How Diversification and Correlation Shape the Minimum-Variance Frontier

# In Mean-Variance Analysis, why not the efficient frontier being pushed to the left near the axis?


# In Mean-Variance Analysis, why not the efficient frontier being pushed to the left near the axis?












I took some classes in portfolio theory, and learnt the Markowitz Mean-Variance Analysis. If only two risky assets, the efficient frontier would be a hyperbola passing through the two points; now if added another asset has some correlation with them two, the efficient frontier would be pushed to the left of these three risky assets, generally not passing through them; so, if we keep adding more assets, it seems intuitively that the efficient frontier would be pushed further to the left of the assets points; wouldn't the efficient frontier be like tangent to the y-axis(y axis is for return), causing the standard deviation to be almost zero? I also read Pennacchi's book: Theory of Asset Pricing. it derived the equation for the minimum variance set using linear algebra. And found the position for the global minimum standard deviation point. And it's positive. I understand the equation derivation, but didn't get through the problem as I just thought of. Any help? Much appreciated.

## Answer by Fab (score 3, accepted)

https://quant.stackexchange.com/a/70865

You have a good point there. Adding an asset will not increase the minimum variance. If the assets are uncorrelated, the minimum will decrease (move left) towards zero. If they're positively correlated, it will decrease towards some minimum above zero.

Consider the case of uncorrelated assets, $N$ of them, and for simplicity assume they all have the same variance $\sigma^2$. Now take an equally weighted portfolio $p = \sum_i \frac 1N X_i$, and remember that Var$[aX+bY] = a^2 \sigma^2_X + 2ab \sigma_X \sigma_Y \rho_{XY} + b^2 \sigma^2_Y$.

Then that portfolio has variance:

$\sigma_p^2 = \sum_i \sum_j \frac 1{N^2} \sigma_i \sigma_j \rho_{ij}$, but the correlation off the diagonals are zero, and you're left with

$\sigma_p^2 = \sum_i \frac 1{N^2} \sigma_i^2 = \frac N{N^2} \sigma_i^2 = \sigma_i^2/N.$

Thus, this portfolio has vol $\sigma/\sqrt N$, which is indeed falling towards zero as you increase $N$.

So, yes, when plotting expected return vs standard deviation of return, and looking at the frontier (minimum standard deviation among them), that is a hyperbola (also for N>2), and its vertex will indeed get closer and closer to "the left" towards zero as you add more assets, if they're sufficiently uncorrelated.

That's the benefit of diversification for you!

Second case: Assume instead that all assets have a constant (or "average") correlation $\rho$ with each other. Then

$\sigma_p^2 = \sum_i \sum_j \frac 1{N^2} \sigma_i \sigma_j \rho_{ij}$, and, for N big enough, we can basically ignore the diagonal, so $\sigma_p^2 \approx \frac {N^2}{N^2} \rho \sigma^2$, or

$\sigma_p \approx \sqrt \rho \sigma.$

So, as you add assets, the vol won't fall to zero, but to $\sqrt \rho \sigma$, which gives you another useful rule of thumb:

`The vol of an index is "square root of the average correlation" times "average vol" of the constituents.`

## Answer by Arne (score 2)

https://quant.stackexchange.com/a/8967

In mean-variance analysis, you combine different assets to minimize variance and maximize expected return. The hyperbola is not a function of the number of assets, but of their mean and variance. If the efficient frontier where a tangent to the y-axis (which can't be) or nearly a tangent, that would mean you would have almost zero portfolio-variance, which won't be the case for risky assets, and clearly not so by adding further assets with variance >0.