Portfolio Variance from a Sample Covariance Matrix

Summary

The document asks whether portfolio variance computed from a fixed vector of weights equals the quadratic form using the return covariance matrix, and whether the same identity holds with sample estimates. The answer applies linearity of expectation: for constant weights, the portfolio return's squared value can be written as a quadratic form in the return vector, with the weights factored outside the expectation. This yields portfolio variance as weights multiplied by the return covariance matrix and then by weights again.

The same algebra carries over to the sample covariance estimate when the variance and covariance are computed consistently from the same return observations and convention. The note gives a compact identity rather than a worked numerical example. It does not discuss estimated weights, degrees-of-freedom conventions, missing observations, or how estimation error in the covariance matrix affects portfolio construction.

Key ideas

• For fixed weights, portfolio variance is the quadratic form of the return covariance matrix.
• The weights can be factored outside the expectation because they are treated as constants.
• The sample portfolio variance identity holds when the portfolio and covariance estimates use consistent data and conventions.
• The identity does not address uncertainty in estimated weights or covariance-matrix estimation error.

Tags

# Sample Variance of Portfolio


# Sample Variance of Portfolio












Let $w$ denote a vector of portfolio weights, $r_i$ denote the $i$th return vector, $\Sigma$ denote the Covariance matrix of $r_i$ and let $\hat{\Sigma}$ denote the sample covariance matrix of $r_i$.

The portfolio variance is given by $$ \mathbf{Var}\left( w' r_i\right) = w' \mathbf{Var}\left( r_i\right) w = w' \Sigma w. $$ Does it hold for the sample portfolio variance that $$ \widehat{\mathbf{Var}}\left( w' r_i\right) = w' \widehat{\mathbf{Var}}\left( r_i\right) w = w' \hat{\Sigma} w? $$

## Answer by André Bittencourt (score 4, accepted)

https://quant.stackexchange.com/a/68840

Yes, indeed. It's a simple Linear Algebra and Expectation result:

Given:

$Var(w'r) = \mathbb{E}[(w'r)^2] = \mathbb{E}[(w'rr'w)]$

With $w$ and $r$ the vectors of weights and returns. As $w$ is constant, it holds:

$\mathbb{E}[w'rr'w] = w'\mathbb{E}[rr']w$

The sample variance, $\hat{\Sigma}$, is a estimator of for $\mathbb{E}[rr']$. Therefore, it holds what you said.

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