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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? Minimum variance line!

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?

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    $\begingroup$ What is Minimum Variance Line? Minimum variance is usually achieved at a single point at the bottom of a en.wikipedia.org/wiki/File:Paraboloid_of_Revolution.svg . Please explain better what you are trying to do. $\endgroup$
    – nbbo2
    May 3, 2020 at 22:41
  • $\begingroup$ I add picture of Minimum variance Frontier. Its a line with the smallest standard deviation on each level of expected return $\endgroup$
    – Mr.Price
    May 3, 2020 at 23:02
  • $\begingroup$ OK you mean Minimum Variance Frontier, that requires knowledge of expected returns $\mu$, covariance matrix $C$ is not sufficient. With only $C$ you could find GMVP point shown in your pic, not the rest. $\endgroup$
    – nbbo2
    May 3, 2020 at 23:04
  • $\begingroup$ Hmmm and if I had a point on this curve could it be done? For exmple portfolio $w=(\frac{1}{2}, \frac{1}{4}, \frac{1}{4})$ lies on Minimum variance Frontier and I want to find portfolio on this line with std. deviation 0.1 $\endgroup$
    – Mr.Price
    May 3, 2020 at 23:33
  • $\begingroup$ covariance loses information about the means. there's your answer $\endgroup$
    – Aksakal almost surely binary
    Jun 3, 2020 at 13:50

1 Answer 1

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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.

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  • $\begingroup$ I edited my post where I try to use LAgrangea Multipliers to solve it. The result is correct but i dont know why it works. I solved it also by another way where I calculated a wallet with a minimum variance and then from this wallet and from the wallet which I have in the command found a solution. However, I would like to understand why this method with Lagrange multipliers returns the correct result, or how to better construct it so that there is no doubt what is being done $\endgroup$
    – Mr.Price
    May 4, 2020 at 9:43

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