ISI INTERVIEW PREP

[neha]

thanks for replying  vasudha:)

[Preet]

Well i find even this line of reasoning to be correct
Sorry i misunderstood it when u earlier made the same point

neha wrote

@preet ,, what you are saying makes sense to me,, but i have one more doubt.

Instead of minimizing the total expenditure we can think in terms of maximizing the utility. Since at Px=0 and Py=1 agent 1 has no income,, so agent 2 cant think of selling any of his y units to him and to maximize his utility he would want 1000 units of x [(1000,1000) is preferred over (999,1000)]  and there is no price attached to it,he can very well afford to do that. so he would want 1000 of x which leads to excess demand of it.. so thats why it can be competitive.  What do you think?

[neha]

@saish ,, if i have understood your doubt correctly, what you are saying is the case when all xi's are less than z is missing out while writing down the DF. I did include this case as well in DF.This is to be included because we are finding out the probability that the second highest is less than z. (There may be case when even the highest x is less than z)

so F(z) = (z/a)^n  + n * (a-z)/a * (z/a)^(n-1)

(z/a)^n = prob that all xis are less than z (including the second highest) or think like prob that the second highest is less than z (including  the highest one is also less than z)

n * (a-z)/a * (z/a)^(n-1) =prob that the second highest is less than z (the highest one is greater than z).

hope it helps better :)

[neha]

never mind,, and thanks a lot:)

[neha]

this is regarding question no 28 of ISI 2007

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/msqe2007.pdf

I think all the options are correct. but the answer keys says option d.

and of the same paper, what do you think p-q would be? i think it will be 1 or -1.

[anon_econ]

neha, statement A is correct. u don't hv to look any further. this makes statement D incorrect. that's what v had to find out- the incorrect statement.
and which other question r u asking?

[neha]

oh yeah,, i didn't read the question carefully,,, and the other one is question 4.

[Saish]

No no... Thats not what I'm saying.... See the method you're following is formulating a df and then differentiating to get the pdf right... I completely agree with each step of this method...
Now I'm trying out a different method of writing the pdf directly with the reasoning that Pr( that a selected no would be the second highest) = (N-1)*Pr(that n-2 nos will be smaller than it)*Pr(that 1 would be greater)... So I guess I'm missing out something that is included when you do it by the first method....
Hope you get my doubt...

[anon_econ]

but saish the pdf does not give u the probability. u cant obtain the pdf directly.

On Thu, Jul 5, 2012 at 1:07 PM, Saish [via Discussion forum] <[hidden email]> wrote:

No no... Thats not what I'm saying.... See the method you're following is formulating a df and then differentiating to get the pdf right... I completely agree with each step of this method...
Now I'm trying out a different method of writing the pdf directly with the reasoning that Pr( that a selected no would be the second highest) = (N-1)*Pr(that n-2 nos will be smaller than it)*Pr(that 1 would be greater)... So I guess I'm missing out something that is included when you do it by the first method....
Hope you get my doubt...

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[komal]

but i find statement c as incorrect. there are atleast two pts. whose is distance is less than or equal to root 2 is true but as statement c says distance is less than root 2 , (which is not necessary) , statement c is false. as not all statements are incorrect so d is incorrect as well . so, should not the answer be both c and d .
please tell me if there is  anything wrong in statement i m making about "Statement c".

[anon_econ]

komal, how do u show that v can always find 2 points whose distance is less than or equal to root 2?

On Thu, Jul 5, 2012 at 1:10 PM, komal [via Discussion forum] <[hidden email]> wrote:

but i find statement c as incorrect. there are atleast two pts. whose is distance is less than or equal to root 2 is true but as statement c says distance is less than root 2 , (which is not necessary) , statement c is false. as not all statements are incorrect so d is incorrect as well . so, should not the answer be both c and d .
please tell me if there is  anything wrong in statement i m making about "Statement c".

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[neha]

@ saish,, thats true i believe,(what vasudha is saying) PDF just gives the ordinates,, not probability.. think about the case when x follows U(0,0.5) where f(x) = 2 which cant be a value of probability.

[anon_econ]

or u saying it's either 1 or -1 bcoz p^2-q^2=(p-q)(p+q)?

[Saish]

GOT IT!!!
I forgot that integral of the pdf should be equal to 1...
So pdf = C*(n-1)*[(x/a)^(n-2)]*[(a-x)/a]
Now integral pdf=1 we get C=n/a!!!
So now when we calculate the expectation we get the same answer.....

[neha]

@vasudha
what i am saying is p-q will either be -1 or 1. what do u think?

[anurag]

p-q wil be 1 right?

[anurag]

ca't be -1 as p=q is alwz =ve so -1 wil lead to a negative no.. tk it as prime no. = 1* no. itself , so p+q is d no. n p-q is 1

[anurag]

i mean p+q is alwaz +ve..

[neha]

it can be -1 too, i think.

[deepak]

Neha, for Q4,

Now it is factorized as (p+q)(p-q) = 1 * A for some A.
so either p+q = 1 or p-q = 1
If p+q = 1, then p = 1-q and this violates the condition that p and q are positive integers.
So p-q = 1

But we could also factorize as (p+q)(p-q) = -|1| * -|A| for some prime A.
Here again, p+q can't be -1 since both are positive integers. p+q can't be -|A| either for the same reason.

So p-q = 1 only seems like an acceptable answer, no?