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Math Help - Condition probability ?

  1. #1
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    Condition probability ?

    I am dealing with a problem written as follows:-

    Suppose that 10% of the people in a certain population have the eye disease glaucoma. For persons who have glaucoma measurements of eye pressure X will have normally distributed X with a mean of 25 and a variance of 1. For persons whithout glaucoma the pressure X is normally distributed with a mean of 20 and a variance of 1. Suppose a person is selected at random from the population and eye pressure X is measured. Determine the probability that the person has glaucoma given that X = a.

    First I note that the probability if having glaucoma + probability of not having it = 1, even at a pressure X = 23, say which is several standard deviations away from each of the means 20 and 25.

    If I write the Gaussian, just for brevity here, with mean 20 as f20(X) and the one with mean 25 as f25(X) then I think the probability of being in the set "has glaucoma" might be expressed as:-

    0.1 f25(a)/[0.9f20(a) + 0.1f25(a)]

    If this is correct, then I'm having trouble expressing myself why this is correct. It seems to have the right behaviour, making a rapid transition between 0 and 1 as we go above the intermediate value a = 22.5

    Comments would be helpful, perhaps I've gotten this entirely wrong.

    HowardF
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  2. #2
    Moo
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    Hello,

    Start from scratch. Let G denote the event : "the person has a glaucoma"

    You're looking for the probability : P(G/(X=a)), that is the probability that the person has a glaucoma given that his eyes' measurement is a.
    Now what is given in the text ?
    For persons who have glaucoma measurements of eye pressure X will have normally distributed X with a mean of 25 and a variance of 1.
    It means that if you pick someone who has a glaucoma, X will follow that normal distribution, which we'll denote as you did : f_{25}(x)
    Once again, this is a conditional probability :
    P((X=a)/G)=f_{25}(a)
    For persons whithout glaucoma the pressure X is normally distributed with a mean of 20 and a variance of 1
    Similarly, we get :

    P((X=a)/\overline{G})=f_{20}(a), where \overline G denotes the event "the person doesn't have a glaucoma". And we indeed have P(G)+P(\overline{G})=1



    Now look at this formula : Bayes' theorem - Wikipedia, the free encyclopedia (derived from Bayes' theorem)


    From this, we can write :

    P(G/(X=a))=\frac{P((X=a)/G)P(G)}{P((X=a)/G)P(G)+P((X=a)/\overline{G})P(\overline{G})}


    Which is :
    P(G/(X=a))=\frac{0.1 \cdot f_{25}(a)}{0.1 \cdot f_{25}(a)+0.9 \cdot f_{20}(a)}
    So you were correct.

    Does it look clear to you ?
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  3. #3
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    Yes, much clearer thank you. I had the answer right, but it is great to know why.
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