Under which conditions the minimum variance portfolio involves no short selling?

Question

If $\rho_{12} < 1$ or $\sigma_1 \not= \sigma_2$ then $\sigma_v^2$ representing the variance of the portfolio with weights $(w_1, w_2) = (s, 1-s)$ as a function of $s$ attains its minimum value at: $$ s_0 = \frac{\sigma_2^2 - \sigma_1\sigma_2\rho_{12}}{\sigma_1^2+\sigma_2^2-2\sigma_1\sigma_2\rho_{12}} $$ Under which conditions on $\sigma_1$, $\sigma_2$, and $\rho_{12}$ does the minimum variance portfolio involve no short selling?

$\rho$ is correlation coefficient and $\sigma$ is standard deviation. Squared is variance. I'm not sure what this question means.

Answer 1

There are two conditions: $W_1=s_0$ has to be non-negative, which means $\sigma_2^2 - \sigma_1\sigma_2\rho \ge 0$, which simplifies to $\sigma_2 \ge \sigma_1 \rho$. (I assumed $\sigma_2 \ne 0$).

The second condition is that $W_2=1-s_0$ also has to be non-negative, i.e. $s_0 \le 1$. So $\sigma_2^2 - \sigma_1\sigma_2\rho \le \sigma_1^2+\sigma_2^2-2\sigma_1\sigma_2\rho$. Which reduces to $\sigma_1 \ge \sigma_2 \rho$. (Again assuming $\sigma_1 \ne 0$).

The two conditions are nicely symmetric, and can be combined into the following statement

$ \rho\le \frac{\sigma_2}{\sigma_1}, \rho\le \frac{\sigma_1}{\sigma_2} $

(Only one of these two conditions, the one with the lower ratio, will be binding. We could say $\rho \le \frac{\sigma_{small}}{\sigma_{big}}$ once we know which vol is bigger and which is smaller).

Answer 2

when the numerator is positive, then you get rho.