The minimum-variance portfolio

Allocation as a bead on a rail; short selling as a repulsive spring.

Harry Markowitz set out mean-variance portfolio selection in 1952, in “Portfolio Selection” in the Journal of Finance. What follows is that idea in one dimension. Hold two assets, fully invested. Put $w$ in asset 1 and $1-w$ in asset 2, with volatilities $\sigma_1, \sigma_2$ and covariance $c = \rho\,\sigma_1\sigma_2$. Half the portfolio variance splits into two spring energies plus a constant that no force ever sees:

$$ \tfrac12\,\mathrm{Var}(w) \;=\; \tfrac12\,(\sigma_1^2 - c)\,w^2 \;+\; \tfrac12\,(\sigma_2^2 - c)\,(1-w)^2 \;+\; \tfrac{c}{2}. $$

So the fully-invested portfolio is a bead on the allocation rail. It is tied to the corner portfolio “all asset 2” ($w=0$) by a spring of stiffness $\sigma_1^2 - c$, and to “all asset 1” ($w=1$) by one of stiffness $\sigma_2^2 - c$. Release it and it settles at the global minimum-variance weight $w^\ast = (\sigma_2^2 - c)\,/\,(\sigma_1^2 + \sigma_2^2 - 2c)$, the bottom of the variance bowl. Note which way the springs pull. Asset $i$'s variance is the stiffness dragging the bead away from it. Risk repels. With uncorrelated assets the equilibrium is $w^\ast = \sigma_2^2/(\sigma_1^2+\sigma_2^2)$, so the weights go as $1/\sigma_i^2$. That is inverse-variance weighting, the centre-of-mass rule yet again.

Now raise the correlation. Both springs slacken together. When $c > \sigma_2^2$, which is $\rho > \sigma_2/\sigma_1$, the spring tied to the riskier corner turns repulsive. Its stiffness is negative, and it is drawn red. The equilibrium is pushed out of $[0,1]$ into the shaded zone. There the minimum-variance portfolio shorts the riskier asset. Short selling needs no special rule here. It is what a repulsive spring does. The total stiffness $\sigma_1^2+\sigma_2^2-2c = \mathrm{Var}(r_1 - r_2)$ stays positive, so the bead always settles.

vol $\sigma_1$ vol $\sigma_2$ correlation $\rho$

$w^\ast$ / min vol stiffnesses ,

Two deeper connections. First, the minimum-variance weights are $w^\ast \propto \Sigma^{-1}\mathbf{1}$, the precision row sums. That makes the minimum-variance portfolio the GLS estimator of a common mean. It uses the same masses as the bid-ask bounce demo, wearing a finance hat. Second, for many assets the corner portfolios become blocks. Eliminating a block is a Schur complement. Damping that complement interpolates hierarchical and minimum-variance allocation. That is the γ-bridge of schur.microprediction.org, whose scalar shadow is the Schur damping demo.

The usual explanation the textbook derivation, in full

For many assets, collect the weights into a vector $w$ and the risk into a covariance matrix $\Sigma$. The portfolio variance is $w^{\mathsf T}\Sigma\,w$, and being fully invested means the weights sum to one. Minimise the variance under that budget:

$$ \min_{w}\; \tfrac12\,w^{\mathsf T}\Sigma\,w \quad\text{subject to}\quad \mathbf{1}^{\mathsf T} w = 1. $$

Attach a Lagrange multiplier $\lambda$ to the constraint and set the gradient with respect to $w$ to zero:

$$ \mathcal{L}(w,\lambda) = \tfrac12\,w^{\mathsf T}\Sigma\,w - \lambda\big(\mathbf{1}^{\mathsf T} w - 1\big), \qquad \Sigma w - \lambda\,\mathbf{1} = 0. $$

So the optimal weights are proportional to $\Sigma^{-1}\mathbf{1}$. Fix the multiplier by feeding this back into the budget constraint:

$$ w = \lambda\,\Sigma^{-1}\mathbf{1}, \qquad \mathbf{1}^{\mathsf T} w = \lambda\,\mathbf{1}^{\mathsf T}\Sigma^{-1}\mathbf{1} = 1. $$

Solve for $\lambda$ and substitute. The minimum-variance portfolio is the normalised precision row sum:

$$ w^\ast = \frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^{\mathsf T}\Sigma^{-1}\mathbf{1}}. $$

With two assets $\Sigma^{-1}$ has row sums $\sigma_2^2 - c$ and $\sigma_1^2 - c$ over the common determinant, and the ratio collapses to $w^\ast = (\sigma_2^2 - c)\,/\,(\sigma_1^2 + \sigma_2^2 - 2c)$, the weight on the rail.

The physics proof the same result, read off the rail

The bead already knows this. It sits on the rail tied to the two corner portfolios by springs of stiffness $k_1 = \sigma_1^2 - c$ and $k_2 = \sigma_2^2 - c$. A bead between two springs settles at their centre of mass, the rule the fusion demo proved once and for all:

$$ w^\ast = \frac{k_1\cdot 0 + k_2\cdot 1}{k_1 + k_2} = \frac{\sigma_2^2 - c}{\sigma_1^2 + \sigma_2^2 - 2c}. $$

Those stiffnesses are proportional to the rows of the precision matrix $\Sigma^{-1}$, so the centre of mass is the precision row sum $\Sigma^{-1}\mathbf{1}$ normalised, the minimum-variance weight itself. No multiplier, no inverse spelled out. Risk repels, and the bead finds the quiet spot on its own.