<p>To get the Hicksian demand function <span class="math-container">$h$</span>, you can use Shephard's lemma, which says that</p>
<p><span class="math-container">$$h_i(p_1,p_2,u)=\frac{\partial e(p_1,p_2,u)}{\partial p_i}$$</span></p>
<p>where <span class="math-container">$e$</span> is the expenditure function.</p>
<p>To get the expenditure function, use the equation:</p>
<p><span class="math-container">$$v\Big(p_1,p_2,e(p_1,p_2,u)\Big)=u$$</span></p>
<p>i.e. solve</p>
<p><span class="math-container">$$v(p_1,p_2,R)=\frac{R}{p_1+p_2}=u$$</span> for <span class="math-container">$R$</span> and call the resulting expression <span class="math-container">$e(p,u)$</span>.</p>

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2023-09-08T17:11:13.500Z
2023-09-08T17:11:13.500Z
smcc
<p>A consumer has an indirect utility function given by</p>
<p><span class="math-container">$$v(p_1, p_2, R)=\frac{R}{p_{1}+p_2}$$</span></p>
<p>Where <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> denote the prices of the two goods consumed by the individual and <span class="math-container">$R$</span> the income.</p>
<p>How do I calculate this consumer's Hicksian demand function for goods 1 and 2?</p>

<microeconomics><utility><demand>
How to calculate direct utility from indirect utility in this exercise?
Arthur Pavezzi
# How to calculate direct utility from indirect utility in this exercise?

(Asked by: username_0 on 2023-09-08)

A consumer has an indirect utility function given by

$$v(p\_1, p\_2, R)=\frac{R}{p\_{1}+p\_2}$$

Where $p\_1$ and $p\_2$ denote the prices of the two goods consumed by the individual and $R$ the income.

How do I calculate this consumer's Hicksian demand function for goods 1 and 2?

**Tags:** microeconomics, utility, demand

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## 1 answers

### Answer by username_1 on 2023-09-08. Score: 2 | Accepted answer

To get the Hicksian demand function $h$, you can use Shephard's lemma, which says that

$$h\_i(p\_1,p\_2,u)=\frac{\partial e(p\_1,p\_2,u)}{\partial p\_i}$$

where $e$ is the expenditure function.

To get the expenditure function, use the equation:

$$v\Big(p\_1,p\_2,e(p\_1,p\_2,u)\Big)=u$$

i.e. solve

$$v(p\_1,p\_2,R)=\frac{R}{p\_1+p\_2}=u$$ for $R$ and call the resulting expression $e(p,u)$.


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