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


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


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:

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


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

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

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

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)

  • $\begingroup$ 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 Feb 12 '17 at 21:51
  • $\begingroup$ @Monir It's a linear transformation, shouldn't it not matter? $\endgroup$ – Grizzly0111 Feb 12 '17 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 Feb 12 '17 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 Feb 12 '17 at 23:47

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.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.