Do perfect complements have to be normal goods? If so, why?

Two goods $x,y$ are perfect complements if they have the utility function $$U(x,y) = \min \lbrace ax,by \rbrace$$ $$a,b \in \Bbb{Q}^+$$ My professor said $x,y$ have to be normal goods but didn't explain why well enough that I could understand.

My Question:

Are perfect complements always normal goods? If so, why?

• Do you know how to derive the demand function? Do you know the definition of normal goods? – Michael Greinecker May 11 '15 at 21:08
• Normal good: $$\frac{\partial x}{\partial m} > 0$$ Perfect Complements Demand Function: $$x = \frac{\bar{U}}{a}$$ Not sure about the latter, but that's what I have in my notes. – Stan Shunpike May 11 '15 at 21:12
• Oh, I see your point. I should just take the derivative of demand wrt to $m$ and I might learn more about its behavior. But my demand function is stated in terms of utility. That's why it didn't occur to me to take the derivative wrt to $m$. I am not sure how to derive the perfect complements demand function wrt $m$. – Stan Shunpike May 11 '15 at 21:17
• Ah, they discuss it here. youtu.be/kjI840VDW5I perhaps i can figure it out now. – Stan Shunpike May 11 '15 at 21:20

A good is normal if its demand is increasing in income. So let $p_x$ and $p_y$ be the price of the goods with quantities $x$ and $y$ and let $m$ be income.
Suppose $ax>by$. Then $\min\{ax,by\}=by$. By slightly reducing $x$ by and spending the saved money on $y$, one gets a better bundle. For an optimal bundle, this cannot be.
Similarly, it cannot be optimal that $by>ax$. So in the optimal consumption bundle, it must be the case that $ax=by$. It is also not that hard to see that the consumer will spend all her income. So rewrite the condition as $$y=\frac{a}{b}x$$ and plug it into the budget equation $$p_x x+p_y y=m$$ to get $$p_x x+ p_y\frac{a}{b}x=m=x\Big(p_x+p_y\frac{a}{b}\Big).$$ Therefore, we get the demand function given by $$x(p_x,p_y,m)=\frac{m}{p_x+p_y\frac{a}{b}},$$ which is clearly increasing in $m$. Similarly, one shows that the other good is normal too.
Pedantic remark: A differentiable function can be increasing at every point without the derivative being strictly positve everywhere. The function given by $x\mapsto x^3$ has derivative $0$ at $0$ but is everywhere increasing.
• How does your last remark apply here? Are you saying $$\frac{\partial x}{\partial m}$$ here is not increasing everywhere in this case? – Stan Shunpike May 11 '15 at 21:41
• @snoram: I suspect it does not have to be too complicated. A utility function like $U(x,y,z)=\sqrt{\min\{a{x},b{y}\}} + cz^2$ might be enough – Henry Dec 18 '17 at 14:26