<p>Monir is correct to point out in the comments that $\ln$ is not a linear transformation. It is an increasing transformation, however, and so should not matter as long as</p>

<p>$$ \sum_{i=1}^n p_i \gamma_i &lt; I$$</p>

<p>In any case, your mistake is in the derivatives you find. Here are the correct expressions:</p>

<p>$$\frac{\partial W}{\partial p_i} = -\gamma_i \sum_{j=1}^n \frac{\beta_j}{I - \sum_{k=1}^n p_k \gamma_k} - \frac{\beta_i}{p_i} = -\frac{\gamma_i}{I - \sum_{j=1}^n p_j \gamma_j} - \frac{\beta_i}{p_i} $$</p>

<p>$$ \frac{\partial W}{\partial I} = \sum_{j=1}^n \frac{\beta_j}{I - \sum_{k=1}^n p_k \gamma_k} = \frac{1}{I - \sum_{j=1}^n p_j \gamma_j} $$</p>

<p>Roy's identity then gives you the Marshallian demand you've initially, and correctly, derived.</p>

CC BY-SA 3.0
2017-02-12T23:51:08.270Z
2017-02-12T23:58:11.300Z
2017-02-12T23:58:11.300Z
Theoretical Economist
<p>I'm having trouble showing Roy's identity for the following Stone-Geary utility function:</p>

<p>$$U(x)=\prod_{i=1}^n\left(x_i-\gamma_i\right)^{\beta_i}$$</p>

<p>where $\sum \beta_i=1$ and $\gamma_i$ is the minimal consumption of $x_i$.</p>

<p>I showed that the Marshallian demand (which I confirmed using multiple sources) is</p>

<p>$$x_i=\gamma_i+\frac{\beta_i\left(I-\sum_{j=1}^np_j\gamma_j\right)}{p_i}$$</p>

<p>Therefore, the indirect utility function is </p>

<p>$$V(x)=\prod_{i=1}^n\left(\frac{\beta_i \left(I-\sum_{j=1}^np_j\gamma_j \right)}{p_i} \right)^{\beta_i}$$</p>

<p>I applied a monotone transformation to simplify the indirect utility function:</p>

<p>$$W(x)=\sum_{i=1}^n\beta_i\left(\left(\ln(\beta_i)+\ln \left(I-\sum_{j=1}^np_j\gamma_j \right)\right)-\ln(p_i)\right)$$</p>

<p>Therefore,</p>

<p>$$\frac{\delta W}{\delta p_i}=\frac{-\beta_i\gamma_i}{I-\sum_{j=1}^np_j\gamma_j}-\frac{\beta_i}{p_i}$$</p>

<p>$$\frac{\delta W}{\delta I}=\frac{\beta_i}{I-\sum_{j=1}^np_j\gamma_j}$$</p>

<p>$$-\frac{\delta W / \delta p_i}{\delta W / \delta I}=\gamma_i+\frac{I-\sum_{j=1}^np_j\gamma_j}{p_i}$$</p>

<p>Basically, as you can see, something doesn't add up: the $\beta_i$ is missing from the numerator. Thus, Roy's inequality is not verified. Where did I mess up?</p>

<p>(Note: I've also tried without the transformation. This is a much more tedious process but it yielded the same answer; the $\beta_i$ was still missing)</p>

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2017-02-12T21:51:53.967Z
Have you noticed that your last expression for $V(x)$ is actually $\ln(V(x))$? Differentiating the latter is not equivalent to differentiating the former.
Elias
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2017-02-12T22:11:26.560Z
@Monir It's a linear transformation, shouldn't it not matter?
Grizzly0111
CC BY-SA 3.0
2017-02-12T22:21:57.540Z
To me it's not linear. For example, $\ln(a+b)\neq\ln(a)+\ln(b)$. Also, the derivative of $\ln(V(x))$ is $V'(x)/V(x)$ which is different from the derivative of $V(x)$, i.e. $V'(x)$.
Elias
CC BY-SA 3.0
2017-02-12T23:47:58.107Z
@Monir You are right: I meant monotone transformation, not linear. Also, I should have changed the name of my transformed utility function. I'll edit the question; [Refer to this](http://economics.stackexchange.com/questions/6982/monotone-transformation-of-utility)
Grizzly0111
<microeconomics><mathematical-economics>
Roy's identities for Stone-Geary utility functions
Grizzly0111
# Roy's identities for Stone-Geary utility functions

(Asked by: username_0 on 2017-02-12)

I'm having trouble showing Roy's identity for the following Stone-Geary utility function:

$$U(x)=\prod\_{i=1}^n\left(x\_i-\gamma\_i\right)^{\beta\_i}$$

where $\sum \beta\_i=1$ and $\gamma\_i$ is the minimal consumption of $x\_i$.

