<blockquote>
<p>Is it possible to show quasiconcavity from its definition, i.e., <span class="math-container">$u(ax_1+(1-a)y_1,ax_2+(1-a)y_2)\geq \textrm{min}\{u(x_1,x_2),u(y_1,y_2)\}$</span>?</p>
</blockquote>
<p>Answer: Yes.</p>
<hr />
<p>A useful trick that can save you some trouble is to perform a monotonic transformation. In preference relation terms you are trying to show</p>
<p><span class="math-container">$$
\left( ax_1+(1-a)y_1,ax_2+(1-a)y_2 \right) \succeq \left[ (x_1,x_2) \text{ OR } (y_1,y_2) \right].
$$</span></p>
<p>If this holds for the utility representation <span class="math-container">$(x_1x_2)^{\alpha}$</span>, it will also hold for monotonic transformations of this function (the ordering of baskets is unchanged).</p>
<p>Clearly there is a more elegant function to represent the relation, making mathematical calculations easier.</p>
<hr />
<p>Alternatively you can power through it and make some assumptions w.o.l., like <span class="math-container">$u(x_1,x_2) \leq u(y_1,y_2)$</span>. The function is clearly strictly monotonic, so that saves you from looking at cases where <span class="math-container">$x_1 \leq y_1$</span> AND <span class="math-container">$x_2 \leq y_2$</span>. All that remains to check (w.o.l.) is the case where <span class="math-container">$x_1 &lt; y_1$</span>, <span class="math-container">$x_2 &gt; y_2$</span>.</p>

CC BY-SA 4.0
2021-08-28T05:15:50.920Z
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?
kékszajkók
CC BY-SA 4.0
2021-08-28T08:05:30.807Z
@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.
Giskard
CC BY-SA 4.0
2021-08-27T16:48:32.577Z
2021-08-27T19:36:21.043Z
2021-08-27T19:36:21.043Z
Giskard
<p>How can I show that the function <span class="math-container">$u(x_1,x_2)=(x_1x_2)^\alpha$</span> is quasiconcave, given <span class="math-container">$\alpha&gt;1,x_i\geq0$</span>?</p>
<p>I managed to find the bordered Hessian, whereby</p>
<p><span class="math-container">$(-1)^1B_1=\alpha^2(x_1x_2)^{2(\alpha-1)}x^2_2\geq0$</span></p>
<p>and</p>
<p><span class="math-container">$(-1)^2B_2=0\geq0$</span>.</p>
<p>Nonetheless, that only provides a necessary condition for quasiconcavity.</p>
<p>Is it possible to show quasiconcavity from its definition, i.e., <span class="math-container">$u(ax_1+(1-a)y_1,ax_2+(1-a)y_2)\geq \textrm{min}\{u(x_1,x_2),u(y_1,y_2)\}$</span>?</p>

<microeconomics><mathematical-economics>
Proof of Quasiconcavity of Utility Function
kékszajkók
# Proof of Quasiconcavity of Utility Function

(Asked by: username_0 on 2021-08-27)

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., $u(ax\_1+(1-a)y\_1,ax\_2+(1-a)y\_2)\geq \textrm{min}\{u(x\_1,x\_2),u(y\_1,y\_2)\}$?

**Tags:** microeconomics, mathematical-economics

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## 1 answers

### Answer by username_1 on 2021-08-27. Score: 3

> 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 $x\_1 < y\_1$, $x\_2 > y\_2$.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @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.  

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