<p>By optimality of <span class="math-container">$(x_1^1, x_2^1)$</span> and concavity of <span class="math-container">$T$</span>, we have,
<span class="math-container">$$
\begin{align*}
0 &amp;\le \pi_1 u_1(x_1^1) + \pi_2 u_1(x_2^1) - \pi_1 u_1(x_0^1) - \pi_2 u_1(x_2^0),\\
&amp;\le \pi_1 T'(u_0(x_1^0))\,\,\,[u_0(x_1^1) -u_0(x_1^0)],\\
&amp; \qquad + \pi_2 T'(u_0(x_2^0))\,\,\,[u_0(x_2^1) - u_0(x_2^0)] 
\end{align*}
$$</span>
Now as <span class="math-container">$x_1^0 &lt; x_2^0$</span>, we have <span class="math-container">$u_0(x_1^0) &lt; u_0(x_2^0)$</span> and as <span class="math-container">$T$</span> is concave, this gives,
<span class="math-container">$$
T'(u_0(x_1^0)) &gt; T'(u_0(x_2^0)).
$$</span>
Next, if, towards a contradiction, <span class="math-container">$x_1^1 &lt; x_1^0$</span> then <span class="math-container">$u_0(x_1^1) &lt; u_0(x_1^0)$</span>. Then,
<span class="math-container">$$
T'(u_0(x_1^0))[u_0(x_1^1) - u_0(x_1^0)] &lt;  T'(u_0(x_2^0))[u_0(x_1^1) - u_0(x_1^0)]
$$</span>
As such,
<span class="math-container">\begin{align*}
0 &amp;\le \pi_1 T'(u_0(x_1^0))\,\,\,[u_0(x_1^1) -u_0(x_1^0)],\\
&amp; \qquad + \pi_2 T'(u_0(x_2^0))\,\,\,[u_0(x_2^1) - u_0(x_2^0)],\\
&amp;&lt; \pi_1 T'(u_0(x_2^0))\,\,\,[u_0(x_1^1) -u_0(x_1^0)],\\
&amp; \qquad + \pi_2 T'(u_0(x_2^0))\,\,\,[u_0(x_2^1) - u_0(x_2^0)],\\
&amp;= T'(u_0(x_2^0))[\pi_1 u_0(x_1^1) + \pi_2 u_0(x_2^1) - \pi_1 u_0(x_1^0) - \pi_2 u_0(x_2^0)],\\
\end{align*}</span>
This implies that,
<span class="math-container">$$
\pi_1 u_0(x_1^1) + \pi_2 u_0(x_2^1) &gt; \pi_1 u_0(x_1^0) + \pi_2 u_0(x_2^0),
$$</span>
contradicting the optimality of <span class="math-container">$(x_1^0, x_2^0)$</span>.</p>

