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#11 (permalink) |
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Within my grasp!
![]() ![]() Join Date: Nov 2004
Posts: 101
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Re: Distance between two random variables
Certainly not... but I begin to feel that I aimed too high
(I don't have the neither the publications nor the contacts which are essential to get admission to a good program). I've got only 2 applications pending and I don't have much hope.What about you? I've read that you didn't have much luck with the Subject Test. Where have you been admitted? |
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#12 (permalink) |
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Socratic realist
![]() ![]() Join Date: Feb 2005
Location: Georgia, USA
Posts: 220
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Re: Distance between two random variables
My practice test was over 100 points (and about 20 percentiles or more) higher than the actual. I panicked on the exam--answered like 4 questions in the last hour. It was going really well too. Oh well. I only applied to local schools as it is. I realize that these tests require speed, and that is something that you lose with age. At 34, my speed isn't going to increase.
Anyway, I am in at Emory with the top University Fellowship. Too bad Emory is only known for graph theory, which I am not interested in at all. I also got into University of Georgia, which has a pretty good Algebra group. Finally, I am waitlisted at Georgia Tech. They are probably my first choice, since they are the best school by far. The only problem there is that they require a minor in some non-math field. I have no interest in applied math, so that will just be a chore for me (although, I recognize the value of not being so narrow minded as I would like to be). C |
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#13 (permalink) |
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Eager!
Join Date: Jan 2005
Location: Minneapolis, MN
Posts: 77
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Re: Distance between two random variables
What if those variables are not uniformly distributed? (simple)
And even better ... what if they're not independent? ![]()
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One of the characteristics of intelligent life is the ability to accommodate...
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#14 (permalink) |
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I JUST got here.
Join Date: Jul 2009
Posts: 1
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Hi:
I have a kind of alternative approach for that problem 1st : I would consider X,Y being in three different subinterval. the distance between X and Y is less than 0.5 if only X,Y belong to either one of these three subinterval [0,0.5],[0.25,0.75],[0.5,1] so :distance of X,Y< 0.5 then X,Y belong to [0,0.5] then p(0<x<0.5& 0<y<0.5)=P(x)P(y)"independent"=0.5*0.5=0.25 or X,Y belong to [0.25,0.75] then p(0.25<x<0.75& 0.25<y<0.75)=P(x)P(y)"independent"=0.5*0.5=0.25 or X,Y belong to [0.5,1] then p(0.5<x<1& 0.5<y<0.1)=P(x)P(y)"independent"=0.5*0.5=0.25 the p(z=distance)=0.25+0.25+0.25=0.75 thanks I did another method if you find this one correct let me know i would love to write the other one |
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#15 (permalink) | ||
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I JUST got here.
Join Date: Jun 2009
Posts: 2
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Quote:
Quote:
A = the event that X and Y are both in [0, .5], B = the event that X and Y are both in [.25, .75], C = the event that X and Y are both in [.5, 1]. Since A, B and C are not disjoint sets of events (e.g., if X = .3 and Y = .4, then A and B are both true), you can't conclude Pr(A or B or C) = Pr(A) + Pr(B) + Pr(C). |
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