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Consider an exchange economy with two consumers A and B, two goods x, y. Consumers have identical preferences $U=xy$. Total available endowment are $E_x$ and $E_y$.

This is the background of a homework question which does not ask but I got this question when I try to draw the Edgeworth Box.

I find the contract curve to be the diagonal of the Edgeworth Box, with slope $\dfrac{E_y}{E_x}$. I also find the price in this economy to be $\dfrac{p_x}{p_y}=\dfrac{E_y}{E_x}$.

Then I got confused that if we start from an endowment point (for example, at the bottom right corner) and trade to the contract curve to get an equilibrium, then the line we trade along is called the priceline right? And its slope should be $-\dfrac{p_x}{p_y}$ right? If I did all these correct, I find my priceline is not orthogonal to the contract curve.

It is more likely I made something conceptually wrong? Is the contract curve in this economy the diagonal?

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You write that the slope of one line is px/py and of another is -px/py. Take a piece of paper and draw the line y=2x and y=-2x are they orthogonal (answer: no)? In particular case the lines y=x and y=-x are orthogonal.

If we regard it more formally, the vectors are orthogonal if the product is 0. Thus,

<1,-p>*<1,p> = 1-p^2 

This is 0 only if p=1 or p=-1, i.e. if the quantities of the 1-st and 2-nd good are equal.

To take a loon on it. Consider the wealth distribution something like 1-st player: [5; 15], 2-nd: [25;5], thus ex=30,ey=20,px/py = 2/3, then it looks like this: Edgeworth Box

p.s. Calculations and plotting were made with pyEdgeworthBox package (I wrote it).

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  • $\begingroup$ Hi, I did not say they are orthogonal, you misunderstood me. Also, I think px/py = 2/3? $\endgroup$ – Bob Aug 4 '15 at 13:54
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    $\begingroup$ yep, you're right, it's 2/3. Where I got it wrong with orthogonality (the title of the question "Does contract curve orthogonal to the priceline?")? $\endgroup$ – Max Li Aug 4 '15 at 15:22
  • $\begingroup$ concerning the rest, yes it's diagonal and you got it right conceptionally $\endgroup$ – Max Li Aug 4 '15 at 20:23
  • $\begingroup$ Hi, I think I saw somewhere that budget line should be perpendicular to the contract curve, but here the budget line (slope is px/py=ey/ex) is not perpendicular to the contract curve (with slope ey/ex also), I got confused. $\endgroup$ – Bob Aug 5 '15 at 4:11

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