CC BY-SA 4.0
2024-12-24T13:31:00.033Z
2024-12-24T13:31:00.033Z
tdm
<p>I have an economics question but it more so interested me mathematically; related to Arrow-Pratt theorem - risk aversion; hence I wanted to ask the following.</p>
<p>Given <span class="math-container">$p \in \mathbb{R^s}_{++}$</span> <span class="math-container">$\pi_s \geq 0$</span>, <span class="math-container">$\sum_{s=1}^{S}\pi_s = 1$</span>, <span class="math-container">$x \in B(p,w) := \{ y \in \mathbb{R}^S_+ \; | \;  p.y = w \}$</span>, for <span class="math-container">$t = 1,2$</span>, <span class="math-container">$u: \mathbb{R} \rightarrow \mathbb{R} $</span> differentiable once, <span class="math-container">$u_t'(.) &gt; 0$</span> where <span class="math-container">$u_1(.) = T(u_0(.))$</span> for <span class="math-container">$T(.)$</span> strictly concave (if not working, can be assumed to be strictly increasing).</p>
<p>Let <span class="math-container">$x^t$</span> be the maximizers of the following:</p>
<p><span class="math-container">$$ max_{x \in B(p,w)} \sum_{s=1}^S \pi_su_t(x_s)$$</span></p>
<p>For <span class="math-container">$S=2, $</span>If <span class="math-container">$x^0 = (x_1^0, x_2^0)$</span> is such that <span class="math-container">$x_1^0 &lt; x_2^0$</span>,</p>
<p>Please help me prove that <span class="math-container">$x_1^1 \geq x_1^0$</span>. The intuition is that strictly concave transformations moves optimizers to more balanced bundles.</p>
<p>The original question:</p>
<ol start="2">
<li>An economy has one physical good and two states of the world. Each consumer has consumption set <span class="math-container">$X=\mathbb{R}_{+}^2$</span> and endowment <span class="math-container">$\omega=\left(\omega_1, \omega_2\right) \gg 0$</span>, where <span class="math-container">$\omega_s$</span> is the consumer's endowment of the good in state <span class="math-container">$s$</span> for <span class="math-container">$s=1,2$</span>. The price of the good is in state <span class="math-container">$s$</span> is <span class="math-container">$p_s&gt;0$</span>. Each consumer's budget constraint is <span class="math-container">$p_1 c_1+p_2 c_2 \leq p_1 \omega_1+p_2 \omega_2=w$</span>.
There are two possible consumers, <span class="math-container">$t=0,1$</span>. Both consumers have subjective-expected-utility preferences with belief <span class="math-container">$\pi=\left(\pi_1, \pi_2\right) \gg 0$</span> about the states, and they have the same endowment. They only differ in their von Neumann - Morgenstern (vN-M) utilities. Consumer <span class="math-container">$t$</span> 's vNM utility is <span class="math-container">$u_t$</span>, a continuously differentiable function with <span class="math-container">$u_t^{\prime}&gt;0$</span>. (Neither is necessarily concave.)
In addition, for every <span class="math-container">$z \geq 0, u_1(z)=T(u(z))$</span>, where <span class="math-container">$u=u_0$</span> and <span class="math-container">$T: \mathbb{R} \rightarrow \mathbb{R}$</span> is a strictly concave function. In words, consumer 1 is strictly more risk averse than consumer 0 .
Let <span class="math-container">$c^t$</span> solve consumer <span class="math-container">$t$</span> 's problem, and, if necessary, relabel the states so that <span class="math-container">$c_1^0 \leq c_2^0$</span>.
Prove that <span class="math-container">$c_1^1 \geq c_1^0$</span> in two steps.</li>
</ol>
<p>(a) First suppose that <span class="math-container">$c_1^0=c_2^0$</span>; prove that <span class="math-container">$c_1^1 \geq c_1^0$</span>.</p>
<p>(b) Next suppose that <span class="math-container">$c_1^0&lt;c_2^0$</span>; prove that <span class="math-container">$c_1^1 \geq c_1^0$</span>. (Hint: consider, as an preliminary step, what consumer <span class="math-container">$t=1$</span> would chose if it were constrained to choose <span class="math-container">$c_1 \leq c_1^0$</span>.)
You might find this notation useful:
<span class="math-container">$$
\bar{c}=\frac{w}{p_1+p_2}
$$</span>
it is the just-affordable consumption in each state when consumption in each state is equal.</p>

<microeconomics><risk><risk-aversion>
Arrow-Pratt Theorem - More risk aversion leads to more balanced bundles
Pluviaum
# Arrow-Pratt Theorem - More risk aversion leads to more balanced bundles

(Asked by: username_0 on 2024-07-25)

I have an economics question but it more so interested me mathematically; related to Arrow-Pratt theorem - risk aversion; hence I wanted to ask the following.

Given $p \in \mathbb{R^s}\_{++}$ $\pi\_s \geq 0$, $\sum\_{s=1}^{S}\pi\_s = 1$, $x \in B(p,w) := \{ y \in \mathbb{R}^S\_+ \; | \; p.y = w \}$, for $t = 1,2$, $u: \mathbb{R} \rightarrow \mathbb{R} $ differentiable once, $u\_t'(.) > 0$ where $u\_1(.) = T(u\_0(.))$ for $T(.)$ strictly concave (if not working, can be assumed to be strictly increasing).

Let $x^t$ be the maximizers of the following:

$$ max\_{x \in B(p,w)} \sum\_{s=1}^S \pi\_su\_t(x\_s)$$

For $S=2, $If $x^0 = (x\_1^0, x\_2^0)$ is such that $x\_1^0 < x\_2^0$,

Please help me prove that $x\_1^1 \geq x\_1^0$. The intuition is that strictly concave transformations moves optimizers to more balanced bundles.

