Beta-Neutral Portfolio Variance

Worked Solution

How to Think About It: This is a constrained portfolio construction problem that every quant PM faces daily. You have two alpha strategies and you want to combine them so that the portfolio has zero market exposure. The tradeoff is straightforward: beta-neutrality removes a degree of freedom, so you cannot simultaneously minimize variance and zero out beta unless you are lucky. The question is how much variance you pay for the hedge.

Quick Estimate: If $\beta_1 = 1$ and $\beta_2 = -1$ (one long-biased, one short-biased strategy), the beta-neutral constraint gives $w_1 = w_2 = 1/2$ -- equal weight. If instead $\beta_1 = 1.5$ and $\beta_2 = 0.5$, the constraint forces $w_1 = -0.5$ and $w_2 = 1.5$, requiring leverage. The general solution follows.

Formal Solution:

Part (a): We have two linear constraints: $w_1 + w_2 = 1$ $w_1 \beta_1 + w_2 \beta_2 = 0$

Substitute $w_2 = 1 - w_1$ into the second equation: $w_1 \beta_1 + (1 - w_1) \beta_2 = 0$ $w_1 (\beta_1 - \beta_2) = -\beta_2$

$\boxed{w_1 = \frac{-\beta_2}{\beta_1 - \beta_2}, \quad w_2 = \frac{\beta_1}{\beta_1 - \beta_2}}$

This requires $\beta_1 \neq \beta_2$ (otherwise the system is either impossible or under-determined). Note: if both betas are positive (common for long-only strategies), then $w_1 < 0$ and $w_2 > 1$ -- you must short strategy 1 and lever up strategy 2, or vice versa.

Part (b): The portfolio variance is: $\text{Var}(R_p) = w_1^2 \sigma_1^2 + w_2^2 \sigma_2^2 + 2 w_1 w_2 \rho \sigma_1 \sigma_2$

Substituting the weights from part (a) with $\Delta\beta = \beta_1 - \beta_2$:

$\boxed{\text{Var}(R_p) = \frac{\beta_2^2 \sigma_1^2 + \beta_1^2 \sigma_2^2 - 2 \beta_1 \beta_2 \rho \sigma_1 \sigma_2}{(\beta_1 - \beta_2)^2}}$

Part (c): The unconstrained minimum-variance portfolio minimizes $w_1^2 \sigma_1^2 + (1-w_1)^2 \sigma_2^2 + 2 w_1(1-w_1) \rho \sigma_1 \sigma_2$ over $w_1$. Taking the derivative and setting it to zero:

$w_1^{\text{mv}} = \frac{\sigma_2^2 - \rho \sigma_1 \sigma_2}{\sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2}$

The beta-neutral portfolio has higher variance than the minimum-variance portfolio whenever the beta-neutral weights $w_1^{\text{bn}}$ differ from the minimum-variance weights $w_1^{\text{mv}}$. Since adding any binding constraint to an optimization problem can only increase (or maintain) the objective, beta-neutrality increases variance whenever:

$\frac{-\beta_2}{\beta_1 - \beta_2} \neq \frac{\sigma_2^2 - \rho \sigma_1 \sigma_2}{\sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2}$

Rearranging, beta-neutrality is "free" (does not increase variance) only when the minimum-variance portfolio happens to already be beta-neutral. This requires:

$\frac{\beta_1}{\beta_2} = -\frac{\sigma_2^2 - \rho \sigma_1 \sigma_2}{\sigma_1^2 - \rho \sigma_1 \sigma_2} = -\frac{\sigma_2(\sigma_2 - \rho \sigma_1)}{\sigma_1(\sigma_1 - \rho \sigma_2)}$

In general this is a knife-edge condition. For almost all parameter configurations, enforcing beta-neutrality strictly increases portfolio variance. The variance penalty is larger when: - The betas are similar in magnitude and sign (forcing extreme leverage) - The strategies are highly correlated (less diversification benefit to offset the constraint) - The unconstrained min-variance weights are far from the beta-neutral weights

Answer: $w_1 = -\beta_2/(\beta_1 - \beta_2)$, $w_2 = \beta_1/(\beta_1 - \beta_2)$. Portfolio variance is $(\beta_2^2 \sigma_1^2 + \beta_1^2 \sigma_2^2 - 2\beta_1 \beta_2 \rho \sigma_1 \sigma_2)/(\beta_1 - \beta_2)^2$. Beta-neutrality increases variance whenever the beta-neutral weights differ from the unconstrained minimum-variance weights -- which is generically always the case unless the ratio $\beta_1/\beta_2$ satisfies a specific relationship involving $\sigma_1, \sigma_2, \rho$.