<p>This
<span class="math-container">$$
a\ln q_1+b\ln q_2=q_1^aq_2^b
$$</span>
is not true, only
<span class="math-container">$$
a\ln q_1+b\ln q_2=\ln(q_1^aq_2^b)
$$</span>
is true.</p>

<p><span class="math-container">$$
u(\lambda q_1, \lambda q_2)=\lambda u(q_1,q_2) 
$$</span>
is a sufficient, but not necessary condition for homotheticity of <span class="math-container">$u$</span>. You can find more on this by searching for questions with the word <a href="https://economics.stackexchange.com/search?q=homothetic">homothetic</a> on this site.</p>

<p>It is indeed true that 
<span class="math-container">$$
\lambda^{a+b}q_1^aq_2^b = \lambda q_1^aq_2^b
$$</span>
if <span class="math-container">$a+b = 1$</span>.</p>

CC BY-SA 4.0
2020-02-15T16:39:42.597Z
2020-02-15T16:39:42.597Z
Giskard
<p>I've been given the Cobb-Douglas utility function:</p>

<p><span class="math-container">$\ u(q_1, q_2)=a\ln q_1+b\ln q_2=q_1^aq_2^b  \ $</span></p>

<p>If I want to prove homothetic preferences, I use the following condition:</p>

<p><span class="math-container">$\ u(\lambda q_1, \lambda q_2)=\lambda u(q_1,q_2)  \ $</span></p>

<p>According to my calculation, this yields:</p>

<p><span class="math-container">$\ u(\lambda q_1, \lambda q_2)=(\lambda q_1)^a(\lambda q_2)^b  \ $</span></p>

<p><span class="math-container">$\ =\ \lambda^{a+b}q_1^aq_2^b   \ $</span></p>

<p>Is it correct to say that as a result, preferences are homothetic if and only if <span class="math-container">$\ a+b=1  \ $</span>?</p>

<microeconomics><utility><consumer-theory><preferences><cobb-douglas>
Cobb-Douglas function homotheticity
Anon
# Cobb-Douglas function homotheticity

(Asked by: username_0 on 2020-02-15)

I've been given the Cobb-Douglas utility function:

$\ u(q\_1, q\_2)=a\ln q\_1+b\ln q\_2=q\_1^aq\_2^b \ $

If I want to prove homothetic preferences, I use the following condition:

$\ u(\lambda q\_1, \lambda q\_2)=\lambda u(q\_1,q\_2) \ $

According to my calculation, this yields:

$\ u(\lambda q\_1, \lambda q\_2)=(\lambda q\_1)^a(\lambda q\_2)^b \ $

$\ =\ \lambda^{a+b}q\_1^aq\_2^b \ $

Is it correct to say that as a result, preferences are homothetic if and only if $\ a+b=1 \ $?

**Tags:** microeconomics, utility, consumer-theory, preferences, cobb-douglas

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## 1 answers

### Answer by username_1 on 2020-02-15. Score: 1

This
$$
a\ln q\_1+b\ln q\_2=q\_1^aq\_2^b
$$
is not true, only
$$
a\ln q\_1+b\ln q\_2=\ln(q\_1^aq\_2^b)
$$
is true.

$$
u(\lambda q\_1, \lambda q\_2)=\lambda u(q\_1,q\_2)
$$
is a sufficient, but not necessary condition for homotheticity of $u$. You can find more on this by searching for questions with the word [homothetic](https://economics.stackexchange.com/search?q=homothetic) on this site.

It is indeed true that
$$
\lambda^{a+b}q\_1^aq\_2^b = \lambda q\_1^aq\_2^b
$$
if $a+b = 1$.


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