# Thread: False positives

1. ## False positives

I'm quite happy to answer part 1. of the following problem.

"You are diagnosed with an uncommon disease. You know that there is only a 1% chance of getting it. Let D stand for the event 'you have the disease' and T stand for 'the test says so'. It is known that the test is imperfect: $\displaystyle P(T | D) = 0.98$ and $\displaystyle P(T^{c} | D^{c}) = 0.95.$

1. Given that you test positive, what is the probability that you really have the disease?
2. You obtain a second opinion: an independent repetition of the test. You test positive again. Given this, what is the probability that you really have the disease?"

So, for part 1. I use Bayes' theorem to get
$\displaystyle P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D^{c})P(D^{c})} = \frac{0.98 \times 0.01}{0.98 \times 0.01 + 0.05 \times 0.99} = 0.165$
which is correct, according to my book. However, when I apply similar reasoning to the second scenario I get an impossible answer.
$\displaystyle P(D|T_{1} \cap T_{2}) = \frac{ P(T_{1} \cap T_{2} | D)P(D) }{ P(T_{1}\cap T_{2}) } = \ldots$
This is where I get stuck. I know that the independence of $\displaystyle T_{1}$ and $\displaystyle T_{2}$ doesn't necessarily imply their conditional independence (conditioned on $\displaystyle D$) but I have no idea how to proceed. If I just assume the conditional independence I get an answer greater than 1, which is obviously wrong.

2. ## Re: False positives

Hey nimon.

If T1 and T2 are independent then

P(D|T1 and T2)
= P(T1 and T2 and D)/P(T1 and T2)
= P(T1 and D)*P(T2 and D)/[P(T1)*P(T2)]
= P(D|T1)*P(D|T2)

This should make sense intuitively since both processes are independent and don't change given data from the other process.

3. ## Re: False positives

Thanks for your input Chiro.

If I understand, then I should have $\displaystyle P(D|T_{1} \cap T_{2}) = P(D|T_{1})P(D|T_{2}) < P(D|T_{1}).$

Now this makes no sense to me: if I get a second positive, shouldn't the probability of my having the disease go up, and not down?

Furthermore, why do you have $\displaystyle P(D\cap T_{1} \cap T_{2}) = P(D \cap T_{1})P(D \cap T_{2})$? This is something I tried and then investigated; I think I can come up with lots of examples where $\displaystyle T_{1}$ and $\displaystyle T_{2}$ are independent but where $\displaystyle T_{1} \cap D$ and $\displaystyle T_{2} \cap D$ are not.

This much is contained in the opening post, where I say that the independence of two events does not imply their independence when conditioned on some third event, and vice-versa.

I seem to be going in circles...

4. ## Re: False positives

You wanted the intersection of events and not the union (in your post).

If you want the union then use the fact P(A OR B) = P(A) + P(B) - P(A and B).

5. ## Re: False positives

This problem is already solved: here

6. ## Re: False positives

Here's how I would do this problem. Imagine a population of 10000 people. "There is only a 1% chance of getting it." so 1% of 10000= 100 people have the disease and 10000-100= 9900 do not. 98% of the people who have the disease test positive so of the 1000 who have the disease 98 test positive and 2 do not. Since 95% of the people who do not have the disease do not test positive, 5% of 9900= 495 people who do NOT have the disease test positive.

1.Given that you test positive, what is the probability that you really have the disease?
Of the 495+98= 593 people who test positive, 98 of them have the disease so the probability that you have the disease, given that the test was positive is 98/593= 16%. The probablity that you do NOT have the disease, given that the test was positive, is 100- 16= 84%.

2.You obtain a second opinion: an independent repetition of the test. You test positive again. Given this, what is the probability that you really have the disease?
The probability is now $\displaystyle 1.00- .84^2= 30%$