The question comes from ‘Mathematics for Finance: An Introduction to Financial Engineering’ by Marek Capiński (Author), Tomasz Zastawniak. The book does not give a complete proof, and I did not find a detailed proof online. But I think this issue is still very important. So posted the question on the stackexchange.

Assume that $-1<\rho_{12}<1$ , and $\mu_1\neq\mu_2$ Then for each asset portfolio V in the feasible set,Prove $x=\sigma_{v}$ and $y=\mu_{v}$ satisfy the hyperbolic equation $$x^2-A^2(y-\mu_0)^2=\sigma_0^2$$ $$A^2=\frac{\sigma_1^2+\sigma_2^2-2c_{12}}{\left(\mu_1-\mu_2\right)^2}>0$$The two asymptotes of this hyperbola are $$y=\mu_0\pm\frac1Ax$$

Background Information:

Then $\sigma_v^2$ as a function of s reaches the minimum value at the following point $s_0$ $$s_0=\frac{\sigma_2^2-c_{12}}{\sigma_1^2+\sigma_2^2-2c_{12}}$$ Because$$\frac{\mathrm{d}(\sigma_V^2)}{\mathrm{d}s}=2s(\sigma_1^2+\sigma_2^2-2c_{12})-2(\sigma_2^2-c_{12})=0$$ Where$$\mu_0=\frac{\mu_1\sigma_2^2+\mu_2\sigma_1^2-(\mu_1+\mu_2)c_{12}}{\sigma_1^2+\sigma_2^2-2c_{12}}$$$$\sigma_0^2=\frac{\sigma_1^2\sigma_2^2-c_{12}}{\sigma_1^2+\sigma_2^2-2c_{12}}$$ $\sigma_1$ is the standard deviation of asset 1’s rate of return

$\sigma_2$ is the standard deviation of asset 2’s rate of return

$\mu_1$ is the expected return on asset 1

$\mu_2$ is the expected return on asset 2

$c_{12}$ is the covariance of the returns on Asset 1 and Asset 2

expected value of portfolio v:$\\;\mu_v=s\mu_1+(1-s)\mu_2$

Variance of portfolio v:$\\;\sigma_v^2=s^2\sigma_1^2+(1-s)^2\sigma_2^2+2s(1-s)c_{12}\\ $ Formula 3.9

The book gives simple proof tips:Substituting $s=\frac{\mu_V-\mu_2}{\mu_1-\mu_2}$ into Equation 3.9.Using simple (or can be considered slightly cumbersome) transformations, we can prove that the proposition is true

geogebra diagram



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