ISI INTERVIEW PREP

[Mr. Nobody]

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[deepak]

Ram, for the SE/IE consideration, I went by this intuition - the labor supply curve is upward sloping because the substitution effect of increased wages outweighs the income effect which tends to dissuade people who earn more from working. So if the substitution effect is exactly offset by the income effect of having more money, the supply curve would be vertical, which is what we have in our case.

[neha]

thanks ram.

[Mr. Nobody]

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[neha]

Has anybody done question 4 (b) ?

[deepak]

I'm nowhere close to being sure about this but here's what I tried :

y = F(x) = P(X<=x)
P(y<= a) = P(F(x) <=a)
Let's consider Finv(x) to be the inverse of F(x). Also, since F is increasing, Finv is also increasing.
so i can write this as
= P(Finv[F(x)] <= G(a) )
Since Finv(F(x) = x,
= P(x<=G(a))
this is the CDF's definition
= F(G(a))
again by definition of inverse
= a
ie P(y<=a) = a.
This must imply that y is a uniform distribution.

[Preet]

Deepak- in the step  P(Finv[F(x)] <= G(a) )

How did u put in the G(a)?
BTW, i agree with ur final answer. it has to be uniform distribution and it is just a transformation qsn i guess..

[deepak]

I'm sorry. I meant Finv. Sorry. I should've used Finv instead of G. My mistake :/

[anon_econ]

Thank u. Got it now :)

[Preet]

Guys, have all of you got ur respective mails for the ISI Delhi interview?

[Mr. Nobody]

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[Mr. Nobody]

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[neha]

@deepak: Thanks for the reply:) Though i have a doubt in the following step

Let's consider Finv(x) to be the inverse of F(x). Also, since F is increasing, Finv is also increasing.
so i can write this as
= P(Finv[F(x)] <= G(a) )

what is the use of F being strictly increasing? Cant we write this = P(Finv[F(x)] <= G(a) ) irrespective of this?

[lovekesh]

@Ram
since the fractional part will always assume values between 0 to 1. So, whether limits are let's say 15 to 16, it effectively means integrating x with limits 0 to 1. so, if limit is between 0 to 200, it just means 200 such integrals each with limit 0 to 1 integrating x.

[Mr. Nobody]

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[neha]

@ram:

Question 1:
I am assuming limits as 0 to 2. For 0<x<1,f(x) =x (because [x] =0 for this range) and For 1<x<2,f(x) =x -1 (because [x] =1 for this range). Now f(x) can be integrated.

Question 3:
Put  h= -x,, you will get f(x,y) = x(1 + y).

[n.saish]

@ Ram:
The answer to Q3 is f(x,y)=x(1+y)
For Q2 any function like f(x,y)= 4x/(-y) or 10*e^x/(-e^y) will satisfy the constraints mentioned. And my answer to the second part would be no. As you change it from non decreasing to non increasing the function would change to f(x,y)= (-x)/y
But they have said a unique function, but here as you can see that you can put plenty of functions.... So there might be some other catch to it...

[deepak]

Neha, the monotonicity of the transformation is important since we are dealing with inequalities here. Let's take this example.
Suppose x < 5. Now a transformation such as 2x is monotonically increasing, and I can say with confidence that 2x <  10. However, if the transformation were to be a decreasing one such as say -2x, then the inequality would have to change from < to > ie, -2x > -10. Hence the importance of the function being strictly increasing.

[Mr. Nobody]

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[anurag]

In a pack of cards there are 52 cards: cards of 13 denominations belonging to each of 4 suits (clubs, diamonds, hearts and spades). 5 cards
are drawn from a pack of 52 cards. What is the probability of getting
3 cards of one denomination and 2 cards of another denomination?

ans. wil it be ( 4C3 * 4C2 / 52C5) * 13 * 12
 * multiplication , C combination