Condition probability ?

HowardF · #1 $Old$ November 30th, 2008, 06:50 AM

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

Moo · #2 $Old$ November 30th, 2008, 07:52 AM

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 ?

Quote:

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 :
$P((X=a)/G)=f_{25}(a)$

Quote:

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 ?

HowardF · #3 $Old$ November 30th, 2008, 08:03 AM

Yes, much clearer thank you. I had the answer right, but it is great to know why.