<p>I'll give a few hints to get you started. First, note that since the preference <span class="math-container">$\succsim$</span> is represented by the utility function <span class="math-container">$U(x_1,x_2)=x_1+\ln x_2$</span>, it follows that
<span class="math-container">\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}</span></p>
<p>Keeping this equivalence in mind, consider:</p>
<ul>
<li><p><strong>Completeness</strong>: <span class="math-container">$\succsim$</span> is complete if for all <span class="math-container">$(x_1,x_2),(x_1',x_2')\in\mathbb R_+^2$</span>,
<span class="math-container">\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}</span>
Using <span class="math-container">$(1)$</span>, we can rewrite <span class="math-container">$(2)$</span> as
<span class="math-container">\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}</span>
Now <span class="math-container">$(2^*)$</span> should be easy to prove using the property that <span class="math-container">$\mathbb R$</span> is an ordered field.</p>
</li>
<li><p><strong>Transitivity</strong>: Use the same trick to translate preference ordering into ordering of real numbers.</p>
</li>
<li><p><strong>Convexity</strong>: Start from the definition that <span class="math-container">$\succsim$</span> is convex if for any <span class="math-container">$\alpha\in[0,1]$</span>,
<span class="math-container">\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}</span>
Again, translate the preference ordering into ordering of real numbers to prove the implication. Since <span class="math-container">$U$</span> is quasi-linear, this way will save you some trouble of dealing with Hessians and so on.</p>
</li>
</ul>

CC BY-SA 4.0
2020-10-29T17:48:57.920Z
Thank you for your explanation!
Alex
CC BY-SA 4.0
2020-10-29T04:55:39.187Z
2020-10-29T05:02:01.230Z
2020-10-29T05:02:01.230Z
Herr K.
<ol>
<li>For preferences to be rational, they must be complete and transitive. Note that since the preferences <span class="math-container">$\succsim$</span> are represented by the utility function <span class="math-container">$u:\mathbb{R}^{2}\to \mathbb{R}$</span> (as defined in the question), we have <span class="math-container">$x\succsim y\iff u(x)\geq u(y)$</span> for any <span class="math-container">$x,y\in \mathbb{R}^{2}$</span>.</li>
</ol>
<p>Completeness: Consider any <span class="math-container">$x, x'\in \mathbb{R}^{2}$</span>. Since the ordering <span class="math-container">$\geq$</span> of <span class="math-container">$\mathbb{R}$</span> is complete(that is, any two real numbers can be <em>compared</em>), we have <span class="math-container">$u(x)\geq u(x')\iff x\succsim x'$</span> or <span class="math-container">$u(x')\geq u(x)\iff x'\succsim x$</span>.
Transitivity: Consider any <span class="math-container">$x,y,z\in \mathbb{R}^{2}$</span> and suppose <span class="math-container">$x\succsim y$</span> and <span class="math-container">$y\succsim z$</span>. Thus we have <span class="math-container">$u(x)\geq u(y)$</span> and <span class="math-container">$u(y)\geq u(z)$</span>. Since the ordering <span class="math-container">$\geq$</span> of <span class="math-container">$\mathbb{R}$</span> is transitive, we have <span class="math-container">$u(x)\geq u(z)$</span> which is equivalent to <span class="math-container">$x\succsim z$</span>.</p>
<p>A side note: If preferences are indeed represented by a utility function, they are rational because the order <span class="math-container">$\geq$</span> is a complete and transitive order on the Reals. So, one doesn't need continuity in the utility function to arrive at rational preferences.</p>
<ol start="2">
<li>The utility function is strictly concave. Here is Martin Osborne's material for this: <a href="https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t" rel="nofollow noreferrer">https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t</a>.</li>
</ol>
<p>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.</p>

