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xEnOnn

  • 2 years ago

Suppose I throw 2 fair dice. Let X be a random variable of the total points from the 2 fair dice thrown. So, the random value X can hold 11 values from 2 to 12: \[X=\{ 2,3,...,12\] As usual in textbooks, there is often another omega in the sample space like this: \[\omega \in \Omega \] In my case, what are the values in my space space of capital Omega? And what are my sampling values?

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  1. xEnOnn
    • 2 years ago
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    In my case, what are the values in my sample space of capital Omega? And what are my sampling values?

  2. ash2326
    • 2 years ago
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    \(\omega \)={2,3,4,5,6,7,8,9,10,11,12} I think Ω is {1,2,3,4,5,6,7,8,9,10,11,12}

  3. xEnOnn
    • 2 years ago
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    Are ω the sampling values? Also, since X is a random variable, which is a function by itself, for this: \[X(\omega )=x\] How would I match the values? Suppose I have X(12), what should the output of X(12) be?

  4. Zarkon
    • 2 years ago
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    X(12) could be 12

  5. xEnOnn
    • 2 years ago
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    When I have X(a)=b, 'a' is the random variable value, which is the total points I get out of the 2 dice, is this right? And what about 'b'? What does it mean for 'b'?

  6. Zarkon
    • 2 years ago
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    a is an element from the sample space...b is a possible value of the random variable

  7. Zarkon
    • 2 years ago
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    You should probably treat it like this... roll two die then the sample space consists of 6*6=36 different ordered pairs let X be the random variable that add the ordered pairs then \(\omega\) is an ordered pair and X is a random variable taking values in {2,3,...,12}

  8. Zarkon
    • 2 years ago
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    so if \(\omega=(2,5)\), \(X(\omega)=7\)

  9. Zarkon
    • 2 years ago
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    X is a measurable function from the sample space to the real line

  10. xEnOnn
    • 2 years ago
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    ohhh..... Thank you so much!! This very clear. Thanks a lot!

  11. xEnOnn
    • 2 years ago
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    Which also means, I cannot write as X(12) because there isn't such a thing. It should be X({6,6})

  12. Zarkon
    • 2 years ago
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    To be completly correct...yes....some times people will say that the experament is to roll two die and add them ... then one could look at X(12) it is better (and correct) to treat the two die rolls as the experament and then get ordered pairs...then let the random variable add them

  13. xEnOnn
    • 2 years ago
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    Now I understand. Thank you so much! It is very clear! :)

  14. Zarkon
    • 2 years ago
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    good ;)

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