We try to analyze the average correlation of a portfolio as it can be found here in section 2 b), the same formula which is also used by the CBOE to calculate implied correlations:

$$ \rho_{av(2)} = \frac{\sigma^2 - \sum_{i=1}^N w_i^2\sigma_i^2}{2 \sum_{i=1}^N \sum_{j>i}^N w_i w_j \sigma_i \sigma_j} $$

EDIT:Assuming that $\sigma^2 = \sum_{i=1}^N \sum_{j=1}^N w_i w_j \sigma_i \sigma_j \rho_{i,j}$, where $\rho_{i,i}=1$, for $i=1,\ldots,N$, the above expression can be written as $$ \rho_{av(2)} = \frac{\sum_{i=1}^N \sum_{j>i}^N w_i w_j \sigma_i \sigma_j \rho_{i,j}}{\sum_{i=1}^N \sum_{j>i}^N w_i w_j \sigma_i \sigma_j}. $$

The following questions arise.

  1. Assuming that $w_i \in \mathbb{R}$, i.e. long/short leverage is allowed, is it possible that $|\rho_{av(2)}|>1 $ ? Note that we don't assume $\sum w_i=1$.
  2. Does there already exist the notion of contribution to average correlation? Meaning that e.g. in a long/short portfolio, where average correlation should be close to zero, I can identify positions that drive the average correlation up (in absolute value).
  • $\begingroup$ For the first question, I tried some extreme weights (adding up to 500% or 0%) and did not see any incorrect average correlations (I may have used a modification of that formula because that didn't seem correct). The main reason is that so long as $\sigma^{2}$ is the correct portfolio variance, then the average correlation should be within normal bounds. For the second point, I'm not familiar with anyone writing about it, but you could presumably take a similar approach using the above formula as they do for contribution to variance (I would not take the absolute value). $\endgroup$
    – John
    Aug 8, 2013 at 17:58

2 Answers 2


Let's start by replacing $\sigma$ by its estimator formula $\sigma^2=\frac{1}{n}\sum^n_{i=1}(x_i-\mu)^2$. Now, by replacing $\mu$ by its estimator $\mu=\frac{1}{n}\sum^n_{i=1}x_i$ in the formula for the variance we obtain:


For the individual asset, the variance will write $\sigma^2_s=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}(x_{s,i}-x_{s,j})^2$, $s=1,2,...,N$. For the portfolio, we can denote the observations by $y_i=\sum_{s=1}^N w_sx_{s,i}$, and so the variance of the portfolio writes

$\sigma^2=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}(y_i-y_j)^2=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}(\sum_{s=1}^N w_sx_{s,i}-\sum_{s=1}^N w_sx_{s,j})^2$

Now, feeding this into your formula we get on the numerator:

$\sigma^2-\sum_{s=1}^N w^2_s\sigma_s^2=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}(\sum_{s=1}^N w_sx_{s,i}-\sum_{s=1}^N w^2_sx_{s,j})^2-\sum_{s=1}^N w_s\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}(x_{s,i}-x_{s,j})^2=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}[(\sum_{s=1}^N w_s(x_{s,i}-x_{s,j}))^2-\sum_{s=1}^N w^2_s(x_{s,i}-x_{s,j})^2]=$

$=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}[\sum_{s=1}^N w_s^2(x_{s,i}-x_{s,j})^2+2\sum^N_{s=1}\sum^N_{t>1} w_s w_t(x_{s,i}-x_{s,j})(x_{t,i}-x_{t,j})-\sum_{s=1}^N w^2_s(x_{s,i}-x_{s,j})^2] $

$=\frac{1}{n^2}\sum^n_{j=1}\sum^n_{i=1}\sum^N_{s=1}\sum^N_{t>1} w_s w_t(x_{s,i}-x_{s,j})(x_{t,i}-x_{t,j})$

On the denominator you have :

$2\sum^N_{s=1}\sum^N_{t>1} w_s w_t\sigma_s\sigma_t=\frac{1}{n^2}\sum^N_{s=1}\sum^N_{t>1} w_s w_t\sqrt{\sum^n_{j=1}\sum^n_{i=1}(x_{s,i}-x_{s,j})^2\sum^n_{j=1}\sum^n_{i=1}(x_{t,i}-x_{t,j})^2}$.