The original question:

2. An economy has one physical good and two states of the world. Each consumer has consumption set $X=\mathbb{R}\_{+}^2$ and endowment $\omega=\left(\omega\_1, \omega\_2\right) \gg 0$, where $\omega\_s$ is the consumer's endowment of the good in state $s$ for $s=1,2$. The price of the good is in state $s$ is $p\_s>0$. Each consumer's budget constraint is $p\_1 c\_1+p\_2 c\_2 \leq p\_1 \omega\_1+p\_2 \omega\_2=w$.
   There are two possible consumers, $t=0,1$. Both consumers have subjective-expected-utility preferences with belief $\pi=\left(\pi\_1, \pi\_2\right) \gg 0$ about the states, and they have the same endowment. They only differ in their von Neumann - Morgenstern (vN-M) utilities. Consumer $t$ 's vNM utility is $u\_t$, a continuously differentiable function with $u\_t^{\prime}>0$. (Neither is necessarily concave.)
   In addition, for every $z \geq 0, u\_1(z)=T(u(z))$, where $u=u\_0$ and $T: \mathbb{R} \rightarrow \mathbb{R}$ is a strictly concave function. In words, consumer 1 is strictly more risk averse than consumer 0 .
   Let $c^t$ solve consumer $t$ 's problem, and, if necessary, relabel the states so that $c\_1^0 \leq c\_2^0$.
   Prove that $c\_1^1 \geq c\_1^0$ in two steps.

(a) First suppose that $c\_1^0=c\_2^0$; prove that $c\_1^1 \geq c\_1^0$.

(b) Next suppose that $c\_1^0; prove that $c\_1^1 \geq c\_1^0$. (Hint: consider, as an preliminary step, what consumer $t=1$ would chose if it were constrained to choose $c\_1 \leq c\_1^0$.)
You might find this notation useful:
$$
\bar{c}=\frac{w}{p\_1+p\_2}
$$
it is the just-affordable consumption in each state when consumption in each state is equal.

**Tags:** microeconomics, risk, risk-aversion


---

## 1 answers

### Answer by username_1 on 2024-12-24. Score: 0

By optimality of $(x\_1^1, x\_2^1)$ and concavity of $T$, we have,
$$
\begin{align\*}
0 &\le \pi\_1 u\_1(x\_1^1) + \pi\_2 u\_1(x\_2^1) - \pi\_1 u\_1(x\_0^1) - \pi\_2 u\_1(x\_2^0),\\
&\le \pi\_1 T'(u\_0(x\_1^0))\,\,\,[u\_0(x\_1^1) -u\_0(x\_1^0)],\\
& \qquad + \pi\_2 T'(u\_0(x\_2^0))\,\,\,[u\_0(x\_2^1) - u\_0(x\_2^0)]
\end{align\*}
$$
Now as $x\_1^0 < x\_2^0$, we have $u\_0(x\_1^0) < u\_0(x\_2^0)$ and as $T$ is concave, this gives,
$$
T'(u\_0(x\_1^0)) > T'(u\_0(x\_2^0)).
$$
Next, if, towards a contradiction, $x\_1^1 < x\_1^0$ then $u\_0(x\_1^1) < u\_0(x\_1^0)$. Then,
$$
T'(u\_0(x\_1^0))[u\_0(x\_1^1) - u\_0(x\_1^0)] < T'(u\_0(x\_2^0))[u\_0(x\_1^1) - u\_0(x\_1^0)]
$$
As such,
\begin{align\*}
0 &\le \pi\_1 T'(u\_0(x\_1^0))\,\,\,[u\_0(x\_1^1) -u\_0(x\_1^0)],\\
& \qquad + \pi\_2 T'(u\_0(x\_2^0))\,\,\,[u\_0(x\_2^1) - u\_0(x\_2^0)],\\
&< \pi\_1 T'(u\_0(x\_2^0))\,\,\,[u\_0(x\_1^1) -u\_0(x\_1^0)],\\
& \qquad + \pi\_2 T'(u\_0(x\_2^0))\,\,\,[u\_0(x\_2^1) - u\_0(x\_2^0)],\\
&= T'(u\_0(x\_2^0))[\pi\_1 u\_0(x\_1^1) + \pi\_2 u\_0(x\_2^1) - \pi\_1 u\_0(x\_1^0) - \pi\_2 u\_0(x\_2^0)],\\
\end{align\*}
This implies that,
$$
\pi\_1 u\_0(x\_1^1) + \pi\_2 u\_0(x\_2^1) > \pi\_1 u\_0(x\_1^0) + \pi\_2 u\_0(x\_2^0),
$$
contradicting the optimality of $(x\_1^0, x\_2^0)$.


---

