<p>Note that <span class="math-container">$(a_1,a_2)$</span> is the utility maximizing bundle at relative price <span class="math-container">$MRS(a) = (= p_1/p_2)$</span> and budget <span class="math-container">$w = MRS(a).a_1 + a_2$</span>.</p>
<ol>
<li><p>First as <span class="math-container">$b$</span> is utility maximising, <span class="math-container">$MRS(b) = \frac{p_1^0}{p_2^0}$</span>. As <span class="math-container">$MRS(a') &gt; MRS(b) = MRS(a)$</span> and the <span class="math-container">$MRS$</span> is unique (if utility is differentiable) we have that <span class="math-container">$a_1' \ne b_1$</span>.</p>
</li>
<li><p>Now for the proof. Assume, towards a contradiction, that <span class="math-container">$b_1 \le a_1'$</span>, which gives by point 1 above that <span class="math-container">$b_1 &lt; a_1$</span>. Then:
<span class="math-container">$$
\begin{align*}
MRS(a') (a_1' - b_1 ) + a_2' - b_2
&amp;= MRS(a') (a_1' - b_1) + w - \frac{p_1^0}{p_2^0} a_1' - w + \frac{p_1^0}{p_2^0} b_1,\\
&amp;= \underbrace{(MRS(a') - MRS(a))}_{&gt; 0} \underbrace{(a_1' - b_1)}_{&gt; 0} &gt; 0
\end{align*}
$$</span>
As such, we get that:
<span class="math-container">$$
MRS(a') a_1' + a_2' &gt; MRS(a') b_1 + b_2.
$$</span>
As <span class="math-container">$(a_1', a_2')$</span> is utility maximising for relative prices <span class="math-container">$MRS(a')$</span> and budget <span class="math-container">$w'' = MRS(a') a_1' + a_2'$</span> and <span class="math-container">$b$</span> was cheaper to buy at that price and budget, we obtain by a revealed preference argument, that <span class="math-container">$u(a_1', a_2') &gt; u(b_1, b_2)$</span>, contradicting the assumption that <span class="math-container">$(b_1, b_2)$</span> was utility maximising at the budget <span class="math-container">$B((p_1^0,p_2^0), w)$</span> and that <span class="math-container">$a' \in B((p_1^0, p_2^0), w)$</span>.</p>
</li>
</ol>

