Apologies in advance if my terms aren't exact, I'm learning "Mathmatical Economoics" in the hebrew language and some of the terms don't translate well.

I was given the following question: Two sellers are a part of an exchange market with two goods, each has the utility function $u(x,y)=\max(x,y)$. Assume the first seller starts with two units of product $x$ and one unit of $y$.

Find $(a,b)$ s.t seller two starts with $a$ units of $x$ and $b$ units of $y$, for which there is a competetive equilebrium.

Any guidence will be of great help! Thanks in advance

  • $\begingroup$ What are you confused about specifically? $\endgroup$ – Art Feb 25 '20 at 10:21

I might be wrong so feel free to point out any mistakes.

Both of them have the same utility function $u(x,y)=max(x,y)$ This means when $Px > Py$ , they'll choose to consume $Y$ instead of $X$ since they get higher satisfaction by selling all units of x and purchasing y.

Take the case of First seller, If he sells all X and purchases Y, His total demand for $Y$ will be $2(Px/Py) + 1$ and since Px/Py >1, total demand will be higher than 3. If he had done the opposite, his total demand will be less than 3. Since the utility function is $max(x,y)$ He will choose the former strategy

Hence, the demand for $X$ will be zero but the supply is $a+2$. Also, there is excess demand of $Y$. Hence, this is not a competitive equilibrium.

Similarly, then $Px <Py$, there will be excess demand for $X$ so that isnt a competitive equilibrium either.

The only case we have left is $Px=Py$ where the seller will be indifferent in consuming either good. Here, each seller consumes all of 1 good each. There will be 2 cases.

Case 1. If seller 1 consumes all $X$, His demand is $2 + (1)Py/Px = 3$ since Px=Py and the supply of X is $a+2$. When we equate this, we get $$a=1$$ Notice that this is the only condition that needs to hold here, B can take any non negative values in this case.

Case 2 is where seller 1 consumes all $Y$ and, His demand is $1+ (2)Px/Py = 3$ and the supply of $Y$ is $b+1$. Since they have to be equal, we get $$b=2$$ And again there's no restriction on $a$ and it can take any non negative values.

  • $\begingroup$ Thank you! I understand it now, finally :) $\endgroup$ – Gil Bar Koltun Feb 25 '20 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.