3
$\begingroup$

In the problem set 2 of Rubinsteins Microeconomics (btw is there a comparably nice written book on macroeconomics?) there is the following question: Let $\succ_n$ be the preference relations defined on $\mathbb{R}^2_+$ by the utility $x_1^n + x_2^n$. Let the preference relations $\succ$ be defined by the utility $\max\{x_1, x_2\}$. Show that $\succ_n$ converge to $\succ$.

Preference relations are said to converge if for $a \succ b$ we have that $a \succ_m b$ for sufficiently large $m$. I am indeed able to prove that.

Now w.l.o.g. $x_1 = \max\{x_1, x_2\}$ and $y_1 = \max\{y_1, y_2\}$. Assuming that $x_1 = y_1$ we have $x \sim y$. But the only case when $x \sim_m y$ is when also $x_2 = y_2$. In other cases we would have either $x \succ_m y$ or $y \succ_m x$ for all $m$ (depending on $x_2$ and $y_2$, the "smaller component". These are actually lexicographic preferences!)

Is there a reason that the convergence of preference relations is done in this way to ignore such subtilities? In my opinion we should have that $\succeq = \lim_{n \to \infty} \succeq_n$ iff $a \succeq b = \lim_{n \to \infty} a \succeq_n b$.

$\endgroup$
  • $\begingroup$ “the only case when $x\sim_m y$ is when $x_2=y_2$”. Did you mean "$x_1=y_1$"? $\endgroup$ – Herr K. Apr 11 at 23:23
  • $\begingroup$ I wanted to assume $x_1 = y_1$ w.l.o.g. but forgot to note it down. I now edited the question to make that part more clear, thank you. $\endgroup$ – Lochend Apr 12 at 6:38
0
$\begingroup$

A source of confusion comes from the distinction between strict and weak preferences. Define $\succ_n$ strict preferences and $\succsim_n$ weak preferences.

1) The sequence of preference relations $\succ_n$ does not converge to preference relation $\succ$, precisely because of the indifference case. If we take your example with $x_1=y_1$, $x_1=\max(x_1,x_2)$ and $y_1=\max(y_1,y_2)$. Assume also that $x_2>y_2$. Then, for any $n>0$, $x\succ_n y$ but we do not have $x\succ y$.

2) However, one can show that the sequence $\succsim_n$ converges to $\succsim$, which is what you suggest at the end of your question. We can check that the counterexample above does not apply here: $x\succsim_n y$ for $n>0$ and $x\succsim y$.

3) Is there an error in Rubinstein's textbook? No. The problem set (2018 update) actually frames the question in terms of weak preferences $\succsim$ (note there is still a mix of notations that is puzzling).

$\endgroup$
1
$\begingroup$

I am assuming you're familiar with the concept of monotonic transformations.

Let $f(z)=z^n$ be a monotonic transformation. $f(z)$ is an order preserving transformation for all positive $n$. Let $v:\mathbb{R_+^2} \to\mathbb{R}$ be an utility function, where $v(x_1,x_2)=(x_1^n + x_2^n)^{1/n}$. The function $f(v(x_1,x_2))=((x_1^n + x_2^n)^{1/n})^n=x_1^n + x_2^n$ will represent the same preferences as $v(x_1,x_2)$. Therefore, $(x_1^n + x_2^n)^{1/n}$ should also converge to $\max\{x_1,x_2\}$.

To prove the same, consider $v(x_1,x_2)=\Big(1+\cfrac{x_1^n }{x_2^n}\Big)^{1/n}x_2$. Assume $x_2 = \max\{x_1,x_2\}. $Now, as $n \to \infty$, two cases arise.

1) If $x_2>x_1$, then $\cfrac{x_1^n }{x_2^n} \to 0$ as $n \to \infty$. In that case, $v(x_1,x_2)=x_2=\max\{x_1,x_2\}$.

2) If $x_2=x_1$, the limit can be calculated by taking log on both sides as follows- \begin{align*} \ln(v(x_1,x_2)) = \cfrac{\ln(1+\cfrac{x_1^n }{x_2^n})}{n} +\ln(x_2) \,\,\,\,-(1) \end{align*} Since $x_1/x_2=1$,we have $(x_1/x_2)^n=1^n$. Using this in equation $(1)$, we get \begin{align*} \ln(v(x_1,x_2)) = \cfrac{\ln(1+1^n)}{n} +\ln(x_2) \end{align*} As $n\to \infty$, the term $\cfrac{\ln(1+1^n)}{n}$ becomes $\infty/\infty$. This limit can be solved using L'Hospital's rule. The value of the limit(you may check it for yourself) comes out to be $0$. Now we have $\ln(v(x_1,x_2)) = 0 +\ln(x_2)$, or, $v(x_1,x_2)=x_2=\max\{x_1,x_2\}$.

$\endgroup$
  • $\begingroup$ The part "If $x_2>x_1$, then $\cfrac{x_1^n }{x_2^n} \to 0$ as $n \to \infty$. In that case, $v(x_1,x_2)=x_2$" is shaky. By the same reasoning $$ \lim_{n \to \infty} \left( \frac{x_1^n}{x_2^n}\right)^{1/n} = 0 $$ if $x_1 < x_2$, but this is of course not true. $\endgroup$ – Giskard Apr 12 at 4:55
  • 1
    $\begingroup$ Also I am not sure if your calculations really answer the OP's question, which seems to be about "why", not "how". $\endgroup$ – Giskard Apr 12 at 4:58
  • $\begingroup$ Yes, thank you for the calculation, but the question is why we define convergence in this way (only requesting that $\lim a \succ_n b = a \succ b$ if $a \succ b$ and not $a \succeq b = \lim_n a \succeq_n b$ for every $a, b$.) $\endgroup$ – Lochend Apr 12 at 6:35

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.