CC BY-SA 4.0
2024-06-24T16:11:13.217Z
Thank you for the reply. Sorry, I still have some doubts: why are we allowed to let $\text{MRS}(b) = \text{MRS}(a) = p_1^0/p_2^0$ without assuming anything about the consumer's utility? If the optimal choice of either good 1 or good 2 is zero, the marginal rate of substitution need not be equal to the price ratio. Also, why is $(a_1', a_2')$ utility-maximising? This is not true neither by assumption nor by construction, no?
Riccardo Iorio
CC BY-SA 4.0
2024-06-25T12:08:03.647Z
Plus, I don't understand the revealed-preference argument you made. Can't a contradiction be equally derived from the result that $\text{MRS}(a')a'_1 + a'_2 > \text{MRS}(b)b_1 + b_2$ by noting that by assumption $a'$ was on the frontier of $B((p_1^0, p_2^0), w')$ and $b$ was optimal (so by monotonicity also on the frontier), and so we would have $w' > w'$ which is clearly a contradiction? This would require $\text{MRS}(a')a'_1 + a'_2 = w'$ though, which relates to my previous question about how you know that $a'$ is optimal.
Riccardo Iorio
CC BY-SA 4.0
2024-06-25T16:00:23.620Z
@Riccardo Iorio For corner solutions, you need to be careful as as derivatives (of the utility function) are not defined on the boundary of the domain. As such, there is no MRS. The hyperplane through the (utility maximising) boundary point with slope $(p_1^0/p_2^0)$ is still tangent to the upper contour set. So one needs convex analysis to analyse such cases.
tdm
CC BY-SA 4.0
2024-06-25T16:07:53.767Z
Revealed preference tells that if one bundle was chosen and a second bundle was also affordable for that budget but cheaper to buy, then the chosen bundle should give a higher utility than the second bundle.
tdm
CC BY-SA 4.0
2024-06-25T16:15:06.980Z
Thanks again. Can you just clarify why $a'$ is optimal? By construction it's on the boundary of $w'$ but is not optimal ($b$ is).
Riccardo Iorio
CC BY-SA 4.0
2024-06-25T16:44:22.513Z
Oh, also, why are marginal utilities not defined for corner solutions? If either optimal solution is zero (considering a two-dimensional commodity space), can't you just define one-sided derivatives?
Riccardo Iorio
CC BY-SA 4.0
2024-06-28T17:54:11.603Z
Sorry if I keep bothering, can you please explain why $a' = (a'_1, a'_2)$ is optimal? Then the contradiction would immediately follow because we'd get $w' > w'$. But why is it utility-maximising? By construction it's on the boundary of $w'$, but that is not sufficient to guarantee optimality.
Riccardo Iorio
CC BY-SA 4.0
2024-06-24T11:38:10.990Z
2024-06-25T16:06:36.893Z
2024-06-25T16:06:36.893Z
tdm
<p>Osborne and Rubinstein (<em>Models in Microeconomic Theory</em>, p. 70) prove the proposition that, if the demand function of a rational consumer has the property that <span class="math-container">$\text{MRS}(x_1, x_2)$</span> is increasing in <span class="math-container">$x_2$</span> for all values of <span class="math-container">$x_1$</span>, then good 1 is normal, as follows:</p>
<blockquote>
<p>Fix <span class="math-container">$p_1^0$</span> and <span class="math-container">$p_2^0$</span> and let <span class="math-container">$w' &gt; w$</span>. Let <span class="math-container">$a$</span> be a solution to the consumer's problem for the budget set <span class="math-container">$B((p_1^0, p_2^0), w)$</span> and let <span class="math-container">$a'$</span> be a bundle on the frontier of <span class="math-container">$B((p_1^0, p_2^0), w')$</span> with <span class="math-container">$a_1' = a_1$</span>. By the assumption on the marginal rate of substitution we have <span class="math-container">$\text{MRS}(a') &gt; \text{MRS}(a)$</span>, and hence the solution <span class="math-container">$b$</span> of the consumer's problem for the budget set <span class="math-container">$B((p_1^0, p_2^0), w')$</span> has <span class="math-container">$b_1 &gt; a'_1 = a_1.$</span></p>
</blockquote>
<p>I don't understand the very last inference. Why does the fact that <span class="math-container">$\text{MRS}(a') &gt; \text{MRS}(a)$</span> imply the desired conclusion?</p>

<microeconomics><mathematical-economics><consumer-theory>
Proof that if the MRS is increasing in one good, then the other is normal
Riccardo Iorio
# Proof that if the MRS is increasing in one good, then the other is normal

(Asked by: username_0 on 2024-06-24)

Osborne and Rubinstein (*Models in Microeconomic Theory*, p. 70) prove the proposition that, if the demand function of a rational consumer has the property that $\text{MRS}(x\_1, x\_2)$ is increasing in $x\_2$ for all values of $x\_1$, then good 1 is normal, as follows:

> Fix $p\_1^0$ and $p\_2^0$ and let $w' > w$. Let $a$ be a solution to the consumer's problem for the budget set $B((p\_1^0, p\_2^0), w)$ and let $a'$ be a bundle on the frontier of $B((p\_1^0, p\_2^0), w')$ with $a\_1' = a\_1$. By the assumption on the marginal rate of substitution we have $\text{MRS}(a') > \text{MRS}(a)$, and hence the solution $b$ of the consumer's problem for the budget set $B((p\_1^0, p\_2^0), w')$ has $b\_1 > a'\_1 = a\_1.$

I don't understand the very last inference. Why does the fact that $\text{MRS}(a') > \text{MRS}(a)$ imply the desired conclusion?

