Answer to Question #122393 in Financial Math for Aakash

Question

Let ρ =( w1,w2) be a portfolio of two securities. Find the value of w1and w2 in the following situations:
i) p12 = -1 and ρ is risk-free.
ii) σ1 = σ2 and variance P is minimum.
iii) Variance on P is minimum and ρ12 = -0.5 , σ1 = 2 and σ2 = 3.

Answer 1

i)

$-1=\mu1\times\omega1+\mu2\times\omega2$

w1=1-w2

$-1=\mu1\times(1-\omega2)+\mu2\times\omega2$

$-1=\mu1-\mu1\omega2+\mu2\times\omega2$

$-1-\mu1=-\mu1\omega2+\mu2\times\omega2$

$\frac{-1-\mu1}{-\mu1+\mu2}=\omega2$

$1-\frac{-1-\mu1}{-\mu1+\mu2}=\omega1$

ii)

σ1 = σ2

solve by the formula

$\sigma^2=(\omega1)^2(\sigma1)^2+2\rho\omega1\omega2\sigma1\sigma2+(\omega2)^2(\sigma2)^2$

$\sigma^2=(\omega1)^2(\sigma1)^2+2\rho\omega1\omega2(\sigma1)^2+(\omega2)^2(\sigma1)^2$

$\sigma^2=(\sigma1)^2((\omega1)^2+2\rho\omega1\omega2+(\omega2)^2)$

$\frac{\sigma^2}{(\sigma1)^2}=((\omega1)^2+2\rho\omega1\omega2+(\omega2)^2)$

$1-\omega2=\omega1$

$\frac{\sigma^2}{(\sigma1)^2}=((1-\omega2)^2+2\rho(1-\omega2)\omega2+(\omega2)^2)$

$\frac{\sigma^2}{(\sigma1)^2}=(1-2\omega2+(\omega2)^2+2\rho\omega2-2\rho(\omega2)^2+(\omega2)^2)$

$\frac{\sigma^2}{(\sigma1)^2}=(1-2\omega2+2(\omega2)^2+2\rho\omega2-2\rho(\omega2)^2)$

$\frac{\sigma^2}{(\sigma1)^2}=(1-2(\omega2+(\omega2)^2)+(-)2\rho(\omega2+(\omega2)^2))$

$\frac{\sigma^2}{(\sigma1)^2}=(1-(\omega2+(\omega2)^2)(2-2\rho)$

$\frac{\frac{\sigma^2}{(\sigma1)^2}-1}{(2-2\rho)}-(\omega2)^2=\omega2$

$1-\frac{\frac{\sigma^2}{(\sigma1)^2}-1}{(2-2\rho)}-(\omega2)^2=\omega1$

iii)

solve by the formula

$\sigma^2=(\omega1)^2(\sigma1)^2+2\rho\omega1\omega2\sigma1\sigma2+(\omega2)^2(\sigma2)^2$

$\sigma^2=(\omega1)^2\times4+2(-0.5)\omega1\omega2\times6+(\omega2)^2\times9$

$\sigma^2=4(\omega1)^2-6\omega1\omega2+9(\omega2)^2$

$\frac{\sigma^2}{4}=(\omega1)^2-1.5\omega1\omega2+2.25(\omega2)^2$

$1−ω2=ω1$

$\frac{\sigma^2}{4}=(1-\omega2)^2-1.5(1-\omega2)\omega2+2.25(\omega2)^2$

$\frac{\sigma^2}{4}=1-2\omega2+(\omega2)^2-1.5\omega2+1.5(\omega2)^2+2.25(\omega2)^2$

$\frac{\sigma^2}{4}=1-3.5\omega2+2.75(\omega2)^2$

$\frac{\frac{\sigma^2}{4}-1-2.75(\omega2)^2}{3.5}=\omega2$

$1-\frac{\frac{\sigma^2}{4}-1-2.75(\omega2)^2}{3.5}=\omega1$