DSE 2010 Disucssion-Please joim

[rongmon]

Thanks guys!

[rongmon]

Q10 and 56, anyone?

[Granpa Simpson]

Q56) the answer will be c this is because it is the only bundle allocation in which transfer can be made in a way such that the utility of A is unchanged while that of B increases..consider the allocation (0.5,1) for A and (3,1) for B in this case A's utility remains same as the initial level i.e. 0.5, however B's utility increases to 1 from 0.5...in other option you can make such an allocation..so c is not Pareto efficient.

[rongmon]

Question 31. I have option d) as a solution using Descartes. The correct answer is b)., how so?

[rongmon]

I don't quite understand.

There are 4 units of good 1 and 2 units of good 2 in total right. Aren't we supposed to look at Pareto optimal allocations using all of these units?

[Granpa Simpson]

See the definition of Pareto Efficient allocation states that it is such an allocation from where no agent can be be made better off without making the other worse off...or in other words you can look at it as an allocation, a deviation from which makes atleast one agent worse off..so if you are able to show that a deviation from a said allocation makes one of the agents better without hampering the utility of the other, your purpose of proving that the allocation not Pareto optimal is done..the same concept is used here..!!!

[rongmon]

Thanks Subhayu.

[Granpa Simpson]

Ur welcome comrade...

[neha:)]

Please explain how to go about Q.9? I know it's a problem of Hoteling Beach... but have no idea how to solve such questions.
http://econdse.org/wp-content/uploads/2012/04/2010-Option-A.pdf

[neha:)]

Guys pls help!
Pls explain how to go about Q.9? I know it's a problem of Hoteling Beach... but have no idea how to solve such questions.
http://econdse.org/wp-content/uploads/2012/04/2010-Option-A.pdf

[rongmon]

Q 38. Series 04.


Could someone share workings?

[neha:)]

Pls explain Q. 50 , series 04..

[Dreyfus]

@neha......u can check the independence of events A and B by using P(A intersection B) = P(A).P(B)
Where P(A intersection B) = p(5<=X<=8)
P(A) = p(X>=5)
P(B) = p(3<=X<=8)
U ll see equality will not hold, these two aren't independent.
Or alternatively this can be seen by the definition of Events A and B. But for 100% surety above method always works!

[xyz123]

Can somebody explain question 2 (series 2)
Thanks