Minimum Variance of a Two-Asset Linear Combination
Summary
The document considers the variance of a weighted linear combination of two random variables, expressed using their individual variances and covariance. Differentiating this quadratic with respect to the weight gives the unconstrained weight that minimizes variance. It then considers the case where the variables have different standard deviations and their correlation exceeds their standard-deviation ratio, placing the minimizing weight beyond the usual long-only range.
The response completes a proof by rewriting the variance as a quadratic and evaluating its minimum, then comparing the resulting expression with the smaller individual standard deviation. This illustrates how covariance and an unrestricted weight can produce a combination with lower volatility than either component. The argument applies to the stated assumptions and permits a weight outside zero to one, which implies a negative exposure to one variable. The posted derivation contains apparent algebra and inequality-direction errors, so its conclusion should be checked using the exact minimum-variance formula before relying on the displayed steps.
Key ideas
- The variance of a two-variable combination depends on both component variances and their covariance.
- Differentiating the variance with respect to the weight yields the unconstrained minimum-variance weight.
- The assumed correlation condition places that weight above one.
- An unconstrained combination can have lower volatility than the less volatile component.
- The displayed proof contains algebraic issues and should be independently verified.
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Full text
# How to prove the inequality for the standard deviation of a linear combination of two random variables
# 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
## Answer by bokabokaboka (score 0, accepted)
https://quant.stackexchange.com/a/78994
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}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.