Average Correlation with Long-Short Portfolio Weights

# Average correlation of index/portfolio


# Average correlation of index/portfolio












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.

- 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 $\sum w_i=1$.

- 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).

## Answer by Richi Wa (score 3, accepted)

https://quant.stackexchange.com/a/8702

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 $0.89$.

## Answer by KAT (score 4)

https://quant.stackexchange.com/a/8711

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:

$\sigma^2=\frac{1}{2n^2}\sum^n_{j=1}\sum^n_{i=1}(x_i-x_j)^2$.

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...