# prove that m(a,y) is non decreasing in a and y and concave in a.

I am a university student specializing mathematics for economists. I am in a preparation for my final exam. My prof gave me some questions that might be on the exam. One of the question dragged me down so hard. I cannot even imagine where to start. If you guys have any idea about this, please help me. If I prepare this, I feel like I will do well on the exam. Thanks a bunch in advance.

The question is,

Let m(α,y) be defined as the minimum value of αx subject to g(x) >= y, where α, x∈ R^n++, y∈ R+, and g(x) is strictly monotonic increasing and quasi-concave. Prove that m(α,y) is (i) non-decreasing in α and y and (ii) concave in α. Then, given that g(x) is homogeneous of degree k, derive the corresponding form of m(α,y)

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(i) Let $\alpha_2\geq \alpha_1$. Then,

$$m(\alpha_2,y)-m(\alpha_1,y)=\min_{g(x)\geq y}(\alpha_2 x)-\min_{g(x)\geq y}(\alpha_1 x)=(\alpha_2-\alpha_1)\min_{g(x)\geq y}(x)\geq 0$$

where we used the fact that $\alpha_2\geq\alpha_1\geq 0$ and $x\geq 0$. This proves nondecreasing in $\alpha$.

For nondecreasing in $y$, if $z>y$, then notice that the set of $x$ such that $g(x)\geq z$ is smaller than the set of $x$ such that $g(x)\geq y$, the latter contains the former. This means that:

$\min_{g(z)\geq x}(\alpha x)\geq \min_{g(y)\geq y}(\alpha x)$

so $m(\alpha,z)-m(\alpha,y)\geq 0$.

(ii)

For concavity, show that minimums are concave functions. That is $\min(ax+(1-a)y)\geq a\min(x)+(1-a)\min(y)$, where the minimum is over whatever.

Homogeneity of degree $k$ means that $g(ax)=a^kg(x)$. Quasi-concavity means, $g(\lambda x+(1-\lambda) z)\geq \min(g(y),g(z))$. Try to do the last part yourself.

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Wow thanks! this is really helpful – user51193 Dec 1 '12 at 6:53
I do no know how to finish the last part，someone could help？ – user51492 Dec 2 '12 at 16:33