3
$\begingroup$

How can I show that the function $u(x_1,x_2)=(x_1x_2)^\alpha$ is quasiconcave, given $\alpha>1,x_i\geq0$?

I managed to find the bordered Hessian, whereby

$(-1)^1B_1=\alpha^2(x_1x_2)^{2(\alpha-1)}x^2_2\geq0$

and

$(-1)^2B_2=0\geq0$.

Nonetheless, that only provides a necessary condition for quasiconcavity.

Is it possible to show quasiconcavity from its definition, i.e., $u(ax_1+(1-a)y_1,ax_2+(1-a)y_2)\geq \textrm{min}\{u(x_1,x_2),u(y_1,y_2)\}$?

$\endgroup$
3
$\begingroup$

Is it possible to show quasiconcavity from its definition, i.e., $u(ax_1+(1-a)y_1,ax_2+(1-a)y_2)\geq \textrm{min}\{u(x_1,x_2),u(y_1,y_2)\}$?

Answer: Yes.


A useful trick that can save you some trouble is to perform a monotonic transformation. In preference relation terms you are trying to show

$$ \left( ax_1+(1-a)y_1,ax_2+(1-a)y_2 \right) \succeq \left[ (x_1,x_2) \text{ OR } (y_1,y_2) \right]. $$

If this holds for the utility representation $(x_1x_2)^{\alpha}$, it will also hold for monotonic transformations of this function (the ordering of baskets is unchanged).

Clearly there is a more elegant function to represent the relation, making mathematical calculations easier.


Alternatively you can power through it and make some assumptions w.o.l., like $u(x_1,x_2) \leq u(y_1,y_2)$. The function is clearly strictly monotonic, so that saves you from looking at cases where $x_1 \leq y_1$ AND $x_2 \leq y_2$. All that remains to check (w.o.l.) is the case where $x_1 < y_1$, $x_2 > y_2$.

$\endgroup$
2
  • $\begingroup$ Under the second strategy, the condition $u(ax_1+(1−a)y_1,ax_2+(1−a)y_2)=\{[ax_1+(1-a) y_1][ax_2+(1-a) y_2 ]\}^\alpha \geq u(x_1,x_2)=(x_1x_2)^\alpha$ must be fulfilled. Assuming that $x_1<y_1$ and $x_2>y_2$, we have that $ax_1+(1-a) y_1\geq x_1$ and $ax_2+(1-a) y_2\leq x_2$, which does not fulfil the above condition. Am I doing something wrong here? $\endgroup$ Aug 28 at 5:15
  • 1
    $\begingroup$ @kékszajkók The above is not a complete answer, as the question is a self-study problem. You will also need to use $u(x1,x2) \leq u(y1,y2)$, and still there are some algebraic manupulations to do. $\endgroup$
    – Giskard
    Aug 28 at 8:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.