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Math Help - problem of random variables

  1. #1
    kin
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    problem of random variables

    Suppose X1;X2 are two independent geometrically
    distributed random variables whose parameters are respectively 1/2 and 1/3 : What
    is P(X1 = X2)?


    i dont know what are the parameters refered to ?
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  2. #2
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    Quote Originally Posted by kin View Post
    Suppose X1;X2 are two independent geometrically
    distributed random variables whose parameters are respectively 1/2 and 1/3 : What
    is P(X1 = X2)?


    i dont know what are the parameters refered to ?
    The parameters are the probability of success for each random variable, that is, the value of p.
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    Quote Originally Posted by kin View Post
    Suppose X1;X2 are two independent geometrically
    distributed random variables whose parameters are respectively 1/2 and 1/3 : What
    is P(X1 = X2)?
    It seems to me that P(X1 = X2)= sum(x=0..+oo) P(X1 = x AND X2 = x)
    Since X1 and X2 are i.i.d, P(X1 = x AND X2 = x) = P(X1 = x) P(X2 = x)
    Now, substituting the expressions for each probability above you should be done
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  4. #4
    kin
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    Quote Originally Posted by tossifan View Post
    It seems to me that P(X1 = X2)= sum(x=0..+oo) P(X1 = x AND X2 = x)
    Since X1 and X2 are i.i.d, P(X1 = x AND X2 = x) = P(X1 = x) P(X2 = x)
    Now, substituting the expressions for each probability above you should be done
    summation(x=0..+oo) [P(X1 = x AND X2 = x)]

    =p(x1=0,x2=0)+ p(x1=1, x2=2)+p(x1=3,x2=3).....

    =p(x1=0)p(x2=0)+ p(x1=1)p(x2=1)+......

    where p(x1=x)=(1/2)( (1-1/2)^x);p(x2=x)= (1/3)( (1-1/3)^x)<--am i right here??
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    Quote Originally Posted by kin View Post
    summation(x=0..+oo) [P(X1 = x AND X2 = x)]

    =p(x1=0,x2=0)+ p(x1=1, x2=2)+p(x1=3,x2=3).....
    =p(x1=0,x2=0)+ p(x1=2, x2=2)+p(x1=3,x2=3).....


    Quote Originally Posted by kin View Post
    =p(x1=0)p(x2=0)+ p(x1=1)p(x2=1)+......

    where p(x1=x)=(1/2)( (1-1/2)^x);p(x2=x)= (1/3)( (1-1/3)^x)<--am i right here??
    No, it should be P(X2=x)=(1/3) (1-1/3)^(x-1)

    Give a look at the definition of geometric distribution.
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  6. #6
    kin
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    Quote Originally Posted by tossifan View Post
    =p(x1=0,x2=0)+ p(x1=2, x2=2)+p(x1=3,x2=3).....




    No, it should be P(X2=x)=(1/3) (1-1/3)^(x-1)

    Give a look at the definition of geometric distribution.
    i start from x2= 0 so that P(X2=x)=(1/3) (1-1/3)^x;

    if i start from x2=1, then i should be P(X2=x)=(1/3) (1-1/3)^(x-1),

    are they not equivalent..?
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