DSE 2011 Queries

[Ayushya Kaul]

Question 9. Could someone explain how we obtain this probability: (1-p1)(1-p2)(1-p3)

Question 25. How is it discontinous at all points for x =/ 0?

Q27-30: Could someone explain these logic questions?

Also question 37.

Any help regarding these questions would be greatly appreciated :)
http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/2011-option-a.pdf

[Darth Vader]

Could anyone help? I have the same doubts as Ayushya.

Also I need help with Q4) and Q5) of DSE 2011.

Any help is welcome people :)

[duck]

Hi Ayushya.. :)

9) P(None of them occurs) = P(ȦḂĊ) = P(Ȧ)*P(Ḃ)*P(Ċ)
[As A,B,C are independent and its easy to show that Ȧ,Ḃ,Ċ will also be independent. ]
Therefore, P(None of them occurs) = (1-p1)(1-p2)(1-p3)
Note: Ȧ is complement of A, Ḃ is complement of B and Ċ is complement of C.

25) Use the definition of continuity.

[duck]

27) C1 ∩ C2 = (C1' ∪ C2')'
Note: C1' denotes complement of C1.
Now, C1' is a tribe and C2' is also a tribe. And union is two tribes is a tribe therefore, (C1' ∪ C2') is a tribe. Hence, complement of (C1' ∪ C2') = (C1' ∪ C2')' will be a club.Hence, option(a)

28) C1 ∪ C2 = (C1' ∩ C2')'
Now, C1' and C2' are tribes. So, their intersection will also be a tribe. And therefore, the complement of
(C1' ∩ C2') = (C1 ∩ C2)' will be a club. Hence, option(c)

29) Society and Empty set are both Club and Tribe.Hence, option(b)

30)Given a tribe T1 and a club C2: (T1 ∩ C2') is a club since C2' is  a tribe and the intersection of two tribes is a tribe. Hence, option(b)

[maahi]

its a problem of multicollinearity . we go by options .
a) can't be as x1x2 will be correlated with x1 and x2.
b) ans as there won't be any correlation b/w d exogenous variables
c)correlation b/w x1x2 and x1 and x2 .
there is perfect collinearity and hence d parameters can't be estimated .

[duck]

4) Given:
a1.x1 + a2.x2 + a3.x3 +....an.xn = b
⇒ a1.x1 + a2.x2 + a3.x3 +....an.xn - b = 0

[Note: RHS = 0 . Here, "0" is a vector , not a real number].

So, its like taking a linear combination of vectors {a1,a2...an,b} and  as "x" is a non-zero vector then there is some x(i) which is not equal to zero.
Therefore, the set of vectors {a1,a2...an,b} is linerly dependent.

[Prerna Rakheja]

Hi Darth,

For question 5, as m>n hence number of equations is more than number of variables. Therefore, if we can find a set of n equations amongst these m equations such that the A matrix is full rank. Then its inverse exists and hence there will be a unique solution.

[Ayushya Kaul]

Hey Duck :)

Thanks a lot! That was really helpful, cleared up a lot of things for me. Couldn't thank you less. Once again- Cheers :D

[Ayushya Kaul]

Thanks Maahi :)

Well that figures, since I havent revised multi variable regression, mulit collinearity and so on yet :/

[kogo]

plz explain question 25
rational no.s are always continuous.. so the answer should be (b)
how can we prove that discontinuous at every x not equal to 0?