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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)

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  • $\begingroup$ 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. $\endgroup$
    – Elias
    Commented Feb 12, 2017 at 21:51
  • $\begingroup$ @Monir It's a linear transformation, shouldn't it not matter? $\endgroup$
    – Grizzly0111
    Commented Feb 12, 2017 at 22:11
  • $\begingroup$ 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. $V'(x)$. $\endgroup$
    – Elias
    Commented Feb 12, 2017 at 22:21
  • $\begingroup$ @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 $\endgroup$
    – Grizzly0111
    Commented Feb 12, 2017 at 23:47

1 Answer 1

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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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