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Thread: probability and expected value

  1. #1 probability and expected value 
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    A have a question whre the random variables x and y have the following joint probability density:



    I resolve the constant a = 6/5, so ,

    What the expected value of 1 / X, given that Y = 3/2.?

    tks


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  3. #2  
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    Your notation is confusing. What is "se"? You should not use the same letters for random variables and function argument.


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  4. #3  
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    Hi Mathman,

    Sorry, "se" = "for".

    tks
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  5. #4  
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    I get a = 1/2. Could you show me your derivation?
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  6. #5  
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    We can say that??

    So,









    a=6/5 (?)
    Last edited by rykardu; November 12th, 2012 at 03:34 AM.
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  7. #6  
    Forum Professor river_rat's Avatar
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    Also check your boundary conditions, the original density function was from 0 to 2 but you are integrating from 1 to 2?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  8. #7  
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    Dear river_rat and mathman,

    Sorry, the original question is from 1 to 2 and not 0 to 2 as:


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  9. #8  
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    Ok, given that I get ?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  10. #9  
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    Hi river_rat

    My development gave my 6/5 (see post of 11 nov). Something wrong in my reasoning?
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  11. #10  
    Forum Professor river_rat's Avatar
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    The bounds on the inner integral look wrong, remember that Y >= X
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  12. #11  
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    a=3/2

    right?
    Last edited by rykardu; November 13th, 2012 at 03:48 PM.
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  13. #12  
    Forum Professor river_rat's Avatar
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    That looks good

    Have you calculated the conditional expectation yet?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  14. #13  
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    Not yet... today. Tks!!
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  15. #14  
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    18/10?
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  16. #15  
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    Now have 2 other topics to resolve:

    (b)Define the marginal probability density function py (Y).
    answer: 18/10 (ok?)

    (c) Determine o valor esperado de 1/X, dado que Y = 3/2.
    answer: ??
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  17. #16  
    Forum Professor river_rat's Avatar
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    Do a sanity check on your answer:

    which implies that and so no matter how we assign positive weight to that interval we cannot get an average less that or greater than 1. So we know there is something wrong with your answer it would seem.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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