I showed that the Marshallian demand (which I confirmed using multiple sources) is

$$x\_i=\gamma\_i+\frac{\beta\_i\left(I-\sum\_{j=1}^np\_j\gamma\_j\right)}{p\_i}$$

Therefore, the indirect utility function is

$$V(x)=\prod\_{i=1}^n\left(\frac{\beta\_i \left(I-\sum\_{j=1}^np\_j\gamma\_j \right)}{p\_i} \right)^{\beta\_i}$$

I applied a monotone transformation to simplify the indirect utility function:

$$W(x)=\sum\_{i=1}^n\beta\_i\left(\left(\ln(\beta\_i)+\ln \left(I-\sum\_{j=1}^np\_j\gamma\_j \right)\right)-\ln(p\_i)\right)$$

Therefore,

$$\frac{\delta W}{\delta p\_i}=\frac{-\beta\_i\gamma\_i}{I-\sum\_{j=1}^np\_j\gamma\_j}-\frac{\beta\_i}{p\_i}$$

$$\frac{\delta W}{\delta I}=\frac{\beta\_i}{I-\sum\_{j=1}^np\_j\gamma\_j}$$

$$-\frac{\delta W / \delta p\_i}{\delta W / \delta I}=\gamma\_i+\frac{I-\sum\_{j=1}^np\_j\gamma\_j}{p\_i}$$

Basically, as you can see, something doesn't add up: the $\beta\_i$ is missing from the numerator. Thus, Roy's inequality is not verified. Where did I mess up?

(Note: I've also tried without the transformation. This is a much more tedious process but it yielded the same answer; the $\beta\_i$ was still missing)

**Tags:** microeconomics, mathematical-economics

> **Comments:**
> **username_1**: Have you noticed that your last expression for $V(x)$ is actually $\ln(V(x))$? Differentiating the latter is not equivalent to differentiating the former.  
> **username_0**: @Monir It's a linear transformation, shouldn't it not matter?  
> **username_1**: To me it's not linear. For example, $\ln(a+b)\neq\ln(a)+\ln(b)$. Also, the derivative of $\ln(V(x))$ is $V'(x)/V(x)$ which is different from the derivative of $V(x)$, i.e. $V'(x)$.  
> **username_0**: @Monir You are right: I meant monotone transformation, not linear. Also, I should have changed the name of my transformed utility function. I'll edit the question; [Refer to this](http://economics.stackexchange.com/questions/6982/monotone-transformation-of-utility)  

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## 1 answers

### Answer by username_2 on 2017-02-12. Score: 7 | Accepted answer

Monir is correct to point out in the comments that $\ln$ is not a linear transformation. It is an increasing transformation, however, and so should not matter as long as

$$ \sum\_{i=1}^n p\_i \gamma\_i < I$$

In any case, your mistake is in the derivatives you find. Here are the correct expressions:

$$\frac{\partial W}{\partial p\_i} = -\gamma\_i \sum\_{j=1}^n \frac{\beta\_j}{I - \sum\_{k=1}^n p\_k \gamma\_k} - \frac{\beta\_i}{p\_i} = -\frac{\gamma\_i}{I - \sum\_{j=1}^n p\_j \gamma\_j} - \frac{\beta\_i}{p\_i} $$

$$ \frac{\partial W}{\partial I} = \sum\_{j=1}^n \frac{\beta\_j}{I - \sum\_{k=1}^n p\_k \gamma\_k} = \frac{1}{I - \sum\_{j=1}^n p\_j \gamma\_j} $$

Roy's identity then gives you the Marshallian demand you've initially, and correctly, derived.


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