# Violation of Monotonicity of preferences

Hi I am reading Jehle and Reny in my master's course and I have come across a problem in one of the exercises. My instructor herself was a bit confused when a student gave her a counter-example and then said that the example illustrates that our logic is faulty. I want to reconcile the example and math logic. Here is the question with the solutions discussed.

Q1.24 Let u($$\textbf x$$) represent some consumer’s monotonic preferences over $$\textbf x ∈ \mathbb{R}^n_+$$. For each of the functions $$f (x)$$ that follow, state whether or not f also represents the preferences of this consumer. In each case, be sure to justify your answer with either an argument or a counterexample.

Part (c) $$f (x) = u(\textbf x) +\Sigma_{i=1}^{n} x_i$$

Now I believe that this function is a monotonic transformation of $$u(\textbf x)$$ and thus is a representation of the inherent "monotonic" preferences. I provide two proofs stating this fact and one counter-example which is creating difficulty in understanding

Proof 1: (OP's proof)

Let $$\textbf x^1 \ge \textbf x^2.$$ (1)

Clearly, $$\textbf x^1 \succsim \textbf x^2$$ (Since preferences are monotonic) (2)

$$\because \textbf x^1 \ge \textbf x^2 \implies \Sigma_{i=1}^{n} x_i^1 \ge \Sigma_{i=1}^{n}x_i^2$$...(3)

$$\therefore u(\textbf x^1)\ge u(\textbf x^2)$$ (by (2))....(4)

$$\therefore f(\textbf x^1)\ge f(\textbf x^2)$$ (by 3 and 4)

Proof (2):Book's solution manual

Counter example: $$u(x_1,x_2)=x_1x_2$$ $$u(1,4)=4$$ and $$u(2,2)=4$$ and $$(2,2)~(1,4)$$ but $$f(1,4)=9$$ and $$f(2,2)=8$$

I believe that we are getting this counter example because we are inherently assuming strict convexity / convexity of preferences. We are not given that preferences are convex/strictly convex.

any thoughts?

In general, it will not represent the same preferences. There seems to be confusion on what "monotonic transformation" means in this context. It does not have much to do with monotonic preferences.

We say that the utility function $$v:X\to\mathbb{R}$$ is a monotonic transformation of the utility function $$u:X\to\mathbb{R}$$ if there exists a strictly increasing function $$g:u(X)\to\mathbb{R}$$ (the domain is the range of $$u$$) such that $$v(x)=g(u(x))$$ for all $$x\in X$$. A monotonic transformation of a utility function does always represent the same preferences, and this has nothing to do with whether these preferences are monotone, convex, or anything like that.

• But by the book's solution manual, it is indeed a monotonic transformation, f' is increasing, isn't it? Nov 14 '21 at 14:10
• @KaranKumar Well, $f$ is increasing in the arguments and, therefore, represents monotone preferences. But it does not represent the same preferences as $u$. And $f$ is not a monotonic transformation of $u$ according to the definition given in the answer. Nov 14 '21 at 14:14
• I disagree. Utility function in regards to representing the preferences are invariant to positive monotonic transformations. Since f' is greater than zero, therefore f and u represent the same preferences. Also, there does not seem to be any ambiguity as to how the function f is not a monotonic transformation as provided in the answer. Perhaps you could illustrate with an example or some argument as I am utterly missing your point. Nov 14 '21 at 15:12
• @KaranKumar If $f$ would be a positive monotonic transformation of $u$, it would represent the same preferences. It does not, as your own example shows. The other "proofs" given by you do not show that it is a monotonic transformation. Where is the function $g$? Nov 14 '21 at 15:50
• I actually found the "solution manual" you quote from. It is not an official solution manual for the book; it is someone typing up their own solutions. And making a mistake, here. Nov 14 '21 at 16:00

We can simplify by assuming that $$u$$ is linear, in which we can treat $$u$$ as the total dollar value of the set of items, and $$f$$ to be the total dollar value plus the count of the items. Or, in other words, $$f$$ adds $$1$$ to the dollar value of each item. Would you prefer $$1000$$ pennies to one diamond worth $$\\\100$$? $$u(1000 \text { pennies}) = 10$$, $$u(\text{diamond})=100$$, so $$u$$ says the diamond is better. $$f(1000 \text { pennies}) = 1010$$, $$f(\text{diamond})=101$$, so $$f$$ says the pennies are better.

The fact that $$f$$ is "monotonic" (a more precise phrasing would be that it's monotonic with respect to each of the components) just means that $$f$$ gives a higher number than $$u$$ for the same basket of goods. It doesn't say anything about what happens when comparing different baskets of goods.

• @Accumulation your logical example is extremely helpful by the way. But shouldn't it apply on say more things. Like if I consume 2 half litre cans of coke and 1 litre can of coke we get the same utility - also assume they are priced so that half a can of coke costs half the price of coke. but by your logic if we add the two cans we will get utility of say 3 and utility of 1 can will be 2! What then? Shouldn't we be indifferent? Nov 19 '21 at 15:14
• @KaranKumar $f$ is given in terms of a particular meaning of $x_i$. Changing what $x_i$ means changes what $f$ is in terms of $x_i$. Nov 20 '21 at 0:55
• Thank you all, we can close this question Nov 27 '21 at 10:19