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Portfolio Variance as Weighted Covariance with the Portfolio

Article Quant Q&A · Author: Alchemy

Summary

The document proves an identity for the variance of a portfolio return. Starting from the portfolio return as a weighted sum of constituent returns, expand its variance into a double sum of weighted pairwise covariances. Covariance is linear in either argument, so the inner sum can be collected into the portfolio return itself. This gives portfolio variance as the sum, across assets, of each asset’s weight multiplied by its covariance with the portfolio.

The result applies to any portfolio, not only the market portfolio. The questioner’s attempted derivation goes wrong by treating the variance of a weighted sum as though it were the variance of one weighted constituent; the answer instead uses covariance linearity across the entire sum. The document also mentions expressing constituent covariance with the market in beta form, but does not develop the beta definition or its assumptions. The identity is algebraic and does not depend on a particular return distribution.

Key ideas

  • A portfolio return is the weighted sum of its constituent returns.
  • The variance of a weighted sum expands into weighted pairwise covariances.
  • Covariance is linear in its first argument, allowing the inner sum to be rewritten as covariance with the portfolio return.
  • Portfolio variance equals the sum of each constituent’s weight times its covariance with the portfolio.
  • The identity holds for any portfolio, not only the market portfolio.

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# Answer by Alex C (score 1)


# Show that the variance of the market portfolio is the weighted average of the ovariances between each constituent and the market portfolio itself












Let us assume that the market portfolio consists of n assets. Given that the return of the market portfolio can be written as $r_m = \sum_{j=1}^{n} w_jr_j$, we have that $\sigma^2_m = E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2$, but how do I show that $$E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2 = \sum_{j=1}^{n} w_jCov(r_j,r_m)$$? If I show that the equation above is true, than I can claim that $$E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2 = \sum_{j=1}^{n} w_jCov(r_j,r_m) = \sum_{j=1}^{n} w_j\beta\sigma^2_m$$

This is how I am trying to prove the result. We know that:

$$\sigma^2_m = E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2= E[(w_jr_j)^2]-E^2[w_jr_j]$$

Accordingly, we may show that:

$$\sum_{j=1}^{n} w_jCov(r_j,r_m) = E[(w_jr_j)^2]-E^2[w_jr_j]$$

Now: $\sum_{j=1}^{n} w_jCov(r_j,r_m) = \sum_{j=1}^{n} w_jE[r_jr_m]-\sum_{j=1}^{n} w_jE[r_j]E[r_m]=\sum_{j=1}^{n} w_jE[r_j\sum_{j=1}^{n} w_jr_j]-\sum_{j=1}^{n} w_jE[r_j]E[\sum_{j=1}^{n} w_jr_j]=\sum_{j=1}^{n} w_jE[\sum_{j=1}^{n} w_jr_j^2]-E[\sum_{j=1}^{n} w_jr_j]E[\sum_{j=1}^{n} w_jr_j]$

It looks like I am not able to prove the result because $$\sum_{j=1}^{n} w_jE[\sum_{j=1}^{n} w_jr_j^2] \neq E[(w_jr_j)^2]$$

Can you help me, please?

## Answer by Alex C (score 1)

https://quant.stackexchange.com/a/46466

The variance of the portfolio is $$ V_p=\sum_i \sum_j w_i w_j Cov(r_i,r_j)$$

because of the properties of $Cov(\cdot,\cdot)$ (namely that Covariance is linear with respect to the second argument: $Cov(x,\alpha y+\beta y)=\alpha Cov(x,y)+\beta Cov(x,z)$ we can rewrite this as $$ V_p=\sum_i w_i Cov(r_i,\underbrace{\sum_j w_j r_j}_{R_P})$$ where $R_p$ is the return on the portfolio. So we have

$$ V_p=\sum_i w_i Cov(r_i,R_p)$$ QED

(And it is true of every portfolio, not just the market portfolio).

## Answer by Hunaphu (score 1)

https://quant.stackexchange.com/a/73544

$$ V[r_m] = Cov(r_m, r_m) = Cov(\sum w_j r_j, r_m) = \sum_j w_j Cov(r_j, r_m). $$ You need the linearity of cov: $$ \begin{align} C[aX + bY, Z] &= E[(aX + bY - am_X - bm_Y)(Z-m_Z)]\\ &= E[(aX - am_X)(Z-m_Z)] + E[(bY - bm_Y)(Z-m_Z)]\\ &= aC(X, Z) + bC(Y,Z). \end{align} $$

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