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?
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 $$e(p^1,u^1)-e(p^1,u^0)=e(p^1,u^1)-e(p^0,u^0)=w^1-w^0.$$