Find restrictions on a>0 and b>0 that ensure that f(x_1,x_2) is concave.

[Dostre]

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.

 

[Ray Vickson]

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.

f must be concave in x1 when x2 is held constant, so 0 < a < 1; f must be concave in x2 when x1 is held constant, so 0 < b < 1. To test whether f is jointly concave we need only look at the determinant of the Hessian, which you have computed already as
\det(H) = x_1^{2a-1} x_2^{2b-1}[a(a-1)b(b-1)-a^2b^2], and [\cdot] = ab(1-a-b). From that you can find the required extra restriction or restrictions on a and b.

RGV

 

[Dostre]

f must be concave in x1 when x2 is held constant, so 0 < a < 1; f must be concave in x2 when x1 is held constant, so 0 < b < 1. To test whether f is jointly concave we need only look at the determinant of the Hessian, which you have computed already as
\det(H) = x_1^{2a-1} x_2^{2b-1}[a(a-1)b(b-1)-a^2b^2], and [\cdot] = ab(1-a-b). From that you can find the required extra restriction or restrictions on a and b.

RGV

Yeah the expression [\cdot] = ab(1-a-b) is exactly what I need, but when I expand [a(a-1)b(b-1)-a^2b^2], I get ab-a-b. I am making some kind of mistake. Can you show me how [a(a-1)b(b-1)-a^2b^2]=ab(1-a-b) .

 

[Ray Vickson]

Yeah the expression [\cdot] = ab(1-a-b) is exactly what I need, but when I expand [a(a-1)b(b-1)-a^2b^2], I get ab-a-b. I am making some kind of mistake. Can you show me how [a(a-1)b(b-1)-a^2b^2]=ab(1-a-b) .

Here it is in Maple:
p:=a*(1-a)*b*(1-b)-a^2*b^2;
2 2
p := a (1 - a) b (1 - b) - a b
expand(p);
2 2
a b - a b - a b

Or, you can just do the algebra yourself.

RGV

 

[Dostre]

Here it is in Maple:
p:=a*(1-a)*b*(1-b)-a^2*b^2;
2 2
p := a (1 - a) b (1 - b) - a b
expand(p);
2 2
a b - a b - a b

Or, you can just do the algebra yourself.

RGV

Ok I see now. Thank you. You was very helpful.