Expectation - June 16

[Amit Goyal]

June16_Expectation.png

[anon_econ]

I'm not at all sure but I'll tell u what I got..
We hv 2 find E(N)-1=E(E(N|X))-1.
N     P(N|X)
1     P(Y1>X)=1-X
2     P(Y1=<X).P(Y2>X)=X(1-X)
3     P(Y1=<X).P(Y2=<X).P(Y3>X)=X^2(1-X)
.      .
.      .
.      .

From here I got E(N|X)= 1/1-X
Could not find expectation of this. I know most probably it's wrong!

[komal]

i am getting answer as n/2   where n is the no. iids drawn(Yn), higher is n , more prize money can be won.

[Rain Man]

@ Vasudha , Komal

It wd be great if you can explain the question in language.
I am having trouble understanding the notations.

iid - independent and identically distributed variables right ?
inf (?)

thx

[anon_econ]

Yeah that's what iid means. And according to me inf means infimum of the set. So i took it to be the lowest n such that Yn>X.

[Amit Goyal]

Excellent Vasudha. You are absolutely right.
E(1/1-X) = ∞

[anon_econ]

Oh. Thanks sir! :)

[Rain Man]

@ Vasudha..

Can you explain your approach in some detail .. sry for the trouble .. i am not able to get it .. :(

[anon_econ]

Rain Man,
we picked a number X and then an infinite sequence of numbers Y1,Y2...Yn. all of them were uniformly distributed on [0,1]. now N is the least value of n for which Yn>X. N could be 1,2,3..and so on. it would be 1 if Y1 itself is g8er than X. it would be 2 if Y2>X and Y1<X. it would be 3 if Y1 and Y2 r less than X while Y3>X and so on. jst write this down..u'll get the probability distribution of N (given X). from here u can get E(N|X). and E(N)=E(E(N|X)). v need to find E(N)-1 bcoz the chap gets N-1 and v hv 2 find the expected value of what he gets.

[Rain Man]

Thanks a ton :) . I get it now :)