Distance between two random variables

[matroid]

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?

[Dragonfinity]

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

[WRusin]

What if those variables are not uniformly distributed? (simple) And even better ... what if they're not independent?

[nartwolf]

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

[mag487]

Originally Posted by nartwolf

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]

That's not true, though. For example, suppose X = .2 and Y = .6. Then the distance between X and Y is only .4, but X and Y don't both belong to any of those three subintervals.

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

In addition to the problem above, the problem here is that the the three different cases you identify are not mutually exclusive. In other words, let

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).

[mag487]

Sorry, double post.