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.


1 Answer 1


Let us consider your maximal correlation case. You are trying to find values such that

$$(\lambda \sigma_x+(1-\lambda)\sigma_y)^2>\lambda\sigma_ x^2 + (1-\lambda)\sigma_y^2$$


$$\lambda^2 \sigma_x^2+2\sigma_x\sigma_y\lambda(1-\lambda)+(1-\lambda)^2\sigma_y^2>\lambda\sigma_ x^2 + (1-\lambda)\sigma_y^2$$


$$\lambda(\lambda-1)\sigma_x^2+2\sigma_x\sigma_y\lambda(1-\lambda)-\lambda(1-\lambda)\sigma_y^2>0 $$


$$\lambda(\lambda-1)(\sigma_x^2+\sigma_y^2)+2\sigma_x\sigma_y\lambda(1-\lambda)>0 $$


$$\lambda(\lambda-1)(\sigma_x-\sigma_y)^2>0 $$

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.

  • $\begingroup$ Thank you so much fesman. This helped a lot! :-) $\endgroup$
    – Wombat
    Aug 15, 2020 at 21:10
  • $\begingroup$ does variance satisfy the triangle inequality? $\endgroup$
    – develarist
    Aug 16, 2020 at 6:33
  • $\begingroup$ No variance doesn't satisfy that in general. $\endgroup$
    – fes
    Aug 16, 2020 at 7:37

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