I'm currently a TA for a class and recently graded a midterm. I gave the answer key back to the teacher, after going over part of the exam in a study hall. I was going to go over the rest of it tomorrow, but while making my own answer key in office hours, I seem to have come to a different answer than the teacher.

We maximize $$u = 2x_1^{1/2} + 4x_2^{1/2}$$ with a normal budget constraint where $p \cdot x \leq w$. We arrive at the Walrasian demand:

$$x^*(p, w) = \left(\frac{p_2 w}{p_1(4p_1 + p_2)} , \frac{4p_1 w}{p_2(4p_1 + p_2)}\right)$$

Suppose $w = 10, \vec p = (1, 4), \vec p' = (3, 2)$.

Thus, $x(p', w) = (\frac{10}{21}, \frac{30}{7})$, and $x(p, w) = (5, \frac{5}{4})$ for our new and old bundles respectively.

So to find compensating variation we find the original utility:

$2 \cdot 5^{1/2} + 4 \cdot (5/4)^{1/2} = 4 \sqrt 5$

and find $w'$ that would get old utility and new prices:

$4 \sqrt 5 = 2(\frac{w'}{21})^{1/2} + 4(\frac{3w'}{7})^{1/2} = 2(\frac{w'}{21})^{1/2} + 12(\frac{w'}{21})^{1/2} = 14(\frac{w'}{21})^{1/2} \implies \\ 4 \sqrt 5 = 14(\frac{w'}{21})^{1/2} \\ 80 = 14^2 \cdot \frac{w'}{21} \\ \boxed{w' = \frac{60}{7}}$

Thus $\boxed{CV = w - w' = 10 - \frac{60}{7} = \frac{10}{7}}$

To find equivalent variation we find the new utility:

$2 \cdot (10/21)^{1/2} + 4 \cdot (30/7)^{1/2} = 2 \cdot (10/21)^{1/2} + 12 \cdot (10/21)^{1/2} = 14 \sqrt {\frac{10}{21}}$

and find $\hat w$ that would get new utility at old prices:

$14 \sqrt {\frac{10}{21}} = 2(\frac{\hat w}{2})^{1/2} + 4(\frac{\hat w}{8})^{1/2} = 4(\frac{\hat w}{2})^{1/2} \\ 14^2 \cdot \frac{10}{21} = 16 \cdot \frac{\hat w}{2} \\ \boxed{\hat w = \frac{70}{6}}$

Thus $\boxed{EV = \hat w - w = \frac{70}{6} - 10 = \frac{5}{3}}$

The problem is if I recall correctly, the CV and EV are supposed to have the opposite sign so that the change in welfare is ambiguous. Where have I gone wrong, if anywhere? (Worth noting that if you do Slutsky decomposition for this question, you find that good 2 is inferior.)


1 Answer 1


Questions with numbers are usually not as good as questions without numbers. If you had written down the formula for CV and EV you would probably have noticed that your premise is false.

CV and EV are not supposed to have opposing signs. You can see this from their definitions where $$ CV = e(p_1,u_1) - e(p_1,u_0), \hskip 20pt EV = e(p_0,u_1) - e(p_0,u_0). $$ Either $u_1 > u_0$ and then a larger income is needed to reach $u_1$ given any price or the opposite is true.

In your final comment you also note that good 2 is inferior. As can be seen from your demand function

$$ x^*(p, w) = \left(\frac{p_2 w}{p_1(4p_1 + p_2)} , \frac{4p_1 w}{p_2(4p_1 + p_2)}\right) $$

this is not true. If income were to increase $x_2^*(p, w)$ would increase as well.

  • $\begingroup$ For CV, EV, that was my intuition as well. For the second good though, once the price of good 2 falls, the income effect is negative, and my impression was that for inferior goods, the income effect moves in the same direction as price. At any rate I'll have to ask my teacher for the answer key again to see if I am just mis-remembering the answers. But I feel like I'm not. $\endgroup$
    – Kitsune Cavalry
    Oct 14, 2016 at 18:03
  • $\begingroup$ It follows from the demand function that if $p_2$ falls the consumer will buy more of $x_2$. You can divide this positive total effect into an income and substition effect, but as the preferences here are of the perfect complement variety the substition effect will be zero. So the income effect has to be positive, like the total effect. $\endgroup$
    – Giskard
    Oct 14, 2016 at 18:54
  • 1
    $\begingroup$ I have to disagree that the preferences are perfect complements. $\min \alpha x_1, \beta x_2$ Maybe they'd be closer to substitutes, but the exponents give the utility different properties. Also it's worth noting that I left numbers in in case my mistake was an arithmetic error. $\endgroup$
    – Kitsune Cavalry
    Oct 14, 2016 at 20:16
  • 1
    $\begingroup$ You are right, I have no either why I thought they were perfect complements... :) $\endgroup$
    – Giskard
    Oct 14, 2016 at 22:04

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