Isi 2016

classic Classic list List threaded Threaded
11 messages Options
Reply | Threaded
Open this post in threaded view
|

Isi 2016

Smith
In the ISI 2016 PEB Cake Division Question
 Q8 iii) a) Does the answer to either 8i) or 8 ii) change?
For 8 ii) shouldn't the utility be such that Utility of A is equal to Utility of B, because at that point, both will not have incentive to deviate, so that alpha= (1-μ)/(2-μ)  and beta= 1/(1-μ).

Please confirm!!!
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

VanHelsing
I don't know if this is correct, but this is how I worked it out:

A's utility is given by uA = alpha + mu(beta). Now since the amount of cake is fixed and taken to be one, we can write beta as beta = 1-alpha. Substituting in Agent A's utility function, we get uA = mu + ( 1- mu)alpha. In part a), it is given that mu<1. In that case, Agent A's utility is increasing in alpha. Agent B's utility function remains the same, it is increasing in beta. This is exactly the same situation we faced in parts i) and ii). So A will still be concerned with maximizing the size of his own piece of cake. In part ii), A manages to do this by dividing the cake into exactly two equal halves. He will do the same here.

Can you tell me what you did about 8 iii) b ?
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Smith
Your answer seems right to me, but i don't get what is the fallacy in my approach. In part ii) also if you notice we get theta= 1/2 by equating alpha and beta and putting it in alpha+beta=1, s.t, alapha+alpha=1 and alpha =beta = 1/2
I have used the same approach for this part, then why is it seeming incorrect.

For μ>1 The only pareto optimal point would be 0,1 i.e B consuming the whole cake, because in this case, A is too altruistic and is giving more weightage to B's consumption than to his own, so even at 0,1 A will manage to get more utility than B and B would be at his maximum too. Hence this point is the equilibrium point as well.
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

VanHelsing
This is just an intuitive answer, I haven't worked out a rigorous proof of it, but I think the mistake you are making is that you are assuming that if both utilities are equal, the agents will have no reason to deviate. I feel that the only reason agent's will not have an incentive to deviate is when they cannot attain higher utility  by moving to a different allocation.

Am still working on a mathematical proof. Will let you know.
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Asd1995
In reply to this post by Smith
Can you please share the question?
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Econenthusiast
8. Consider a cake of size 1 which can be divided between two individuals, A and B. Let α (resp. β) be the amount allocated to A (resp. B), where α+β = 1 and 0 ≤ α,β ≤ 1. Agents A’s utility function is uA(α) = α and that of agent B is uB(β) = β.
(i) What is the set of Pareto optimal allocations in this economy?
(ii) Suppose A is asked to cut the cake in two parts, after which B can choose which of the two segments to pick for herself, leaving the other segment for agent A. How should A cut the cake?
(iii) Suppose A is altruistic, and his utility function puts weight on what B obtains, i.e. uA(α,β) = α+μβ, where μ is the weight on agent B’s utility. (a) If 0 < μ < 1, does the answer to either 8(i) or 8(ii) change?
(b) What if μ > 1?
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Asd1995
1. Everything is PO if the whole cake is utilized.
2. Cut it in half, otherwise you'll always get the smaller part.
3. mu <1 means that A still gets more marginal utility from having the cake herself than giving it to B. Notihng changes. Mathematically A optimises alpha plus mu - mu*alpha, where coefficient of alpha remains positive if mu <1

4. If mu>1 then A gets mroe marginal utility from giving the cake to B than having it himself, so he will never eat any cake himself ( he can always gain by giving it to B). So PO allocation is where there is nothing given to A ie B gets all of the cake.

A should not cut the cake: He is happiest if B gets the whole portion.
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Econenthusiast
Thanks Asd!!
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Asd1995
In reply to this post by Smith
The fallacy is that UA=UB does not determine anything. Utility functions are just a way to represent preferences: and pareto optimality etc. are dependent on indifference curve structures and slopes- which are independent of how you give your utility function. Eg. U=x and U=5x represent the same preferences.

Consider, for example, utility of A being 10*alpha instead of alpha. This is just a monotonic transformation of the original utility function, hence, it represents the same preferences as before. The pareto optimal allocation should not change, since preferences are the same. Your logic would equate 10*alpha=beta, changing your original answer.


Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Smith
That true! Thanks asd! So basically we never equate the utility functions. We could, however, equate the MRS because that doesn't change after monotonic transformation. But here finding the MRS is not possible because we aren't substituting for anything and/or there are no two goods.That's the point you're making, right?
Reply | Threaded
Open this post in threaded view
|

Re: Isi 2016

Asd1995
If you want to think in the MRS way, here you go:

The utility function can be written as U1= my_share + 0*other_person_share and U2= my_share+0*other_person_share, so there are two "goods" per se, not in the conventional sense. The MRS for both these utility functions is infinity, and hence always "equal", not in a conventional sense either :P. That is why everything is Pareto Optimal.

Otherwise just use the definition of PO:

Only way to make A happier at any division of cake is to take some away from B and give it to him, which makes B worse off. So, any division is Pareto optimal.