Dse 2012

[rohan]

Consider the following two-player game. The players simultaneously
draw one sample each from a continuous random variable X,
which follows Uniform[0; 100]. After observing the value of her own
sample,which is private information (that is, opponent does not
observe it), players simultaneously and independently choose one of
the following:SWAP,RETAIN.If both the players choose SWAP then they
exchange their initially drawn numbers. Otherwise, if at least one
person chooses RETAIN,both of them retain their numbers. A player
earns as many Rupees as the number she is holding at the end of the
game.Find the probability that the players will exchange their
initially drawn numbers.

i know its been discussed earlier but somebody plz explain again properly??.(the answer is 0)

[Erika]

Swap is two sided..means both HAVE to swap (exchange cards) for a swap to occur. But retaining is one sided. If one chooses to retain, it is sufficient for him to retain as the other doesnt have the option but to also retain.

so, I am player 2 and you are player 1 suppose.

I choose a card, you choose a card.
Now, suppose, I decide to swap with yours. this would mean that i believe my score is lower than yours  (in a uniform distn, I would say this when I have a number lesser than 50)and by swapping i can have your larger number. This would lead to two cases:

1. you really do have a number greater than 50. In this case, you would CALL to retain and NO SWAP OCCURS.

2. you have a number lower than 50. In this case, you wonder about the probability of me getting a number higher than 50. This itself is contradictory because I am swapping only because I have a number less than 50. Therefore in this case also, swap does not seem to promise you a benefit.

thus, no swap occurs .

P(swap)=0


NOTE: this is an intuitive answer. Plz let me know in case anyone figures out the mathematical ans.

[Sinistral]

sorry I didnt get your point 2.

what if both the players choose a sample whose value is less than 50?

[Sinistral]

do u have the link to the thread where this ques was discussed earlier? I cant seem to find that discussion on this forum.

[anonymouse]

@ Sinistral

Here.

[Sinistral]

Oh. I think the explanation provided by Vasudha is quite satisfactory. No doubt she got rank 7 or something (if I am not wrong)

thanx anonymouse and thanx a lot Vasudha

[Erika]

I was just going to quote Vasudha's soln :)

Incase both get less than 50...you'd swap only if you expect the other to have more than 25...and so on..

[Devika]

Heyyy
Could you explain Q40,2011?
Thanks :)

[Sinistral]

http://discussion-forum.2150183.n2.nabble.com/JNU-2010-tp7580642p7580652.html

courtesy Amit Sir.

[Erika]

for Q 31, How do we prove option a. and c. to be true?

non math ans for a. would do.
But for c., the formula for MSE otherwise(with intercept) is sum of (y-y bar )^2 / n. Using n-1 here ..why?

Also, for proving d. false, I wish to confirm:

We need Beta1 cap here...which equals sum of [(Xi-Xbar)(Yi-Ybar)/ sum of (Xi-Xbar)^2.
Thus the given formula is wrong. correct?

[Erika]

for Q 33, I understood all others, but the ans itself, which is option c.

I get var (beta* hat ) = (1/w2)^2 . Var (beta2 hat) -----

from Var (Beta2 hat) = sigma sq/sum xi ^2
xi^2 xi^*/w2.

w1 is not getting included :(

[Erika]

Q 34. Just a hint on how to measure changes using given parameters. Tnx

[Erika]

so many ecotrix questions in 2012 paper!!

In Q38, I didnt quite understand what the squared experience stands for. And how do we solve the question and come to the conclusion about returns being diminishing?

[maahi]

34 )   line of reg of y on x : y-mean y = b (x - mean x ) where b = cov(x,Y) / var x

line of reg of x on y : x-mean x = B ( y - mean y ) where B = cov(x,y) / var y
 

[Erika]

Thanks maahi :)

For Q 31,33,38..i still need help to clear up doubts ..anyone?