Calculating Covariance Between a Weighted Index and an Asset

Summary

The document shows how to compute the covariance of a weighted return index with one of its component assets or with another asset. Using the linearity of covariance, the covariance between a weighted sum and a variable is the sum of each component’s covariance with that variable, multiplied by its corresponding weight.

For the index and a component asset, the calculation includes that component’s variance as its self-covariance, along with the cross-covariances between the other assets and the component. For an external asset, it uses the known covariance of each index constituent with that asset. Thus, given the weights and required covariance entries, both quantities can be computed directly. The explanation assumes compatible return definitions and periods and does not discuss estimation error, changing weights, or whether the weights are normalized; the stated weights sum to one but linearity itself does not require that condition.

Key ideas

• Covariance is linear in either argument when that argument is a weighted sum.
• The covariance of an index with a constituent combines the constituent’s variance and its covariances with the other holdings.
• The covariance of an index with an external asset is the weighted sum of constituent covariances with that asset.
• The calculation requires consistent return data, index weights, and the relevant covariance estimates.

Tags

# Combining covariances?


# Combining covariances?












Consider an economy with assets with return processes $A$, $B$, $C$, $D$. Consider a weighted index with return process $I=aA + bB + cC + dD$ where $a,b,c,d$ are coefficients, and $a+b+c+d = 1$.

Suppose I want to find $cov(I,A)$. Is this possible given that I know the covariance between all possible pairs of $A,B,C,D$?

Also, suppose I have some asset $E$. Suppose I know $cov(A,E),cov(B,E),cov(C,E),cov(D,E)$. How do I find $cov(I,E)$.

## Answer by Phil H (score 1, accepted)

https://quant.stackexchange.com/a/4263

Wikipedia gives:

$\sigma(x,y) = E[xy] - E[x]E[y]$

and

$\sigma(ax+by,cz) = ac\, \sigma(x,z) + bc\, \sigma(y,z)$

(paraphrasing the $\sigma(ax+by,cW+dV)$ rule).

So

$\sigma(I,A) = \sigma([aA+bB+cC+dD],A)$ $\sigma(I,A) = a\,\sigma(A,A) + b\,\sigma(B,A) + c\,\sigma(C,A) + d\,\sigma(D,A)$ $\sigma(I,A) = a\,\sigma^2(A) + b\,\sigma(B,A) + c\,\sigma(C,A) + d\,\sigma(D,A)$

Since you know the covariances between all the pairs and presumably the variance ($\sigma^2$) of A, you can thus calculate $\sigma(I,A)$.

The same holds for $\sigma(I,E)$, only you won't get a variance term: $\sigma(I,E) = a\,\sigma(A,E) + b\,\sigma(B,E) + c\,\sigma(C,E) + d\,\sigma(D,E)$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.