DSE 2009

[ritu]

yar vasudha...see visualize a function f as it is defined,
initially a 45 degree line to say x=1/3,then f(x) staying constant at 1/3,nd then increasing again from x=3/4 till it reaches the point 1,1....

ie...f(x)=x when x belongs to [0,1/3)
            1/3 when x belongs to [1/3,3/4)
            8x-5/3 when x belongs to [3/4,1]

this satisfies f(0)=0 and f(1)=1,its continous and not decreasing anywhere...
now till x=1/3 g(x)=0 ...then from there onwards it wud be constant at 1/3 and then increasing till1....though it wud not be continuous in this case it wud still be non decreasing(its a step function kind) so a is also possible....and as you said b is completely true....continuity of g means f is strictly increasing....so answer is d....:)
           

[anon_econ]

ritu vaise it's not right to come 2 any conclusion based on any 1 function bcoz it has to be true for all such functions..but in this case even i'm getting d..i dont know..v cud be making some mistake w/o realizing it!

[Amit Goyal]

For problem 3,
g(y)= min {x belongs to [0,1] | f(x)>=y }
First we will show g is non decreasing.
Let y and y' be two points in
[0,1]. Without loss of generality, let y'>y. We will show that
g(y')>=g(y). Now y'>y implies that {x belongs to [0,1] | f(x)>=y' } is
a subset of {x belongs to [0,1] | f(x)>=y }. [If for some x, f(x) is bigger than y' then for that x f(x) is also bigger than y because y' is bigger than y] This implies that
g(y')=min{x belongs to [0,1] | f(x)>=y' }>=min{x belongs to [0,1] |
f(x)>=y }=g(y). [Smallest value in a set is less than or equal to the smallest value of any of its subset] Hence proved.

If g is continuous,then f is strictly increasing is not true.
Consider f(x) = min{2x, 1}. This is not a strictly increasing function.
However, the associated g(y) = (0.5)y is continuous.

[Amit Goyal]

Regarding the micro problem:
In situation (b) the govt. supplies fixed quantity of Y through ration shops free of cost. We assume that in this situation he cannot buy (additional) Y from the free markets. We will show that in none of the three cases he is necessarily better off.
Case 1: Suppose
utility function, u(x, y) = xy
price of X, p(x) = 1
price of Y, p(y) = 1
Income = 10
Lets say govt. supply 2 units of Y free of cost.
Situation (a): In optimum he buys (5, 5) and derives u = 25.
Situation (b): In optimum he buys (10, 2) and derives u = 20.

Case 2: Suppose
utility function, u(x, y) = x + y
price of X, p(x) = 2
price of Y, p(y) = 1
Income = 10
Lets say govt. supply 2 units of Y free of cost.
Situation (a): In optimum he buys (0, 10) and derives u = 10.
Situation (b): In optimum he buys (5, 2) and derives u = 7.

Case 3: Suppose
utility function, u(x, y) = min{x, y}
price of X, p(x) = 1
price of Y, p(y) = 1
Income = 10
Lets say govt. supply 2 units of Y free of cost.
Situation (a): In optimum he buys (5, 5) and derives u = 5.
Situation (b): In optimum he buys (10, 2) and derives u = 2.

[anon_econ]

Thank u sir!!