# Preference relations defined by $x_1^n + x_2^n$ converge to $\max\{x_1, x_2\}$

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 $$a \succeq b = \lim_{n \to \infty} a \succeq_n b$$.

• “the only case when $x\sim_m y$ is when $x_2=y_2$”. Did you mean "$x_1=y_1$"? – Herr K. Apr 11 '19 at 23:23
• 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. – Lochend Apr 12 '19 at 6:38

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).

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, $$v(x_1,x_2)=x_2=\max\{x_1,x_2\}$$.

• 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. – Giskard Apr 12 '19 at 4:55
• Also I am not sure if your calculations really answer the OP's question, which seems to be about "why", not "how". – Giskard Apr 12 '19 at 4:58
• 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 $a, b$.) – Lochend Apr 12 '19 at 6:35