How to prove the inequality for the standard deviation of a linear combination of two random variables

The variance of the linear combination V of random variables X₁ and X₂ is given by the following formula:

$$\sigma_{V}^{2} = s^{2} \sigma_{1}^{2}+(1-s)^2 \sigma_{2}^{2}+2 s(1-s) c_{12}$$ where s and (1-s) are the coefficients of the linear combination, $$c_{12}$$ represents the covariance of two random variables!

When the variance of the combination V with respect to s has a derivative of zero, the expression for s can be written as:

$$s_{0} = \frac{\sigma_{2}^{2}-c_{12}}{\sigma_{1}^{2}+\sigma_{2}^{2}-2 c_{12}}$$

if $$\frac{\sigma_1}{\sigma_2}<\rho_{12}<1$$ then $$s_{0}>1$$

$$\rho_{12}$$ is the linear correlation coefficient between random variable 1 and random variable 2.

Given $$\sigma_1 <\sigma_2$$ $$\sigma_{V}^{2} = s_{0}^{2} \sigma_{1}^{2}+(1-s_{0})^2 \sigma_{2}^{2}+2 s_{0}(1-s_{0}) c_{12}$$ prove

$$\sigma_v <\sigma_1$$

geogebra diagram

• You are being asked to prove that the minimum variance combination of two assets has a smaller variance than either of the two assets (why is it stated in such a confusing way?). You know what $s_0$ is (second equation), just plug that into the last equation and see what the resulting $\sigma^2_V$ is in terms of the problem data $\sigma_1,c_{12},\sigma_2$ Commented Apr 13 at 16:12
• The expression obtained when d( $σ^2_v$ )/ds=0 is $s_0$ Commented Apr 13 at 16:52

First, transform $$σ_{v}^2$$ into the form of a standard quadratic function.

\begin{aligned} &\sigma _{v}^{2}=s^{2}\sigma _{1}^{2}+\left( 1-s\right) ^{2}\sigma _{2}^{2}+2s\left( 1-s\right) c_{12}\\ &=s^{2}\sigma _{1}^{2}+\left( 1-2s+s^{2}\right) \sigma _{2}^{2}+2s\rho \sigma _{1}\sigma _{2}-2s^{2}\rho \sigma _{1}\sigma _{2}\\ &=s^{2}\left( \sigma _{1}^{2}+\sigma _{2}^{2}-2\rho \sigma _{1}\sigma _{2}\right) +s\left( 2\rho \sigma _{1}\sigma _{2}-2\sigma _{2}^{2}\right) +\sigma _{2}^{2} \end{aligned}

Consider ‘s’ as the independent variable and the rest as coefficients.

\begin{aligned} &y=ax^{2}+bx+c\\ & \because a >0\\ & \therefore y\geq \dfrac{4ac-b^{2}}{4a}\\ & so\ \ \ \sigma _{v}^{2}\geq \dfrac{4\left( \sigma _{1}^{2}+\sigma _{2}^{2}-2\rho \sigma _{1}\sigma _{2}\right) \sigma _{2}^{2}-\left( 2\rho \sigma _{1}\sigma _{2}-2\sigma _{2}^{2}\right) ^{2}}{4\left( \sigma _{1}^{2}+\sigma _{2}^{2}-2\rho \sigma _{1}\sigma _{2}\right) } \end{aligned}

\begin{aligned} & \sigma_v^2 \geqslant \frac{\left(1-\rho^2\right) \sigma_1^2 \sigma_2^2}{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2} \\ & \because \rho<1 \\ & \therefore \sigma_v^2 \leqslant \frac{\left(\rho^2-1\right) \sigma_1^2 \sigma_2^2}{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2} \\ & \because \sigma_v>0 \\ & \therefore \sigma_v \leqslant \frac{\sqrt{\left(1-\rho^2\right)} \sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1^2 \sigma_2}} \\ \end{aligned}

\begin{aligned} & \because \rho<1 \\ & 2 \rho\sigma_1 \sigma_2<2 \sigma_1 \sigma_2 \\ & \\ & \therefore \frac{\sqrt{1-\rho^2} \sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2}} < \frac{\sqrt{1-\rho^2} \sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2-2 \sigma_1 \sigma_2}}=\frac{\sqrt{1-\rho^2} \sigma_1 \sigma_2}{\sigma_1+\sigma_2} \\ & \because 1-\rho^2<1 \\ & \therefore \sqrt{1-\rho^2} \sigma_2<\sigma_2<\sigma_1+\sigma_2 \\ & \therefore \frac{\sqrt{1-\rho^2} \sigma_2}{\sigma_1+\sigma_2}<1 \\ & \therefore \frac{\sqrt{1-\rho^2} \sigma_2 \sigma_1}{\sigma_1+\sigma_2}<\sigma_1 \\ &So \ \ \sigma _{v}<\sigma _{1} \end{aligned}