# 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.

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.

• Thank you so much fesman. This helped a lot! :-) – Wombat Aug 15 '20 at 21:10
• does variance satisfy the triangle inequality? – develarist Aug 16 '20 at 6:33
• No variance doesn't satisfy that in general. – fesman Aug 16 '20 at 7:37