Deriving Equal-Weight Portfolio Variance from Mean Variance and Covariance
Summary
For an equally weighted portfolio of n shares, the document expands portfolio variance into diagonal variance terms and off-diagonal covariance terms. The diagonal contains n individual variances, so their contribution is 1/n times the mean variance across shares. The covariance matrix has n²−n off-diagonal entries when ordered pairs are counted, giving the covariance contribution a multiplier of (n²−n)/n², or 1−1/n.
Together these yield portfolio variance as 1/n times mean share variance plus (1−1/n) times mean covariance between distinct shares. The argument is a direct identity for equal weights; it does not assume that all shares have identical variances or covariances. The meaning of “mean covariance” matters: it averages ordered off-diagonal matrix entries. For a symmetric covariance matrix this equals the average over distinct unordered pairs, since each pair appears twice in the ordered sum.
Key ideas
- Expand portfolio variance as a double sum over all pairs of asset returns.
- The covariance matrix contributes n diagonal variance terms and n²−n off-diagonal terms.
- With equal weights, the variance component is 1/n times the average individual variance.
- The covariance component is weighted by 1−1/n times the mean covariance of distinct shares.
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# Show that the following result holds true for the variance of the return of a portfolio of shares
# Show that the following result holds true for the variance of the return of a portfolio of shares
Start with a portfolio $p$ of $n$ shares, each with weight $x_i = \dfrac{1}{n}$ (for $i$ ranging from $1$ to $n$, discretely). Its return is given by: $$R_p=x_1R_1+\ldots+x_nR_n=\sum_{i=1}^{n}=x_iR_i\tag{1}$$ Additionally, the variance of its return is given by: $$Var\left(R_p\right)=Cov\left(R_p, R_p\right)=Cov\left(\sum_{i=1}^n x_i R_i, R_p\right)=\sum_{i=1}^n x_i Cov \left(R_i, R_p\right)\tag{2}$$ Show that: $$Var (R_p)=\frac{1}{n}\times\text{mean variance of shares}+\left(1-\frac{1}{n}\right)\times\text{mean covariance between shares}\tag{3}$$
I tried to show that in the following way:
$$Var(R_p)=\sum_i\sum_jx_ix_j Cov(R_i, R_j)\\=\sum_{i=j=1}^n x_i^2 Var(R_i)+\sum_{i\neq j}x_i x_jCov\left(R_i, R_j\right) \\=\dfrac{1}{n^2}\sum_{i=j=1}^n Var(R_i)+\dfrac{1}{n^2}\sum_{i\neq j}Cov (R_i, R_j)\\=\underbrace{\dfrac{1}{n}\times\dfrac{1}{n}\sum_{i=j=1}^n Var(R_i)}_{\dfrac{1}{n}\times\text{mean variance of shares}}+\dfrac{1}{n^2}\sum_{i\neq j}Cov (R_i, R_j)$$ Hence, I might have managed to show the first part. Could you please confirm me that and, above all, give me some hint so as to show the second part as well? (I was thinking about the fact that so as to get to mean covariance between shares I have to compute the number of all the possible combinations of paired shares, given by $\dfrac{n\times (n-1)}{2}$; however, I am not sure that this is a good starting point). Thank you in advance for your precious support
## Answer by Pleb (score 4, accepted)
https://quant.stackexchange.com/a/65473
### The variance part is correct.
For the covariance part we can observe the following: There are $n$ variance terms in the $n \times n$ covariance matrix. This implies that there must be $n^2-n$ covariance terms (ie. lower and upper triangular matrix subtracted from the diagonal). As such, you can rewrite the last expression by dividing and multiplying with $n^2-n$:
\begin{align} \frac{1}{n^2} \sum_{i\neq j} Cov(R_i,R_j) &= \left(\frac{n^2-n}{n^2}\right) \cdot \left(\frac{1}{n^2-n}\sum_{i \neq j} Cov(R_i, R_j)\right)\\ &= \left(1- \frac{1}{n}\right) \cdot (\text{"mean covariance"}) \end{align}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.