Binary Relations for Cobb-Douglas

Question

I am reviewing old midterms to prepare for my upcoming midterm and ran across this question:

Let $\alpha , \beta \in (0,1)$. Now, let $f_{\alpha}$ and $f_{\beta}$ on $\mathbb{R^2}$ be defined as $f_{\alpha}(x)=x_1^{\alpha} x_2^{1 - \alpha}$ and $f_{\beta}(x)=x_1^{\beta} x_2^{1 - \beta}$

Now, let R be a binary relation on $\mathbb{R_x^2}$. Let ${x,y} \subset R_+^2$ We have that: $$xRy \leftrightarrow f_{\alpha}(x) \geq f_{\alpha}(y) \land f_{\beta}(x) \geq f_{\beta}(y)$$

For which combinations of $\alpha$ and $\beta$ is this binary relation complete, for which combinations is it transitive and for which combinations is it continuous.

My thoughts

• It seems this can only be complete if $\alpha=\beta$ but I can't quite finish the proof whenever I proceed WLOG with $\alpha < \beta$ Can anyone here offer an attempt at formally proving that this is complete iff $\alpha = \beta$ ?

• I think that anytime we have xRy, yRz we will necessarily have xRz. That is. And so, I think this is transitive for all combinations of $\alpha,\beta$. My proof involves using the well-ordering of the reals and the definition given for this particular relation. If anyone thinks that this is not true for all $\alpha,\beta$ please let me know why/how.

• I know what continuity is and how to prove it. However, I am not sure for which combinations of $\alpha,\beta$ this relation is continuous. I suspect it is continuous for all combinations of $\alpha,\beta$. Is this true? If so, can you prove it?

Answer 1

I will be using the following definition of continuity for binary relations.

Definition 1: A binary relation $\mathcal{R}$ on $\mathbb{R}^2_+$ is continuous if for every $x$, the following sets are closed. $$ USC_x = \{y\in \mathbb{R}^2_+|y\mathcal{R}x\} $$ $$ LSC_x = \{y\in \mathbb{R}^2_+|x\mathcal{R}y\} $$

Definition 2: A set $Z$ is closed if $z_n$ is a sequence in $Z$ and $z_n\rightarrow z$, then $z\in Z$.

Now, let $\mathcal{R}$ be defined as in your question, i.e. $$ x\mathcal{R} y\iff f_\alpha(x) \geq f_\alpha(y)\text{ and }f_\beta(x) \geq f_\beta(y) $$ Let $x\in \mathbb{R}^2_+$. Let $x_n$ be a sequence in $USC_x$ converging to $x^\ast$. Then, for each $n$, $$ f_\alpha(x_n) \geq f_\alpha(x)\text{ and }f_\beta(x_n) \geq f_\beta(x). $$ Since $f_\alpha$ and $f_\beta$ are continuous, we have $$ f_\alpha(x^\ast) \geq f_\alpha(x)\text{ and }f_\beta(x^\ast) \geq f_\beta(x). $$ implying that $x^\ast\in USC_x$; hence, $USC_x$ is closed. We can show also that $LCS_x$ is closed by using identical arguments. This concludes that $\mathcal{R}$ is continuous.

Remark 1: The arguments above provide a general strategy of analyzing statements regarding $\mathcal{R}$. Transitivity of $\mathcal{R}$ can be proven in a couple of simple steps. Your statement regarding this property is a bit vague, and should be more rigorous imho.

Remark 2: Your explanation about completeness needs a clearly written argument as well. While it is true that $\alpha=\beta$ implies completeness of $\mathcal{R}$, the reasoning that makes it true is not exactly what you are proposing. It is also necessary to show that $\alpha\neq \beta$ implies that $\mathcal{R}$ wouldn't be complete (I believe that this should be the case). This would require a bit more work on your part. Let me know if you run into troubles in this.

Bottomline: In my sincerely humble opinion, you should exercise a lot with proving things. Maybe grab a book on introduction to abstract mathematics and such.

Answer 2

I solved this:

note - I use $R$ to denote the binary relation as defined above.

Completeness:

I claim that this $R$ is not complete for $\alpha \neq \beta$. To illustrate, I use an extreme case.

Let $\alpha \to 1$ and $\beta \to 0$ s.t. $f_{\alpha} \approx x_1$ and $f_{\beta} \approx x_2$

Now, let [x,y] $\subset \mathbb{R}^2$ s.t. $x_1 >> y_1$ and $x_2<<y_2$

Now: $f_{\alpha}(x) \approx x_1 >> y_1 \approx f_{\alpha}(y) $

and: $f_{\beta}(x) \approx x_2 << y_2 \approx f_{\beta}(y)$

so $f_{\alpha}(x) >> f_{\alpha}(y)$ but $f_{\beta}(x)<<f_\beta(y) $

I have used extreme case reasoning to demonstrate, WLOG, that anytime $\alpha \neq \beta \space \exists [x,y] \subset \mathbb{R}_+^2 s.t.$ neither $xRy$ or $yRx$

note_1 - The intuition here is very similar to a delta/epsilon proof. For any combination of alpha and beta (where the two are not equal) I can pick x,y so that the above proof always hold.

note_2 - That the case of $\alpha = \beta$ is complete is trivial. I assume its inclusion here is not necessary.

Transitivity:

Let $[x,y,z] \subset \mathbb{R}^2$ s.t. $xRy, yRx$

I claim $xRz$

since $Xry$, $yRx$ and by the properties of $\mathbb{R}_+^2$

$$f_{\alpha}(x) \geq f_{\alpha}(y) \geq f_{\alpha}(z)$$ $$f_{\beta}(x) \geq f_{\beta}(y) \geq f_{\beta}(z)$$

$$\implies xRz$$

and $R$ is transitive.

Continuity

proven above so I will not repeat this part of the proof.

Answer 3

A few thoughts:

For completeness, I think you can make a slightly stronger case that for any $x_1 \geq y_1, x_2 \geq y_2$, this satisfies completeness $\forall \alpha, \beta$.

$$x_1^\alpha x_2^{1-\alpha} \geq y_1^\alpha y_2^{1-\alpha}$$ $$x_1^\beta x_2^{1-\beta} \geq y_1^\beta y_2^{1-\beta}$$

For continuity, there are two definitions you can use, one with sequences and one with epsilon-balls around a point in your binary relation.