CC BY-SA 4.0
2020-10-29T05:10:50.897Z
Thank you for your explanation!
Alex
CC BY-SA 4.0
2020-10-29T05:19:15.350Z
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
Alex
CC BY-SA 4.0
2020-10-29T15:23:13.253Z
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.
soslow
CC BY-SA 4.0
2024-09-03T09:16:39.367Z
$u$ is not strictly concave, here is the proof: consider two bundles $(0,1)$ and $(2,1)$, and observe that $u(\frac{1}{2}(0,1)+\frac{1}{2}(2,1))=u(1,1)=1=\frac{1}{2}u(0,1)+\frac{1}{2}u(2,1)$.
Amit
CC BY-SA 4.0
2020-10-29T04:38:45.100Z
2020-10-29T04:38:45.100Z
soslow
<p>Preferences represented by <span class="math-container">$u:\mathbb{R}^2_+\rightarrow\mathbb{R}\cup\{-\infty\}$</span> defined as <span class="math-container">$u(x_1,x_2)=x_1+\ln x_2$</span> are rational because there is utility representation. That is a sufficient condition for rationality.
Preferences represented by <span class="math-container">$u(x_1,x_2)=x_1+\ln x_2$</span> are convex, because <span class="math-container">$u$</span> is quasi-concave, and that is true because <span class="math-container">$u$</span> is concave (as it is the sum of two concave functions - <span class="math-container">$f_1(x_1,x_2)=x_1$</span> and <span class="math-container">$f_2(x_1,x_2)=\ln x_2$</span>).
Preferences represented by <span class="math-container">$u(x_1,x_2)=x_1+\ln x_2$</span> are not strictly convex, because <span class="math-container">$u$</span> is not strictly quasi-concave. To prove that it is not strictly quasi-concave, consider two bundles <span class="math-container">$(0,0)$</span> and <span class="math-container">$(2,0)$</span>, and observe that <span class="math-container">$u(\frac{1}{2}(0,0)+\frac{1}{2}(2,0))=-\infty=u(0,0)=u(2,0)$</span>. In words, since we have horizontal axis <span class="math-container">$x_2=0$</span> as an indifference curve for satisfaction level <span class="math-container">$-\infty$</span>, therefore preferences are not strictly convex or <span class="math-container">$u$</span> is not strictly quasi-concave.</p>

CC BY-SA 4.0
2024-09-03T08:28:01.313Z
2024-09-03T13:54:33.300Z
2024-09-03T13:54:33.300Z
Amit
<p>Consider a consumer with preferences relation <span class="math-container">$\succsim$</span> over non-negative commodities <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> such that their utility U = <span class="math-container">$x_1$</span> + <span class="math-container">$\ln(x_2)$</span></p>
<p>Are these preferences rational and are they convex/strictly-convex?</p>
<p>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?</p>
<p>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?</p>
<p>Online, it says a function is quasi-concave if <span class="math-container">$f(\lambda x+(1-\lambda )y)\geq \min {\big \{}f(x),f(y){\big \}}$</span>, but I'm having a tough time understanding this in relation to a utility that has both an <span class="math-container">$x_1$</span> and an <span class="math-container">$x_2$</span> value.  When I'm looking at the above function, I only understand it for like <span class="math-container">$f(a) = a^2$</span> and there's not a second variable in there.</p>
<p>Thanks!</p>

<mathematical-economics><utility><preferences>
Showing utility function gives preferences that are rational and convex
Alex
# Showing utility function gives preferences that are rational and convex

(Asked by: username_0 on 2020-10-29)

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.

Thanks!

**Tags:** mathematical-economics, utility, preferences


---

## 3 answers

### Answer by username_1 on 2020-10-29. Score: 4

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.

> **Comments:**
> **username_0**: Thank you for your explanation!  

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### Answer by username_2 on 2020-10-29. Score: 3

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.

2. 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.

> **Comments:**
> **username_0**: Thank you for your explanation!  
> **username_0**: 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  
> **username_2**: Yes, we just need the strict inequality. The solution by username_1 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.  
> **username_3**: $u$ is not strictly concave, here is the proof: consider two bundles $(0,1)$ and $(2,1)$, and observe that $u(\frac{1}{2}(0,1)+\frac{1}{2}(2,1))=u(1,1)=1=\frac{1}{2}u(0,1)+\frac{1}{2}u(2,1)$.  

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### Answer by username_3 on 2024-09-03. Score: 0

Preferences represented by $u:\mathbb{R}^2\_+\rightarrow\mathbb{R}\cup\{-\infty\}$ defined as $u(x\_1,x\_2)=x\_1+\ln x\_2$ are rational because there is utility representation. That is a sufficient condition for rationality.
Preferences represented by $u(x\_1,x\_2)=x\_1+\ln x\_2$ are convex, because $u$ is quasi-concave, and that is true because $u$ is concave (as it is the sum of two concave functions - $f\_1(x\_1,x\_2)=x\_1$ and $f\_2(x\_1,x\_2)=\ln x\_2$).
Preferences represented by $u(x\_1,x\_2)=x\_1+\ln x\_2$ are not strictly convex, because $u$ is not strictly quasi-concave. To prove that it is not strictly quasi-concave, consider two bundles $(0,0)$ and $(2,0)$, and observe that $u(\frac{1}{2}(0,0)+\frac{1}{2}(2,0))=-\infty=u(0,0)=u(2,0)$. In words, since we have horizontal axis $x\_2=0$ as an indifference curve for satisfaction level $-\infty$, therefore preferences are not strictly convex or $u$ is not strictly quasi-concave.


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