# Standard deviation of large equal-weighted portfolios

Say I've got a portfolio of shares with the following parameters: Let $$n$$ be the number of shares in the portfolio, let $$\bar\sigma$$ be the average standard deviation (volatility/risk) for each share, let $$\bar\rho$$ be the average correlation between each pair of shares.

I would like to determine the portfolio's volatility. For $$n=1$$, this is simple, $$\bar\sigma=\sigma_P$$.

For $$n=50$$, I did the following, which I know is wrong, but I can't spot my mistake:

Let $$R_P$$ be the portfolio's return, $$R_i$$ an individual asset's return and let $$x_i$$ be an asset's weight in the portfolio.

$$Var(R_P) = Cov(R_P, R_P) = Cov(\Sigma x_iR_i, R_P)=\Sigma x_iCov(R_i, R_P)=\Sigma_i\Sigma_jx_ix_jCov(R_i, R_j)=\Sigma_i\Sigma_jx_ix_j\rho_{ij}\sigma_i\sigma_j.$$

Now, for equal-weighted portfolios, instead of $$x_i, x_j$$, we can write $$1/n$$; also, we can write $$\sigma_i=\sigma_j=\bar\sigma$$ and $$\rho_{i,j}=\bar\rho$$:

$$\Sigma_i\Sigma_j\frac{1}{n^2}\bar\rho\bar\sigma^2=\frac{1}{n^2}\bar\rho\bar\sigma^2n^2=\bar\rho\bar\sigma^2 = Var(R_P).$$

Thus, we'd have $$\sigma_P=\sqrt{\bar\rho\bar\sigma^2}$$. This is true for $$n=1$$ but it doesn't make sense for multiple shares, does it? I should have forgotten $$n$$ somewhere.

Why is this wrong?

Please don't be too harsh with me, I'm new to this topic.

• Is it behause $\rho_{ij}$ for $i=j$ is one which I forgot? Feb 21 at 23:30

$$Var(R_P)=\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n^2} Cov(R_i,R_j)$$
You then have to consider two cases $$i=j$$ and $$i \neq j$$. For $$i=j$$, there are $$n$$ terms of $$\frac{1}{n^2}\overline{\sigma}^2$$. For $$i\neq j$$, there are $$n(n-1)$$ terms of $$\frac{1}{n^2}\overline{\rho}\overline{\sigma}^2$$.
This results in $$Var(R_P)=\frac{1}{n}\overline{\sigma}^2+\frac{n-1}{n}\overline{\rho}\overline{\sigma}^2$$.