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Math Help - Independence rules

  1. #1
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    Question Independence rules

    Hello,

    It is a simple probability question but as it has been too long since I took a statistics course, I am not able to solve the problem. I will be appreciated if you can give an idea for the problem.

    I have a joint probability distribution on the variables x and y given the parameters a and b:

    p(x,y|a,b)

    x and y are assumed to be independent. p(x|a) and p(y|b) are multinomial distributions separately. Therefore, a is defined for x and b is defined for y. Therefore the equation can be written as given below (because of the independence between x and y):

    p(x|a,b) p(y|a,b)

    Is it also possible to derive this equation as given below by discarding b for x, and a for y: ?

    p(x|a) p(y|b)

    Because b does not mean anything for x, and a does not mean anything for x.

    Thanks a lot in advance!
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  2. #2
    MHF Contributor
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    the short answer is yes. Your intuition that irrelevant parameters can be removed is correct.



    More formally, for continuous variables (without loss of generality):

    p(x|a) = \int p(x|a,b)p(b) db

    since you have said that p(x|a,b) is does not actually vary with b, it can go outside the integral:

    p(x|a) = p(x|a,b) \int p(b) db


    but the integral of a pdf is always 1:
    p(x|a) = p(x|a,b) \times 1

    p(x|a) = p(x|a,b)
    which is the result you wanted to show. You can do the same thing for p(y|b,a)
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  3. #3
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    Very good proof. Thank you very much!
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