Quasi-concavity of $f(x_1,x_2)=x_1^2 x_2^2$ for non-negative $x_1,x_2$

Question

Show that $f(x_1, x_2) = x_1^2 x_2^2$ is quasiconcave for non-negative $x_1,x_2$?

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This is a coursework question. I tried showing the upper contour set is convex by picking $\alpha \in [0,1]$, $x=(x_1,x_2)$, $w =(w_1,w_2)$ with $f(x)=x_1^2x_2^2>r, f(w)=w_1^2w_2^2>r$ and computing $f$ of their combination:

$$ \begin{align} f(\alpha x + (1-\alpha)) &= f(\alpha x_1 + (1-\alpha)w_1,\alpha x_2 + (1-\alpha)w_2) \\ &= (\alpha x_1 + (1-\alpha)w_1)^2(\alpha x_2 + (1-\alpha)w_2)^2 \end{align} $$

However, once I expand and distribute this expression, I am unable to compare to $f(x)$ and $f(w)$ (and subsequently $r$) because of the $\alpha^n$ terms, which are smaller than one, getting in the way of the comparison. Any guidance on how to proceed with this proof?

Answer 1

Let $f(x_1,x_2,...,x_n)\in C^2$. Then $f$ is $\color{red}{\text{quasiconcave}}$ ($\color{blue}{\text{quasiconvex}}$) if: $$\color{red}{(-1)^rB_r(x)\ge 0} \quad \left(\color{blue}{B_r(x)\le 0}\right) \quad \text{for} \ r=1,2,...,n, \\ B_r=\begin{pmatrix}0&f'_1&f'_2&\cdots&f'_r\\ f'_1&f''_{11}&f''_{12}&\cdots &f''_{1r}\\ \vdots\\ f'_r&f''_{r1}&f''_{r2}&\cdots&f''_{rr}\end{pmatrix}.$$ Source: Knut Sydsaeter, Peter Hammond, Arne Strom, Essential Mathematics for Economic Analysis (4th Edition). pp. 95-96.

The bordered Hessian: $$ B_r=\begin{bmatrix} 0 & f_{x_1} & f_{x_2} \\ f_{x_1} & f_{x_1x_1} & f_{x_1x_2} \\ f_{x_2} & f_{x_2x_1} & f_{x_2x_2} \end{bmatrix}= \begin{bmatrix} 0 & 2x_1x_2^2 & 2x_1^2x_2 \\ 2x_1x_2^2 & 2x_2^2 & 4x_1x_2 \\ 2x_1^2x_2 & 4x_1x_2 & 2x_1^2 \end{bmatrix};\\ (-1)^1B_1=-\begin{vmatrix}0&2x_1x_2^2\\ 2x_1x_2^2&2x_2^2\end{vmatrix}=4x_1^2x_2^2\ge 0\\ (-1)^2B_2=\begin{vmatrix}0 & 2x_1x_2^2 & 2x_1^2x_2 \\ 2x_1x_2^2 & 2x_2^2 & 4x_1x_2 \\ 2x_1^2x_2 & 4x_1x_2 & 2x_1^2\end{vmatrix}=16x_1^4x_2^4\ge 0.$$ Hence, the function is quasiconcave.

Answer 2

Since $x^2y^2 \ge a>0$ is equivalent to $xy \ge \sqrt a$ for nonnegative $x,y$ showing quasiconcavity for the original function is equivalent to showing it for $xy$ so is equivalent to showing that the "inside" of the hyperbola $xy=b>0$ in the first quadrant is convex and that is geometrically obvious but can be easily proved by noticing that $x_1y_2+x_2y_1 \ge 2b$ by the mean inequality when $x_1y_1, x_2y_2 \ge b$ in the first quadrant and then it trivially follows $(tx_1+(1-t)x_2)(ty_1+(1-t)y_2) \ge b$ by expliciting the terms out

Answer 3

We will prove that a more general familiy of functions that include yours: $x^2y^2$ is quasiconcave.

Proposition 1: Any convex function is also quasiconvex. Any concave function is also quasiconcave.

proof: By using the inequality of the definition of convex and quasiconvex function.

Proposition 2: If $f$ is monotone increasing and $g$ is quasiconvex/quasiconcave then $f \circ g$ is also quasiconvex/quasiconcave.

Proof: By definition of quasiconvex/quasioncave function.

Remark: You can state and prove a similar proposition for $f$ monotone decreasing using the fact that $h$ is quasiconvex/quasiconcave iff $-h$ is quasiconcave/quasiconvex, and also that $h$ is decreasing iff $-h$ is increasing.

Exercise: Prove that the function $f(x,y) := A x^a y^b$ defined in the domain $D := \left \{ (x,y) : x>0,y>0 \right \}$ is quasiconcave, for all $A,a,b >0$.

Solution: Take $n \in \mathbb{N}$ odd and big enough such that $a/n + b/n \leq 1$. Then consider the functions $g(t) := t^n$ and $h(x,y) := A^{1/n}x^{a/n}y^{b/n}$. We have that $f = g \circ h$, whatsmore, $g$ is increasing and $h$ is a Cobb–Douglas function, i.e. $h$ is of the form $Bx^\alpha y^\beta$ with $B,\alpha,\beta >0$. We also know that the exponents of $h$ satisfying the condition $\alpha + \beta \leq 1$ by our choice of $n$. It is known (you can google it or use the hessian criterion for differentiable concave functions) that Cobb-Douglas functions whose exponents satisfy the last inequality are concave, then $h$ is concave. By our Proposition 1 we have that $h$ is quasiconcave. In summary, $f$ is the composition of an increasing function and a quasiconcave function, thus by our Proposition 2 it must be quasiconcave.

Answer 4

There is a subtletly with this exercise. If we wanted to check quasiconcavity in the positive orthant, that would be MUCH easier. We can use the second derivative test. $$ B = \begin{bmatrix} 0 & 2x_1 x_2^2 & 2x_1^2 x_2 \\ 2x_1 x_2^2 & 2x_2^2 & 4 x_1 x_2 \\ 2x_1^2 x_2 & 4x_1 x_2 & 2x_1^2 \end{bmatrix} $$

The determinant is (as farruhota calculated) $|B| = 16x_1^4 x_2^4$. And it is strictly positive for $x_1>0$ and $x_2>0$. Hence the function is quasiconcave in the positive orthant. All the calculus criteria for quasiconcavity work on open sets. So any of these second order tests won't work (at least in a rigurous way) in the non-negative orthant, which is a closed set.

An alternative -and very common- way to test quasiconcavity is using the monotonic transformation property (quasiconcavity is preserved under monotonic transformations) and testing the concavity of a simpler function. Usually we take that monotonic transformation to be $\ln(x)$, which is also a concave function. So $2\ln(x_1) + 2\ln(x_2)$ is the sum of two concave functions, hence it is concave, and that proves that $x_1^2 x_2^2$ is quasiconcave on the positive orthant. But we cannot use this to test quasiconcavity on the non-negative orthant because $\ln(0)$ is not defined.

Finally. For this example I would draw the level curves, which are hyperbolas in the positive orthant, and the zero level curve are the $x_1$ and $x_2$ axis. Hence, the upper level sets $S^{\geq}$ are convex sets, and the function is quasiconcave on the non-negative orthant.