Finding Minimum-Variance Portfolios from a Covariance Matrix
Summary
The document asks whether a covariance matrix alone is enough to identify portfolios on the minimum-variance line at a specified volatility. It describes the variance as a quadratic form in portfolio weights and imposes the usual constraint that weights sum to one. In the example, the questioner uses Lagrange conditions to find a relationship between weights, then combines that relationship with the target variance to obtain two portfolios.
The answer explains that the frontier is a hyperbola arising from a quadratic optimization problem and that two points are needed to define it. The example illustrates why selected first-order conditions can help locate frontier portfolios, but it does not explain the derivation in detail or establish that this shortcut works generally. The result also depends on the full setup, including the budget constraint and the covariance matrix; expected returns would be needed to describe the usual risk-return frontier, rather than a variance-only locus.
Key ideas
- Portfolio variance can be expressed as a quadratic form using the covariance matrix and portfolio weights.
- The example assumes that portfolio weights sum to one.
- The questioner combines a first-order relationship among weights with a target variance to find candidate portfolios.
- The answer characterizes the frontier as a hyperbola and says two points are needed to define it.
- The brief answer does not fully derive when the shortcut applies.
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Full text
# is it possible to get minimum variance line having only covariance matrix?
# is it possible to get minimum variance line having only covariance matrix?
Hey I have covariance matrix:
$$C=\begin{pmatrix} 0,01 & 0.01 & 0\\ \\ 0.01 & 0,02 & -0.01 \\ \\ 0 & -0.01 & 0,03 \end{pmatrix}$$
So the variance of porfolio is:
$$\sigma_w^2=\begin{pmatrix}w_1 & w_2 & w_3 \end{pmatrix} \begin{pmatrix}0.01 & 0.01 & 0\\0.01 & 0.02 & -0.01\\0 & -0.01 & 0.03 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix}=0.01 w_1^2+ 0.02 w_2^2 + 0.03 w_3^2 + 0.02 w_1 w_2-0.02w_2w_3$$
And i want to find portoflios on MVL wchich standard deviation is $$\sigma_w=\frac{1}{10}$$. IS it possible to find this porfolios having only this information? !
EDIT:
I tried to do it using the Lagrange multiplier method, which gave the same result, I don't know why it works :( Here's what I do: I use lagrange multipliers to minimalize function $f(w_1,w_2,w_3)=0.01 w_1^2+ 0.02 w_2^2 + 0.03 w_3^2 + 0.02 w_1 w_2-0.02w_2w_3$ under condition $g_1(w_1,w_2,w_3)=w_1+w_2+w_3-1=0$ I get equation system $\left \{\begin{array}{lr}0.02w_1+0.02w_2-\lambda=0\\0.04w_2+0.02w_1-0.02w_3-\lambda=0 \\0.06w_3-0.02w_2-\lambda=0 \\w_1+w_2+w_3-1=0 \\\end{array} \right.$ I know that if I solved it to the end I would get a wallet with minimal variance, but I don't want to do it, so not using the equation of 3 gets: $-0.02w_2+0.02w_3=0 \Rightarrow w_2=w_3$ (from 1 and 2) and $w_1=1-2w_2$ from 4 now I put these for equation $\sigma_{w}^{2}=0.01 w_1^2+ 0.02 w_2^2 + 0.03 w_3^2 + 0.02 w_1 w_2-0.02w_2w_3$ where $ \sigma_{w}= \frac{1}{10}$ and solve it $0.01=0.01(1-2w_2)^2+0.02w_2^2+0.03w_2^2+0.02w_2(1-2w_2)-0.02w_2^2$ Finally I get 2 solutions $w_1=(-\frac{1}{3},\frac{2}{3},\frac{2}{3})$ and $w_2=(1,0,0)$ which are correct.My question is why it works? Can anyone explain me?
I didn't solve the equations to the end, I just determined relationships between the weights and just put in a specific variance. Why can I do it, and if I can't, why did I get a good result?
## Answer by Kermittfrog (score 1)
https://quant.stackexchange.com/a/53840
The Frontier is a hyperbola (it’s underlying problem is a quadratic one). To fully define it, we need at least two of its points.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.