dse

[ritu]

1.suppose x and y are independent rv that fillow uniform distributn on [0,1]
Let Z=min{x,y}..Find
a.Pr(z<0.5)
b.pr(z<0.5/x=0.75)


2.which of the parameters cant be estimated using ols...
Y=a+bx1+cx2+d(x2-x1)+ex1x2+u

3.an increase inexpected rate of inflation will
a.shift lm up
b.shift lm down
c.leave lm unaffected
d.have an ambiguous effect on lm
i think it shud be c..

Thanks..

[ritu]

Let x=r be the mode of the distributn with probability mass functn
 p(x)=nCx p^x (1-p)^n-x.then which of the following holds..
A.(n+1)p-1<r<(n+1)p
B.r<(n+1)p-1
C.r>(n+1)p
D. r<np

[ritu]

Pls someone reply....:-(:-(

[anon_econ]

which year are these questions from?

[ritu]

First three are from dse 2011 nd the last one is from isi 2008...

[anon_econ]

in the isi one the PMF is maximum for x=r so u can get option a from p(r)>p(r+1) and p(r)>p(r-1).
i'm trying not to look at questions from dse 2011 bcoz i want to do it as a sample paper in 3 hours later on..

[ritu]

Oh ya..its ok...nyway thanx...dnt kno wot wud i have  done widout u:-*

[Amit Goyal]

(a) Pr(Z < 0.5) = 1 - Pr(Z > 0.5) = 1 - Pr(X > 0.5, Y > 0.5) = 1 - Pr(X > 0.5)Pr(Y > 0.5) = 1 - 0.25 = 0.75.
(b) Pr(Z < 0.5| X = 0.75) = Pr(Y < 0.5) = 0.5

[Amit Goyal]

If you plot LM in (Y, i) plane (where Y is output and i is nominal interest rate) then change in expected inflation will leave the LM curve unchanged.

[Amit Goyal]

Including X2-X1 along with X1 and X2 results in multicollinearity. To be able to estimate using OLS, we need to remove X2-X1 and then we can estimate the remaining parameters.

[ritu]

Thank u so much sir...just wanted to clarify one thing....in that uniform distn question part (b) why the probability isnt undefined??

[Amit Goyal]

Dear Ritu,

Probability isn't undefined because it is defined :)
and this is how you do it:
Pr(Z < 0.5| X = 0.75) = Pr(min{X, Y} < 0.5| X = 0.75) = Pr(min{0.75, Y} < 0.5| X = 0.75) = Pr(Y < 0.5| X = 0.75) = ? (We need to find the conditional density of Y given X = 0.75 but since X and Y are independent its the same as marginal density of Y which is given to be U[0, 1])
Therefore,
Pr(Y < 0.5| X = 0.75) =  Pr(Y < 0.5) = 0.5

(PS: Please don't confuse it with the Bayes' Rule for the discrete sample space. I suggest read about continuous distributions.)

[ritu]

Oh yes...thank u sir. Now i kno where i was wrong...:-):-)

[ritu]

If it wud have been the case that"" x nd y were dependent" then the conditional probability wud have been undefined coz x being a continous rv..has zero prob of assuming a fixed value...denominator being zero wud have meant that fraction is undefined...right sir??

[Amit Goyal]

No, That's also not true. Consider the extreme form of dependence where X = Y distributed as U[0,1] (Note that X and Y are identically distributed but not independent)
Pr(Z < 0.5| X = 0.75) = Pr(min{X, Y} < 0.5| X = 0.75) = Pr(min{0.75, Y} < 0.5| X = 0.75) = Pr(Y < 0.5| X = 0.75) = Pr(X < 0.5| X = 0.75) = 0 (Since X = Y)
Its still defined.

[ritu]

Oh...yes sir...nw theres no scope of any doubt...all clear now...
No words to express my gratitude:-):-)

[SHIKHA]

sir can u explain both the cases wen X and Y follow some relation
P(z<0.5)
P(z<0.5|X=0.75)

[ritu]

Shikha..for the second case sir has already provided a full explanation..pls read it...in earlier comments...:-)

[SHIKHA]

i mean to say give one more example for the same..
like X+Y=1
or XY=1 ...something like that...what will be the approach...

[Amit Goyal]

Consider the dependence of the form X + Y = 1
Pr(Z < 0.5| X = 0.75) = Pr(min{X, Y} < 0.5| X = 0.75) = Pr(min{0.75, Y} < 0.5| X = 0.75) = Pr(Y < 0.5| X = 0.75) = Pr(X > 0.5| X = 0.75) = 1 (Since X = 1 - Y)