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Equality of distributions

  1. Jul 13, 2012 #1
    Hello ,

    frustrated with my lecturer assignments , i need your help with this :

    if X,Y and are two different random variable , is it possible that X,X+Y have equally distributed.

    if it can be give an example , if not prove it.

    thanks,
     
  2. jcsd
  3. Jul 13, 2012 #2

    Ray Vickson

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    Show your work---those are the Forum rules.

    RGV
     
  4. Jul 13, 2012 #3
    sorry but i don't have any ,
    i don't remember that we learnt who to prove equality in distribution.
    all i thought of is that if i want to prove it right i have to prove that :
    P(X[itex]\leq[/itex]t)=P(X+Y[itex]\leq[/itex]t) for all t,

    maybe there other conditions that lead to equality in distribution, but i don't know them.

    generally i think it is possible , so i am trying to think on distribution that will satisfy the answer
     
  5. Jul 13, 2012 #4

    Ray Vickson

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    Equality in distribution is equivalent to equality of the characteristic functions,or Laplace transforms, or moment-generating functions, etc.

    RGV
     
  6. Jul 13, 2012 #5
    those subjects you mentioned aren't in the topics of the course , haven't learnt them...

    maybe a simple case that answer this problem ?
     
  7. Jul 13, 2012 #6

    jbunniii

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    Are there any constraints at all on X and Y?

    For example, if X is equal to some constant with probability 1, can you find a solution in that case?
     
  8. Jul 13, 2012 #7
    jbunniii , if i understood you ,
    if x is a constant with probability 1 and y is also the same constant with probability 1 then
    x,x+y has the same distribution.

    is it right ?

    but the only constraint about x and y that they will be different
     
  9. Jul 13, 2012 #8

    jbunniii

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    Well, it depends on which constant you choose. If x and y are constants, and you need x = x + y, then what does this imply?
     
  10. Jul 15, 2012 #9
    going back to the start , if X,Y are different random variable they can't be constants.
    no ? i can't see where is the randomness of X or Y if the are constants.
     
  11. Jul 16, 2012 #10

    jbunniii

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    Certainly a random variable can be constant. Simply assign it a constant value with probability 1. Yes, it's a somewhat trivial example, but that's why I asked if there are any constraints on what kind of distribution can be assumed.

    To find out if there is a less trivial example, consider this: if X and X + Y have the same distribution, then they must have the same moments. Try using this fact for the first and second moments, and see what you can conclude.
     
  12. Jul 19, 2012 #11
    if i take a fair coin where , X=1 for heads , X=0 for tail so P(X=1)=P(X=0)=0.5 ,
    and i will define Y=1-X so P(Y=1)=P(Y=0)=0.5 ,
    can i say that P(X=0)=P(X=1)=P(X+Y=1)=0.5 ?
    this means X,X+Y have equal distribution and X,Y are not constants ?
     
  13. Jul 19, 2012 #12

    jbunniii

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    I don't think this example works. If Y = 1 - X, then X + Y cannot equal 0, whereas X can, with probability 0.5. Therefore X and X + Y do not have the same probability distribution.
     
  14. Jul 19, 2012 #13
    o.k got it , with slight modification - Z=1-X , Y=1-2*X then Z=X+Y, where X is Bernoulli
    => X,Z equal in distribution.
     
  15. Jul 19, 2012 #14

    jbunniii

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    Looks good to me.
     
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