Standard probability space and random variables.











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Problem :



Let $Omega:=[0,1]$



How to define two independent random variables, which describe coin toss on the above omega?



My idea:



$Omega=[0,1]$



$Sigma = mathbb{B}([0,1])$



Consider the Lebesgue measure



1-reverse coin
2-obverse coin



$X(w)= begin{cases} 0 &text{when } w in [0,1/2) \ 1 &text{when } win [1/2,1) end{cases} $



$Y(w)= begin{cases} 1 &text{when } w in [0,1/4) cup [1/2,3/4) \ 0 &text{when } win [1/4,1/2)cup [3/4,1) end{cases} $



It seems, that $P(X=1)=P(X=0)=P(Y=0)=P(Y=1)=1/2$



And, for example $P(X=0)P(Y=1)=1/4=P(X=0,Y=1)$



What do you think about this?






probability-theory random-variables







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    up vote
    1
    down vote

    favorite












    Problem :



    Let $Omega:=[0,1]$



    How to define two independent random variables, which describe coin toss on the above omega?



    My idea:



    $Omega=[0,1]$



    $Sigma = mathbb{B}([0,1])$



    Consider the Lebesgue measure



    1-reverse coin
    2-obverse coin



    $X(w)= begin{cases} 0 &text{when } w in [0,1/2) \ 1 &text{when } win [1/2,1) end{cases} $



    $Y(w)= begin{cases} 1 &text{when } w in [0,1/4) cup [1/2,3/4) \ 0 &text{when } win [1/4,1/2)cup [3/4,1) end{cases} $



    It seems, that $P(X=1)=P(X=0)=P(Y=0)=P(Y=1)=1/2$



    And, for example $P(X=0)P(Y=1)=1/4=P(X=0,Y=1)$



    What do you think about this?






    probability-theory random-variables







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Problem :



      Let $Omega:=[0,1]$



      How to define two independent random variables, which describe coin toss on the above omega?



      My idea:



      $Omega=[0,1]$



      $Sigma = mathbb{B}([0,1])$



      Consider the Lebesgue measure



      1-reverse coin
      2-obverse coin



      $X(w)= begin{cases} 0 &text{when } w in [0,1/2) \ 1 &text{when } win [1/2,1) end{cases} $



      $Y(w)= begin{cases} 1 &text{when } w in [0,1/4) cup [1/2,3/4) \ 0 &text{when } win [1/4,1/2)cup [3/4,1) end{cases} $



      It seems, that $P(X=1)=P(X=0)=P(Y=0)=P(Y=1)=1/2$



      And, for example $P(X=0)P(Y=1)=1/4=P(X=0,Y=1)$



      What do you think about this?






      probability-theory random-variables







      share|cite|improve this question















      Problem :



      Let $Omega:=[0,1]$



      How to define two independent random variables, which describe coin toss on the above omega?



      My idea:



      $Omega=[0,1]$



      $Sigma = mathbb{B}([0,1])$



      Consider the Lebesgue measure



      1-reverse coin
      2-obverse coin



      $X(w)= begin{cases} 0 &text{when } w in [0,1/2) \ 1 &text{when } win [1/2,1) end{cases} $



      $Y(w)= begin{cases} 1 &text{when } w in [0,1/4) cup [1/2,3/4) \ 0 &text{when } win [1/4,1/2)cup [3/4,1) end{cases} $



      It seems, that $P(X=1)=P(X=0)=P(Y=0)=P(Y=1)=1/2$



      And, for example $P(X=0)P(Y=1)=1/4=P(X=0,Y=1)$



      What do you think about this?






      probability-theory random-variables




      probability-theory random-variables






      share|cite|improve this question















      share|cite|improve this question













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      edited yesterday

























      asked yesterday









      PabloZ392

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      1356






















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          Yes, your idea works just fine.






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            Yes, your idea works just fine.






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              Yes, your idea works just fine.






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                up vote
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                Yes, your idea works just fine.






                share|cite|improve this answer












                Yes, your idea works just fine.







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                answered yesterday









                Nate Eldredge

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