<p>Use Roy's identity to find demand functions:
$$\displaystyle x_1=-\frac{\frac{\partial v}{\partial p_1}}{\frac{\partial v}{\partial y}} =-\frac{\alpha yp_1^{\alpha-1} p_2^\beta}{p_1^\alpha p_2^\beta}=-\frac{\alpha y }{p_1} $$</p>

<p>$$\displaystyle x_2=-\frac{\frac{\partial v}{\partial p_2}}{\frac{\partial v}{\partial y}} =-\frac{\beta yp_1^{\alpha} p_2^{\beta-1}}{p_1^\alpha p_2^\beta}=-\frac{\beta y }{p_2} $$</p>

<p>Since $p_1x_1 + p_2x_2 = y$, from the above demand expressions we get $-\alpha -\beta = 1$.
Demand equations above can also be rewritten as:</p>

<p>$$x_1^{-\alpha}  = (-\alpha y)^{-\alpha}p_1^\alpha$$
$$x_2^{-\beta}  = (-\beta y)^{-\beta}p_2^\beta$$</p>

<p>Multiplying them we obtain the following:
$$x_1^{-\alpha}x_2^{-\beta} = (-\alpha y)^{-\alpha}p_1^\alpha (-\beta y)^{-\beta}p_1^\beta = (-\alpha)^{-\alpha}(-\beta)^{-\beta}yp_1^\alpha p_2^\beta $$</p>

<p>Comparing it with the indirect utility function, we get the direct utility function as:</p>

<p>$$u(x_1, x_2) = \frac{x_1^{-\alpha}x_2^{-\beta}}{(-\alpha)^{-\alpha}(-\beta)^{-\beta}}$$</p>

CC BY-SA 3.0
2017-05-12T08:32:20.557Z
2017-05-12T08:32:20.557Z
Amit
<p>Derive the consumer’s direct utility function if his indirect utility function has the form</p>

<p>$$v(\boldsymbol{\mathbf{p}},y)=yp_1^\alpha p_2^\beta$$</p>

<p>where $\beta ,\alpha$ are negative. </p>

<microeconomics><utility>
How to derive utility function from indirect utility in this exercise
Andrea F.L.
# How to derive utility function from indirect utility in this exercise

(Asked by: username_0 on 2017-05-12)

Derive the consumer’s direct utility function if his indirect utility function has the form

$$v(\boldsymbol{\mathbf{p}},y)=yp\_1^\alpha p\_2^\beta$$

where $\beta ,\alpha$ are negative.

**Tags:** microeconomics, utility


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## 1 answers

### Answer by username_1 on 2017-05-12. Score: 13

Use Roy's identity to find demand functions:
$$\displaystyle x\_1=-\frac{\frac{\partial v}{\partial p\_1}}{\frac{\partial v}{\partial y}} =-\frac{\alpha yp\_1^{\alpha-1} p\_2^\beta}{p\_1^\alpha p\_2^\beta}=-\frac{\alpha y }{p\_1} $$

$$\displaystyle x\_2=-\frac{\frac{\partial v}{\partial p\_2}}{\frac{\partial v}{\partial y}} =-\frac{\beta yp\_1^{\alpha} p\_2^{\beta-1}}{p\_1^\alpha p\_2^\beta}=-\frac{\beta y }{p\_2} $$

Since $p\_1x\_1 + p\_2x\_2 = y$, from the above demand expressions we get $-\alpha -\beta = 1$.
Demand equations above can also be rewritten as:

$$x\_1^{-\alpha} = (-\alpha y)^{-\alpha}p\_1^\alpha$$
$$x\_2^{-\beta} = (-\beta y)^{-\beta}p\_2^\beta$$

Multiplying them we obtain the following:
$$x\_1^{-\alpha}x\_2^{-\beta} = (-\alpha y)^{-\alpha}p\_1^\alpha (-\beta y)^{-\beta}p\_1^\beta = (-\alpha)^{-\alpha}(-\beta)^{-\beta}yp\_1^\alpha p\_2^\beta $$

Comparing it with the indirect utility function, we get the direct utility function as:

$$u(x\_1, x\_2) = \frac{x\_1^{-\alpha}x\_2^{-\beta}}{(-\alpha)^{-\alpha}(-\beta)^{-\beta}}$$


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