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I'm solving a problem where wealth changes and I got that $EV=0$ and $CV=1$. This seems impossible to me; is it?

I'm given $e(p,u)=u/(p_1^{-1}+p_2^{-1})$.

I'm also given $(p_1^0,p_2^0,w^0)=(1,1,1), (p_1^1,p_2^1,w^1)=(1,1,2)$.

First, I used duality to solve for $v(p,w)$, getting that $v(p,w)=w(p_1^{-1}+p_2^{-1})$.

I found $EV=v(p^1,w^1)/(1/p_1^0+1/p_2^0)-w^0=0$.

I then found $CV=w^1-v(p^0,w^0)/(1/p_1^1+1/p_2^1)=1$.

Did I do something wrong?

leecarvallo
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    Could you please type the question and your attempt at solving it. It helps getting your question more attention. Also, try to type it in Math Jax. – Stackman Oct 02 '18 at 20:24
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    Have you made a typo? The price vectors are the same before and after: $p^0=p^1=(1,1)$ (yet your question says both prices and wealth change). This means the EV and CV are both equal to the change in wealth (i.e. $1$), as calculated below. – smcc Oct 02 '18 at 22:07
  • @smcc Yes, corrected! – leecarvallo Oct 02 '18 at 22:17

1 Answers1

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We have $$e(p,v(p,w))=w$$ which in this particular case gives $$\frac{v(p,w)}{p_1^{-1}+p_2^{-1}}=w \iff v(p,w)=w(p_1^{-1}+p_2^{-1})$$ as you claimed.

Here $u^0=v(p^0,w^0)=1(1+1)=2$ and $u^1=v(p^1,w^1)=2(1+1)=4$.

The equivalent variation is $$e(p^0,u^1)-e(p^0,u^0)=e(p^0,u^1)-w^0=\frac{4}{1+1}-1=1$$

The compensating variation is $$e(p^1,u^1)-e(p^1,u^0)=w^1-e(p^1,u^0)=2-\frac{2}{1+1}=1$$

Just to add: none of the calculation above is necessary when prices do not change, since then the equivalent variation and compensating variation are both equal to the change in wealth:

When $p^0=p^1$, the equivalent variation is $$e(p^0,u^1)-e(p^0,u^0)=e(p^1,u^1)-e(p^0,u^0)=w^1-w^0$$ and the the compensating variation is $$e(p^1,u^1)-e(p^1,u^0)=e(p^1,u^1)-e(p^0,u^0)=w^1-w^0.$$

smcc
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