How many stocks are enough

Idea. Diversification removes the part of the risk that is specific to each asset, but not the part they share. The variance tends to the common covariance, and the question is how fast.

Step 1: the variance. With weights $\frac{1}{n}$, the variance is the average of all $n^2$ entries of the covariance matrix: $${\sigma }_n^2=\frac{n\cdot 1+n(n-1)\cdot \rho }{n^2}=\rho +\frac{1-\rho }{n}.$$

Step 2: the limit. As $n\to \infty$, ${\sigma }_n^2\to \rho =0.2$. The specific part, $\frac{1-\rho }{n}$, vanishes; the common part stays.

Step 3: the condition. We need $\frac{1-\rho }{n}\le 0.1\rho$: $$\frac{0.8}{n}\le 0.02⟺n\ge 40.$$ At $n=40$ the variance is exactly $0.2+0.02=0.22$, so the smallest $n$ is 40.

The trap. Dropping the factor $1-\rho$ and solving $\frac{1}{n}\le 0.02$ gives 50.

In volatility. The volatility of a large equal-weight portfolio tends to $\sigma \sqrt{\rho }$. With $\sigma =20\%$ and $\rho =0.3$, no amount of diversification brings it below about $11\%$.