ISI INTERVIEW PREP

[anurag]

YA U R RYT , THNX

[neha]

@deepak ,,,Yeah ,,, my answer was [(x+y) / n] -y,,,, wrote [(x-y) / n] -y by mistake ,,

[anurag]

@ram - how wil d indiffernce curve look like for U= x^1/2 + y (in relation to RE2 Q.2 2011)

[Mr. Nobody]

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[Mr. Nobody]

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[deepak]

Ram, just a thought.
Wouldn't it suffice to say that any allocation of the form
(a2/a2+b2, y1) for P1 and (b2/a2+b2, y2) for P2 such that y1 + y2 = 2 is a P.O solution? Unless asked for a competitive equilibrium, any allocation of the above form would be optimal?

[anurag]

Suppose that you are testing H0 : p =1/2
against H1 : p =2/3 for a binomial variable X with n = 3. What values of X would you assign
to the critical region if you wish to have Prob[Type I Error] ≤1/8
and you wish to minimize Prob[Type II Error]?


how to solve this?

[Mr. Nobody]

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[Mr. Nobody]

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[deepak]

Cool :)

And any ideas about Anurag's question on Hypothesis testing?

[Mr. Nobody]

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[Amit Goyal]

You will minimize the Probability of Type II Error if you choose to reject the null if X takes value 3 and accept the null if X takes the value either 0, 1 or 2. Now think about it why? And provide a Mathematical proof.

[anurag]

f(x) = x^1/x hv a max at x=e ,ryt?

[Mr. Nobody]

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[deepak]

P(Type 1 error) = P(Reject Ho|p = 1/2) = P(X > Some critical value | given p = 1/2) and this is <= 1/8
ie P(X > Xc | p=1/2) <= 1/8
For the present binomial distribution with n = 3, this is satisfied only for X = 3.
Hence Xc = 3 and accept NULL for X = 0,1,2. Is this correct? I'm not sure I understand how Type II error features here, sir.

[anurag]

n wt abt d max/min fr x^3-6x^2+24x ? i guess both exist at +/- infinity..

[deepak]

It won't have local optima!

[anon_econ]

I'm amazed at the beauty of this proof.

[anon_econ]

guys there was a question in the dse entrance in which 2 players had the options 'swap' and 'retain'. does anyone remember the exact question? can we plz discuss it?

[anon_econ]

First i got the probability distribution of X in both situations-when p=1/2 and when p=2/3.
When p=1/2,
X   P(X)
0   1/8
1   3/8
2   3/8
3   1/8

From here it's clear that to have P(type I error)=<1/8 we must reject the null when x=0, or alternatively, when x=3. The choice b/w the two critical values will be determined by the second consideration, that of minimizing P(type 2 error).

When p=2/3
X   P(X)
0   1/27
1   6/27
2   12/27
3   8/27

From here it's clear that the probability of type 2 error is minimized when we decide to reject the null when x=3.