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I'm struggling with determining whether a function is convex/concave and quasi-convex/quasi-concave using Hessians and Bordered Hessians respectively.

As far as I understand it, a function is convex if all Hessian determinants are larger or equal to $0$, while it is concave if this is smaller or equal to $0$. For quasi, we look at the determinants of bordered hessians, and a function is quasi convex if all determinants are larger or equal to $0$, quasiconcave if $|H_1| \geq 0$, $|H_2| \leq 0$ etc.

Is this correct?

Using this, I can see why $x_1 x_2$ is quasi-concave, but it seems to me that it is concave as well. The Hessian is: $$ \begin{pmatrix} 0 &1 \\ 1 &0 \end{pmatrix} $$ so $H_1 = 0$, $H_2 = -1$, hence concave?

What's going wrong?

Cheers, Calcer

Siminore
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Calcer
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2 Answers2

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You have

$$u( t(0,1)+(1-t)(1,0) ) = u(1-t,t) = t(1-t)$$

$$tu(0,1)+(1-t)u(1,0) = 0$$

So clearly, there exists $t=\frac{1}{2}$ such that

$$u( t(0,1)+(1-t)(1,0) ) < tu(0,1)+(1-t)u(1,0)$$

So $u$ is not concave (by the definition of concavity).

Tryss
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If you look along the line $x_1 = x_2$ you will see the function is actually convex along this line for $x_1 \geq 0$. As far as 'what's going on', your Hessian is negative-semidefinite, not negative definite. When you have semidefinite instead of definite, you can't draw any global conclusions about convexity or concavity.

user2566092
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  • Thanks for your reply. Is the reason that the above goes wrong due to the fact that the function u is not twice differentiable? – Calcer Sep 27 '15 at 14:32
  • @Calcer No, your function is smooth and twice differentiable. See https://en.wikipedia.org/wiki/Hessian_matrix#Second_derivative_test where they state that a convex/concave function has positive/negative semidefinite Hessian, but the converse is only guaranteed to be true if you have strict positive/negative definite Hessian -- semidefinite is not always enough, as you have found. – user2566092 Sep 27 '15 at 14:36
  • Thanks. My lecture notes say: a function is concave iff the second derivative is everywhere negative semi-definite. Is this incorrect then? – Calcer Sep 27 '15 at 14:43