The fraction looks like :

$\rho=\frac{\sum^N_{s=1}\sum^N_{t>1} w_s w_t\sum^n_{j=1}\sum^n_{i=1}(x_{s,i}-x_{s,j})(x_{t,i}-x_{t,j})}{\sum^N_{s=1}\sum^N_{t>1} w_s w_t\sqrt{\sum^n_{j=1}\sum^n_{i=1}(x_{s,i}-x_{s,j})^2(x_{t,i}-x_{t,j})^2}}=\frac{\sum^N_{s=1}\sum^N_{t>1}A}{\sum^N_{s=1}\sum^N_{t>1}\sqrt{B}}$

Now we look at the relation between A and B, by Cauchy-Schwarz inequality $A^2\leq B$ which translates into $|\rho|\leq1$. Hopefully I didn't make to many mistakes...

  • $\begingroup$ Thanks for your detailed calculation. Unfortunately I had a typo up there. The second sum in the denominator should not be with $j>1$ but $j>i$. I try to check if your calculations still hold. However I prefer to work with probabilistic expressions (expectations, variances) instead if the statistical ones. This should reduce the number of sums and still yield the result. $\endgroup$
    – Richi Wa
    Sep 19, 2013 at 14:56
  • $\begingroup$ I am sorry but I can not follow all these sums. I really appreciate your efforts to do all those calculations, but using the statistical estimators with all these sums does not make the picture clearer to me. Below in our answer you find a counter example ... $\endgroup$
    – Richi Wa
    Sep 19, 2013 at 15:40

I did some calculations in mathematica in the 3 asset case. Assume we have exposures $w_i,i=1,2,3$ and volatilities $\sigma_i,i=1,2,3$ and correlations $\rho_{1,2},\rho_{1,3},\rho_{2,3}$. Let's assume $\sigma_1=\sigma_2=\sigma_3=\sigma$ for some arbitrary positive $\sigma$. For the weights we assume $w_2=w_3 = 0.5$ and we have a short in asset 1 of $w_1 = -0.5$. Then the above formula becomes $$ \rho_{av(2)} = \rho_{1,2}+\rho_{1,3}-\rho_{2,3}. $$ Then the question is whether we can find valid (pos.definite correlation matrix) values for the correlations such that the above formula delivers a results out side of the unit interval. A possible choice is $\rho_{1,2}=0.95, \rho_{1,3}=0.95$ and $\rho_{2,3}=0.89$ with the result $1.01$!

The mathematica code is the following:

pfvar[w1_, w2_, w3_] := w1^2*[Sigma]1^2 + w2^2*[Sigma]2^2 + w3^2*[Sigma]3^2 + 2*([Sigma]1*[Sigma]2*[Rho]12*w1*w2 + [Sigma]1*[Sigma]3*[Rho]13*w1*w3 + [Sigma]3*[Sigma]2*[Rho]23*w3*w2)

impliedCorr[w1_, w2_, w3_] := (pfvar[w1, w2, w3] - (w1^2*\[Sigma]1^2 + w2^2*\[Sigma]2^2 
   + w3^2*\[Sigma]3^2))/(  2*(\[Sigma]1*\[Sigma]2*w1*w2 + \[Sigma]1*\[Sigma]3*w1*
       w3 + \[Sigma]3*\[Sigma]2*w3*w2)  )

impliedCorr[w1, w2, w3] /. w2 -> w3 /. [Sigma]2 -> [Sigma]3 /. [Sigma]3 -> [Sigma]1 /. w3 -> 0.5 /. w1 -> -0.5 // Simplify

[Rho]12 + [Rho]13 - [Rho]23 /. [Rho]12 -> 0.95 /. [Rho]13 -> 0.95 /. [Rho]23 -> 0.89

EDIT: Thanks to @John I found a mistake and corrected $\rho_{2,3}$ to $0.89$.

  • $\begingroup$ You have $w_2 = w_3 = 0.5$ and $w_3 = -0.5$, this does not seem right to me. $\endgroup$
    – Bob Jansen
    Aug 9, 2013 at 9:31
  • $\begingroup$ You are right - a typo and I make the code shorter, thanks! $\endgroup$
    – Richi Wa
    Aug 9, 2013 at 9:43
  • $\begingroup$ $w_1=-0.5$ is correct. $\endgroup$
    – Richi Wa
    Aug 9, 2013 at 9:48
  • $\begingroup$ I don't think it matters for the general argument but you don't have $w'\iota = 1$? $\endgroup$
    – Bob Jansen
    Aug 9, 2013 at 11:07
  • 1
    $\begingroup$ When I find the eigenvalues for your proposed correlation matrix, I get a negative value for the smallest. This suggests your choice is not positive definite and thus not a valid correlation matrix. Setting $\rho_{2,3}=0.89$ would result in all positive eigenvalues, but I can confirm that the average correlation by this formula would be greater than 1. So your problem remains. $\endgroup$
    – John
    Aug 9, 2013 at 14:56

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