor) I'll correct the faulty remark in my answer, too. $\endgroup$ – Gottfried Helms Sep 7 '16 at 21:16

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