DSE 2015 answer

[varnika1880]

Thank u so so much ....

[JMKeynes]

Hey there Qwert, remember, Correlation does not imply Causation. It is simply a degree of the linear relationship between two variables. According to A, it is implying that income causes savings to change which is wrong. Similar argument for B. And since r=.2, very low, the plot will not lie on a straight line. So D is out of the question.

[JMKeynes]

Anytime Varnika and harshdua

[Qwerty]

Thanks JMKeynes got your point...could you throw some light on Q35 and Q36 too..

[JMKeynes]

For 35, if your read the question, clearly R(P) is the range space of P and N(P) is the nullspace of P. Also from condition (b) of a projector, P(y+z)=y. But since P is a linear transformation, P(y+z) can be written as Py+Pz. {Pz=0, by definition}. Therefore LHS becomes Py.
Equating LHS with RHS we get,
Py=y.........(1)
Now, since y belongs to the range space, there exist an x belonging to R^n, such that Px=y.......(2)
 Putting this in the LHS of (1),
 we get P(Px)=y, or,
P^2 x=y.......(3)
From (2) and (3), we get P^2=P
(P.S. I'm hoping this is the right solution)

[varnika1880]

Hey can u clear one thing plz in the 2 image u have equated wages =mpl ...shouldn't it be wages =mpl *L as total wages are considerd

[JMKeynes]

Hi. No, the question states that each factor is paid the mpl. That is, "w" is the wage received per unit quantity of labour supplied. "wL" would be total wage earnings, from the economy as a whole.

[Qwerty]

Kudos JMKeynes

[urvashi]

How do we attempt Q56 ? I cannot get an explicit expression for the steady state value when both savings rates are positive

[JMKeynes]

[JMKeynes]

Hi Urvashi, just refer to the previous post for the values of w and r and the steady state equation. :)

[urvashi]

Thanks so much

[Ansh malhotra]

Please help with question number 31 of this paper (2015)

[Mike]

31) The first condition is simply MRS = Px/Py
For well Behaved prefrences we use the above condition. Also, we have to look at the bordered hessian matrix to check if the prefrences are covex or not.
 bordered hessian matrix = 0    Px    Py
                                      Px   Uxx  Uxy
                                      Py   Uyx  Uyy
The determinant is equal to the determinant of the matrix given in the question. AN for convex prefrences the determinant of the bordered hessian matrix should be <=0.

Uxx = D11U(X1,X2)
similarly for others
so answer is (d)

[Ansh malhotra]

thank you mike. ii think bordered hessian shouldn't be positive. i am bit  confused help

[Mike]

You can write Hessian in two ways
Either 0   px   py
           Px uxx uxy.   For this det<=0
           Py uyx uyy

Or.      Uxx uxy  px
           Uyx  uyy py.  For this det >=0
            Px   py   0
           

[Eco16]

Can somebody help me with q 48?

[knowpraveen]

There are three options for each of the four testers to rank a particular kind of chocolate. The chocolate A gets the sum of ranks equal to four only for 1 combination i.e (1,1,1,1). On the whole, there are 3x3x3x3=81 options available to the for testers. Therefore, P=1/81

[Eco16]

Thanks!

[Ansh malhotra]

Question no 17 and no 19 of this paper please help(2015)