**Tags:** microeconomics, mathematical-economics, consumer-theory


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## 1 answers

### Answer by username_1 on 2024-06-24. Score: 1 | Accepted answer

Note that $(a\_1,a\_2)$ is the utility maximizing bundle at relative price $MRS(a) = (= p\_1/p\_2)$ and budget $w = MRS(a).a\_1 + a\_2$.

1. First as $b$ is utility maximising, $MRS(b) = \frac{p\_1^0}{p\_2^0}$. As $MRS(a') > MRS(b) = MRS(a)$ and the $MRS$ is unique (if utility is differentiable) we have that $a\_1' \ne b\_1$.
2. Now for the proof. Assume, towards a contradiction, that $b\_1 \le a\_1'$, which gives by point 1 above that $b\_1 < a\_1$. Then:
   $$
   \begin{align\*}
   MRS(a') (a\_1' - b\_1 ) + a\_2' - b\_2
   &= MRS(a') (a\_1' - b\_1) + w - \frac{p\_1^0}{p\_2^0} a\_1' - w + \frac{p\_1^0}{p\_2^0} b\_1,\\
   &= \underbrace{(MRS(a') - MRS(a))}\_{> 0} \underbrace{(a\_1' - b\_1)}\_{> 0} > 0
   \end{align\*}
   $$
   As such, we get that:
   $$
   MRS(a') a\_1' + a\_2' > MRS(a') b\_1 + b\_2.
   $$
   As $(a\_1', a\_2')$ is utility maximising for relative prices $MRS(a')$ and budget $w'' = MRS(a') a\_1' + a\_2'$ and $b$ was cheaper to buy at that price and budget, we obtain by a revealed preference argument, that $u(a\_1', a\_2') > u(b\_1, b\_2)$, contradicting the assumption that $(b\_1, b\_2)$ was utility maximising at the budget $B((p\_1^0,p\_2^0), w)$ and that $a' \in B((p\_1^0, p\_2^0), w)$.

> **Comments:**
> **username_0**: Thank you for the reply. Sorry, I still have some doubts: why are we allowed to let $\text{MRS}(b) = \text{MRS}(a) = p_1^0/p_2^0$ without assuming anything about the consumer's utility? If the optimal choice of either good 1 or good 2 is zero, the marginal rate of substitution need not be equal to the price ratio. Also, why is $(a_1', a_2')$ utility-maximising? This is not true neither by assumption nor by construction, no?  
> **username_0**: Plus, I don't understand the revealed-preference argument you made. Can't a contradiction be equally derived from the result that $\text{MRS}(a')a'_1 + a'_2 > \text{MRS}(b)b_1 + b_2$ by noting that by assumption $a'$ was on the frontier of $B((p_1^0, p_2^0), w')$ and $b$ was optimal (so by monotonicity also on the frontier), and so we would have $w' > w'$ which is clearly a contradiction? This would require $\text{MRS}(a')a'_1 + a'_2 = w'$ though, which relates to my previous question about how you know that $a'$ is optimal.  
> **username_1**: @username_0 For corner solutions, you need to be careful as as derivatives (of the utility function) are not defined on the boundary of the domain. As such, there is no MRS. The hyperplane through the (utility maximising) boundary point with slope $(p_1^0/p_2^0)$ is still tangent to the upper contour set. So one needs convex analysis to analyse such cases.  
> **username_1**: Revealed preference tells that if one bundle was chosen and a second bundle was also affordable for that budget but cheaper to buy, then the chosen bundle should give a higher utility than the second bundle.  
> **username_0**: Thanks again. Can you just clarify why $a'$ is optimal? By construction it's on the boundary of $w'$ but is not optimal ($b$ is).  
> **username_0**: Oh, also, why are marginal utilities not defined for corner solutions? If either optimal solution is zero (considering a two-dimensional commodity space), can't you just define one-sided derivatives?  
> **username_0**: Sorry if I keep bothering, can you please explain why $a' = (a'_1, a'_2)$ is optimal? Then the contradiction would immediately follow because we'd get $w' > w'$. But why is it utility-maximising? By construction it's on the boundary of $w'$, but that is not sufficient to guarantee optimality.  

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