# Why is EV<CV if the good is inferior?

Through the Slutsky equation I know that if the good is inferior the marshallian demand function is steeper than the hicksian demand but I cannot understand why the Compensating variation is higher than the equivalent variation. Is this because the increase in utility shifts the hicksian to the left instead of shifting it to the right (in a graph with prices as y and demand as x)?

• No, EV>(ConsumerSurplus>)CV if the good is normal and the other way around if it is inferior. I am just struggling with justifying this last assertion. Nov 1, 2016 at 20:55
• I believe you are mistaken. In the symmetric Cobb-Douglas case $U(x,y) = x \cdot y$ none of the goods are inferior. Yet if you have income $m > 0$, prices $p_x = p_y = 1$ and altered price $p_x' = 4$ you will have $CV = m > \frac{m}{2} = EV$. Nov 1, 2016 at 22:04
• I think it's possible to have EV=CV=CS, when there is no wealth effect. Let me mess around with a utility function that's quasilinear in one good to double-check. Dec 2, 2016 at 0:49

I believe Marshallian demands are less steep than Hicksian demands because we reverse the y and x axis in economics. Thus a larger derivative of x with respect to p will be less steep since p is on the vertical and x is on the horizontal. (image from here)

• You are completely right but does that imply a shift in the welfare inequalities if the good is inferior? (Turning EV>CS>CV into CV>CS>CV) Nov 2, 2016 at 14:23

The answer comes from looking at the Hicksian compensated demand:

Since, WLOG, EV for only one good price change can be written as $$EV(p_1,p_0,u) = \int_{p_1}^{p_0} h(p,u_1)dp$$ and CV as $$CV(p_1,p_0,u) = \int_{p_1}^{p_0} h(p,u_0)dp,$$ and we know that $$\frac{\partial h}{\partial u} \leq 0$$ when the good is inferior (using the fact that $$\frac{\partial x}{\partial w}=\frac{\partial h}{\partial u}\frac{\partial v}{\partial w} \leq 0$$ and $$\frac{\partial v}{\partial w}\geq0$$, coming from the identity $$x(p,w)=h(p,v(p,w))$$), we can use the monotonicity of the integral to establish that $$EV(p_1,p_0,u) = \int_{p_1}^{p_0} h(p,u_1)dp \leq \int_{p_1}^{p_0} h(p,u_0)dp = CV(p_1,p_0,u)$$ when $$p_1.

For the case when $$p_1 > p_0$$ just notice that $$u_1 \leq u_0$$, where $$u_i = v(p_i,w)$$, $$\implies$$ $$h(p,u_1) \geq h(p,u_1)$$, so that $$-EV(p_1,p_0,u) = \int_{p_0}^{p_1} h(p,u_1) dp \geq \int_{p_0}^{p_1} h(p,u_0) dp = - CV(p_1,p_0,u).$$

$$\square$$