0
$\begingroup$

Let's say you have a Cobb-Douglas utility function, U = x^.1*y^.9

This will result in Marshallian demand functions x* = .1I / Px, and y* = .9I / Py.

If you perform a Slutsky decomposition for x* as Px changes; the income effect will be 10% of the total effect.

Can someone help me understand why the effect exactly matches the percent of income spent on the good?

$\endgroup$

1 Answer 1

0
$\begingroup$

The income effect is given by: $$ x^\ast \frac{\partial x}{\partial I} = \frac{0.1 I}{P_x} \frac{0.1}{P_x} = 0.1\left(0.1\frac{I}{(P_x)^2}\right). $$ The Price effect is: $$ \frac{\partial x}{\partial P_x} = -0.1 \frac{I}{(P_x)^2}.$$ The first is 10% of the second (in absolute terms).

Another way to see this, is to start from the fact that for the Cobb-Douglass, the expenditure share is constant: $$ \frac{p x(p,I)}{I} = \alpha. $$ Taking derivative with respect to $p$ gives: $$ \begin{align*} &\frac{x(p,I) + p \frac{\partial x}{\partial p}}{I} = 0,\\ \to &\frac{\partial x}{\partial p} = -\frac{x(p,I)}{p} \end{align*} $$ Taking the partial derivative with respect to $I$ gives: $$ \begin{align*} &\frac{p \frac{\partial x(p,I)}{\partial I} I - p x(p,I)}{I^2} = 0,\\ \to &\frac{\partial x(p,I)}{\partial I} = \frac{p x(p,I)}{pI} = \frac{\alpha}{p} \end{align*} $$ So the income effect is: $$ \frac{\partial x(p,I)}{\partial I} x(p,I) = \alpha \frac{x(p,I)}{p} $$ This is $\alpha$ times the absolute value of the price effect.

$\endgroup$
2
  • $\begingroup$ This makes sense, but is there any specific mathematical reason as to why the price effect is exactly the same as the share of income spent on the good? Or does it just happen to work that way? $\endgroup$
    – Aa Z
    Nov 13 at 19:39
  • $\begingroup$ @Aa Z I added some additional explanation. $\endgroup$
    – tdm
    Nov 14 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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