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Let $f:\mathbb{R}_{+}^2 \rightarrow \mathbb{R}$ be $f(x_1,x_2)=x_1^a x_2^b$ for $a>0$ and $b>0$. Find restrictions on $a>0$ and $b>0$ that ensure that $f(x_1,x_2)$ is concave.

I tried solving this by finding the principal minors of the Hessian of this function and making first principal minor be less or equal to zero and second principal minor be greater or equal to zero(conditions for a function to be concave). It is easy with the first principal minor, but I cannot workout the second principal minor to have a strong inequality.

$$H(x)=\left| \begin{array}{cc} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right|$$

$$f_{xx}=\frac{\partial^2 f}{\partial x_1^2}=a(a-1)x_2^b x_1^{a-2}$$

$$f_{xy}=\frac{\partial^2f}{\partial x_1 \partial x_2}=ab x_1^{a-1} x_2^{b-1}$$

$$f_{yx}=\frac{\partial^2f}{\partial x_2 \partial x_1}=ab x_1^{a-1} x_2^{b-1}$$

$$f_{yy}=\frac{\partial^2f}{\partial x_2^2}=b(b-1)x_1^a x_2^{b-2}$$

First principal minor:

$f_{xx}=a(a-1)x_2^b x_1^{a-2} \leq 0$; restriction on $a$ is $a<1$.

Second principal minor:

$f_{xx} f_{yy} - f_{xy} f_{yx} = ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2} - a^2 b^2 x_1^{2a-2} x_2^{2b-2} \geq 0$; restriction on b?

I might have computed the second partial derivatives for the Hessian wrong, so if you can please try solving this problem from scratch.

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    $\begingroup$ 1. The restriction on $a$ for $f_{xx}$ to be nonpositive is not what you write. 2. The second condition is that the Hessian is nonnegative, not that is is nonpositive. 3. The same powers of $x_1$ and $x_2$ are in every term of the Hessian, which simplifies the determination of its sign. $\endgroup$ – Did Apr 22 '12 at 7:59
  • $\begingroup$ @Didier: 1. Oh it was a typo. Restriction is a<1. 2. Typo again. 3. I know the powers are the same. When I cancel out terms I get ab>=a+b. This inequality is weak. I wonder if there is a way to make b be greater or less than some number. $\endgroup$ – Koba Apr 22 '12 at 14:44
  • $\begingroup$ My professor says I should show that a+b<=1. $\endgroup$ – Koba Apr 22 '12 at 14:55
  • $\begingroup$ About 3.: no, one does not get $ab\geqslant a+b$. $\endgroup$ – Did Apr 22 '12 at 15:22
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To find restrictions on $a>0$ and $b>0$ that ensure that $f(x_1,x_2)=x_1^ax_2^b$ is concave I first calculated its Hessian and made the first principal minor be $\leq0$ and second principal minor be $\geq0$(the sufficient conditions for a function to be concave):

$$H(x)=\left| \begin{array}{cc} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right|$$

$$f_{xx}=\frac{\partial^2 f}{\partial x_1^2}=a(a-1)x_2^b x_1^{a-2}$$

$$f_{xy}=\frac{\partial^2f}{\partial x_1 \partial x_2}=ab x_1^{a-1} x_2^{b-1}$$

$$f_{yx}=\frac{\partial^2f}{\partial x_2 \partial x_1}=ab x_1^{a-1} x_2^{b-1}$$

$$f_{yy}=\frac{\partial^2f}{\partial x_2^2}=b(b-1)x_1^a x_2^{b-2}$$

First principal minor:

$f_{xx}=a(a-1)x_2^b x_1^{a-2} \leq 0$

for the above inequality to be true the restriction on $a$ should be $a<1$.

Second principal minor:

$f_{xx} f_{yy} - f_{xy} f_{yx} = ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2} - a^2 b^2 x_1^{2a-2} x_2^{2b-2} \geq 0$

This inequality is not so obvious and I had to expand all expressions in parenthesis and divide both sides by the same terms: $$ ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2} - a^2 b^2 x_1^{2a-2} x_2^{2b-2} \geq 0 $$ $$ ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2}\geq a^2 b^2 x_1^{2a-2} x_2^{2b-2} $$ $$ ab(a-1)(b-1)x_1^{2a-2}x_2^{2b-2}\geq a^2 b^2 x_1^{2a-2} x_2^{2b-2} $$ $$ (a-1)(b-1)\geq ab $$ $$ ab-a-b+1-ab\geq 0 $$ $$ -a-b+1\geq 0 $$ $$ 1\geq a+b $$

So here is the second restriction: $a+b\leq 1$

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