Consider a consumer with preferences relation $\succsim$ over non-negative commodities $x_1$ and $x_2$ such that their utility U = $x_1$ + $\ln(x_2)$

Are these preferences rational and are they convex/strictly-convex?

I'm a bit confused on how to do this. So first, I know preferences need to be complete and transitive to be rational, but for a utility function, it just needs to be continuous right? Is there another property it needs? With that said, how would I mathematically prove this function is actually continuous? If I graph it, it's continuous, but is there a mathematical proof for this?

For the second part, if preferences are (strictly) convex, then the preferences must be (strictly) quasi-concave right? How would I mathematically prove that this function above is quasi-concave?

Online, it says a function is quasi-concave if $f(\lambda x+(1-\lambda )y)\geq \min {\big \{}f(x),f(y){\big \}}$, but I'm having a tough time understanding this in relation to a utility that has both an $x_1$ and an $x_2$ value. When I'm looking at the above function, I only understand it for like $f(a) = a^2$ and there's not a second variable in there.



I'll give a few hints to get you started. First, note that since the preference $\succsim$ is represented by the utility function $U(x_1,x_2)=x_1+\ln x_2$, it follows that \begin{equation} (x_1,x_2)\succsim (x_1',x_2')\quad\Leftrightarrow\quad U(x_1,x_2)\ge U(x_1',x_2') \tag{1} \end{equation}

Keeping this equivalence in mind, consider:

  • Completeness: $\succsim$ is complete if for all $(x_1,x_2),(x_1',x_2')\in\mathbb R_+^2$, \begin{equation} \text{either }(x_1,x_2)\succsim (x_1',x_2'), \quad\text{or }(x_1',x_2')\succsim(x_1,x_2). \tag{2} \end{equation} Using $(1)$, we can rewrite $(2)$ as \begin{equation} \text{either }U(x_1,x_2)\ge U(x_1',x_2'), \quad\text{or }U(x_1',x_2')\ge U(x_1,x_2). \tag{2*} \end{equation} Now $(2^*)$ should be easy to prove using the property that $\mathbb R$ is an ordered field.

  • Transitivity: Use the same trick to translate preference ordering into ordering of real numbers.

  • Convexity: Start from the definition that $\succsim$ is convex if for any $\alpha\in[0,1]$, \begin{multline} (x_1,x_2)\succsim(x_1'',x_2'') \text{ and } (x_1',x_2')\succsim (x_1'',x_2'') \\\Rightarrow \quad \alpha(x_1,x_2)+(1-\alpha)(x_1',x_2')\succsim(x_1'',x_2'') \end{multline} Again, translate the preference ordering into ordering of real numbers to prove the implication. Since $U$ is quasi-linear, this way will save you some trouble of dealing with Hessians and so on.

  • $\begingroup$ Thank you for your explanation! $\endgroup$ – Alex Oct 29 '20 at 17:48
  1. For preferences to be rational, they must be complete and transitive. Note that since the preferences $\succsim$ are represented by the utility function $u:\mathbb{R}^{2}\to \mathbb{R}$ (as defined in the question), we have $x\succsim y\iff u(x)\geq u(y)$ for any $x,y\in \mathbb{R}^{2}$.

Completeness: Consider any $x, x'\in \mathbb{R}^{2}$. Since the ordering $\geq$ of $\mathbb{R}$ is complete(that is, any two real numbers can be compared), we have $u(x)\geq u(x')\iff x\succsim x'$ or $u(x')\geq u(x)\iff x'\succsim x$. Transitivity: Consider any $x,y,z\in \mathbb{R}^{2}$ and suppose $x\succsim y$ and $y\succsim z$. Thus we have $u(x)\geq u(y)$ and $u(y)\geq u(z)$. Since the ordering $\geq$ of $\mathbb{R}$ is transitive, we have $u(x)\geq u(z)$ which is equivalent to $x\succsim z$.

A side note: If preferences are indeed represented by a utility function, they are rational because the order $\geq$ is a complete and transitive order on the Reals. So, one doesn't need continuity in the utility function to arrive at rational preferences.

  1. The utility function is strictly concave. Here is Martin Osborne's material for this: https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t.

You can, in general, look at the Hessian for multivariable twice-differentiable functions like this one. The utility function that you mention is (strictly) concave. Strict concavity implies strict quasi-concavity and hence the preferences are (strictly) convex.

  • $\begingroup$ Thank you for your explanation! $\endgroup$ – Alex Oct 29 '20 at 5:10
  • $\begingroup$ I calculated the Hessian and got $\begin{bmatrix} 0 & f'_{1}(x) & f'_{2}(x)\\ f'_{1}(x) & f''_{11}(x) & f''_{12}(x) \\ f'_{2}(x) & f''_{21}(x) & f''_{22}(x) \\ \end{bmatrix}$ = $\begin{bmatrix} 0 & 1 & \frac{1}{x_2} \\ 1 & 0 & 0 \\ \frac{1}{x_2} & 0 & -\frac{1}{x_{2}^{2}} \\ \end{bmatrix}$ gives me the determinant of $\frac{1}{x_{2}^{2}}$ for $D_2$ and a determinant of -1 for $D_1$. The notes mention that if $D_1 <= 0$ and $D_2 >=0$, it is quasi-concave (so proves it). But when is it strictly quasi-concave? When it's < and > rather than equality? Or do we need something else? Thanks $\endgroup$ – Alex Oct 29 '20 at 5:19
  • $\begingroup$ Yes, we just need the strict inequality. The solution by Herr to convexity is more general, that is, it does not depend on the smoothness of the utility function. You could probably try that out too. $\endgroup$ – soslow Oct 29 '20 at 15:23

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.