Why Diversification Can Lower Minimum-Variance Portfolio Risk

# Why isn't the asset with minimum variance given a 100% portfolio weight?


# Why isn't the asset with minimum variance given a 100% portfolio weight?












The maximum expected return portfolio is the one that assigns a 100% weight to the asset with the highest expected return amongst all assets under consideration.

Shouldn't then the asset with the lowest variance in the candidate pool likewise be assigned a 100% weight in the minimum-variance portfolio, using the mean-variance model? Why not?

does it have to do with the different nature of the moments, or with the fact that portfolio variance is a function of covariance, whereas the portfolio return/mean is only a function of itself?

(What can be said about the asset that has maximum skew or minimum kurtosis in the max skewness or minimum kurtosis portfolios? Are they not weighted 100%)

## Answer by Kermittfrog (score 7, accepted)

https://quant.stackexchange.com/a/59699

Diversification is key.

The clear cut answer is diversification. A weighted combination of assets will more often than not show a lower return variance than even the asset with the lowest variance across the asset universe.

The setup

Without loss of generality, let us assume there exist two assets $a$ and $b$ with variance $\sigma_a^2=\alpha^2<\sigma_b^2=1$. These assets are correlated with parameter $\rho \in [0,1]$.

From basic portfolio theory we know that asset weights in the minimum-variance-portfolio are $$ w_{MVP}=\frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}} $$ with $\mathbf{1}$ a vector of ones. In our setup, $ \Sigma = \begin{pmatrix}\alpha^2 & \alpha\rho \\ \alpha\rho & 1\end{pmatrix} $ and thus the optimal weight on asset $a$ is

$$ w_a\equiv w_{MVP,a}=\frac{1-\alpha\rho}{1+\alpha^2-2\alpha\rho} $$

We can now answer some questions.

When is the weight on the asset with lower risk exactly equal to 100%?

The weight on asset $a$ is 100% if $$ \begin{align} 1&\stackrel{!}{=}w_a\\ &=\frac{1-\alpha\rho}{1+\alpha^2-2\alpha\rho}\\ \Rightarrow 1+\alpha^2-2\alpha\rho &= 1-\alpha\rho\\ \Rightarrow \alpha&=\rho \end{align} $$ i.e. when asset $a$'s volatility in relationship to asset $b$'s volatility (conveniently, $\alpha$) equals its correlation with asset $b$.

When will total variance be smaller than the smallest asset variance, $\sigma_a^2=\alpha^2$?

The total variance of the MVP equals $$ \sigma_{MVP}^2=\frac{1}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}}=\frac{\alpha^2(1-\rho^2)}{1+\alpha^2-2\alpha\rho} $$

It is smaller than the smallest asset variance $\sigma_a^2=\alpha^2$ if:

$$ \begin{align} \alpha^2 &\stackrel{!}{>} \sigma_{MVP}^2\\ &=\frac{\alpha^2(1-\rho^2)}{1+\alpha^2-2\alpha\rho}\\ \Rightarrow 1+\alpha^2-2\alpha\rho &> 1-\rho^2\\ \Rightarrow \alpha^2-2\alpha\rho+\rho^2 &> 0 \\ \Rightarrow \left(\alpha-\rho\right)^2 &> 0 \end{align} $$

The last expression is a quadratic form in $\alpha,\rho$. Hence, any combination of $\alpha,\rho$ with $\alpha \neq \rho$ will result in a decrease in total variance.

## Answer by Dimitri Vulis (score 1)

https://quant.stackexchange.com/a/59695

If you start out with a matrix specifying the covariance of every pair of assets, and an alpha for every asset (because people usually do), and define an objective function that maximizes the alpha and minimizes the variance of the portfolio, and run a quadratic optimizer, but don't specify a lot of constraints, then you may well end up with 100% of a long-only portfolio in a single asset with lowest variance or highest alpha. Or you may end up with -1000% of one asset and +1100% of another - a highly leveraged pair bet if you allow negative weights.

To avoid this, you specify linear constraints. For example, you might between 0% and 5% in each asset (long only), between 0% and 50% in each industry, etc.

In fact, it would be hard to get the optimizer to diversify without such linear constraints.

Sorry, I actually don't know how to use higher moments in portfolio optimization in classical Markowitz (modern portfolio theory). Maybe "even more modern" techniques do?