Why Variance Is Convex and the Correlation Bound
Summary
The discussion asks whether variance is a convex risk measure and tests the question by expanding the variance of a weighted combination of two random variables. The key method is to bound covariance by its maximum possible value, attained when the variables have perfect positive correlation, then compare the resulting expression with the weighted average of their variances.
The algebra reduces the proposed convexity violation to a term proportional to the negative quantity λ(λ−1) times the squared difference in standard deviations. Since λ lies between zero and one, that term cannot be positive. The answer therefore concludes that variance is convex. This addresses convexity alone; it does not establish that variance satisfies every other axiom commonly required of a coherent risk measure.
Key ideas
- Variance of a weighted sum depends on both individual variances and covariance.
- Covariance is bounded above by the product of the standard deviations.
- At perfect positive correlation, the variance comparison reduces to a squared difference in standard deviations.
- For mixture weights between zero and one, the proposed convexity violation cannot occur.
- Convexity alone does not show that variance satisfies all coherent risk measure axioms.
Tags
Full text
# Variance convex risk measure # Variance convex risk measure I hope you can help me with this question that I really struggle with. Is variance a convex risk measure? I guess not, but I find it really hard to find a counter example. Here are my thoughts. I tried to find an example where: $var(\lambda X+(1-\lambda)Y))>\lambda var(X)+(1-\lambda)var(Y)$. I know that $var(\lambda X+(1-\lambda) Y)= \lambda^2var(X)+(1-\lambda)^2var(Y)+2\lambda (1-\lambda)cov(X,Y)$ $=\lambda^2var(X)+(1-\lambda)^2var(Y)+2\lambda (1-\lambda)corr(X,Y)sd(X)sd(Y)$. Now, if the correlation is maximal, in which case $corr(X,Y)=1$ then:$\lambda^2var(X)+(1-\lambda)^2var(Y)+2\lambda (1-\lambda)corr(X,Y)sd(X)sd(Y)=\lambda^2var(X)+(1-\lambda)^2var(Y)+2\lambda(1-\lambda)sd(X)sd(Y)=(\lambda sd(X)+(1-\lambda)sd(Y))^2$. But I still can't find any example where this is greater than $\lambda var(X)+(1-\lambda)var(Y)$. Can you give me any hints? I appreciate it a lot. ## Answer by fes (score 2, accepted) https://quant.stackexchange.com/a/57371 Let us consider your maximal correlation case. You are trying to find values such that $$(\lambda \sigma_x+(1-\lambda)\sigma_y)^2>\lambda\sigma_ x^2 + (1-\lambda)\sigma_y^2$$ or $$\lambda^2 \sigma_x^2+2\sigma_x\sigma_y\lambda(1-\lambda)+(1-\lambda)^2\sigma_y^2>\lambda\sigma_ x^2 + (1-\lambda)\sigma_y^2$$ or $$\lambda(\lambda-1)\sigma_x^2+2\sigma_x\sigma_y\lambda(1-\lambda)-\lambda(1-\lambda)\sigma_y^2>0 $$ or $$\lambda(\lambda-1)(\sigma_x^2+\sigma_y^2)+2\sigma_x\sigma_y\lambda(1-\lambda)>0 $$ or $$\lambda(\lambda-1)(\sigma_x-\sigma_y)^2>0 $$ which is clearly never true for any $0\leq\lambda\leq 1.$ Because LHS is greatest at the maximal correlation case: $$Var(\lambda x+(1-\lambda)y)\leq \lambda Var( x)+(1-\lambda)Var(y)$$ and variance is a convex risk measure.
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.