# How to deal with correlations not statistically significantly different from zero?

I often find parts of financial correlation matrices not statistically significantly different from zero. Sometimes, these correlations have a tangible effect on results - low correlations lead to high diversification benefit. Also, the sign of correlations can skew the results.

Would you consider it appropriate to make these correlations equal to zero to at least get rid of the directionality and, if necessary, fix the negative eigenvalues that might arise due to such corrections?

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I often find parts of financial correlation matrices not statistically significantly different from zero

You might want to be careful with that. First of all, statistically significant and large enough to be meaningful are not the same thing. Second, be careful that the test you use to assess the significance of the correlation coefficients do not assume the underlying data to be Gaussian because if you are dealing with financial series, they are most likely fat-tailed (and this is a case where it would make a difference, i.e. render your test void). Third, testing whether a single correlation coefficient is statistically significant is not the same as testing whether some of your correlation coefficients are statistically significant.

Also, the sign of correlations can skew the results.

Presumably, not if the correlations are small to begin with (i.e. if you only fudge with small correlation coefficients this should not be a concern).

see Edward's comment below.

fix the negative eigenvalues that might arise due to such corrections?

decreasing the absolute value of a correlation coefficient (as you consider doing) will never make a PSD matrix become indefinite (<=> it will not change the sign of any of your eigenvalues).

see Whuber's comment below.

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@ kwak - that you! The sign of a small correlation can become important in certain optimisation problems. For example, a small negative correlation (which, statistically might have been positive just as likely due to insignificance) can be combined with 2 large volatilities (and as a result large negative covariance), which will lead to misleading optimisation results. Had the sign been positive, the results would have been very different. Zero correlation would at least not give preference to either. –  Edward Sep 13 '10 at 7:34
By the way, I agree that the matrix might become indefinite. However, wouldn't it be feasible to simply make the negative eigenvalues = 0 and recalculate and rescale the matrix so that the diagonal values are 1? Example of the procedure can be found here: comisef.wikidot.com/tutorial:repairingcorrelation –  Edward Sep 13 '10 at 7:39
Edward:> your second point: i need to think about this but i have the impression that what you propose amounts to shrinking all your correlation coefficients by a small amount towards 0. –  user603 Sep 13 '10 at 10:57
@kwak: Have I misinterpreted that last statement? The eigenvalues of a 3 x 3 correlation matrix, for instance, are determined by three correlation coefficients (a,b,c). For (a,b,c) = (-0.9, 0.5, -0.1), the ev's are (2.07415, 0.91532, 0.0105346). Changing c from -0.1 to -0.01 changes the ev's to (2.03383, 0.99151, -0.0253402). Thus, a relatively small decrease in absolute value of a single coefficient can destroy the positive definiteness of the matrix. –  whuber Sep 13 '10 at 15:16
+1 counter example: i had assumed that making the ellipse more spherical was a transformation that preserved the convexity of the associated quadratic form. Corrected (i.e. strike in the response). –  user603 Sep 13 '10 at 16:19
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I am not sure if this is feasible in your context but one thing you can do to avoid these issues is to use bayesian estimation and to compute the expected values of different investing decisions based on the posterior distribution of the parameters.

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