An Application of Conditional Probability
D
D*
Total
T
Disease Present, Screens +
Disease Absent, Screens +
Screens +
T*
Disease Present, Screens -
Disease Absent, Screens -
Screens -
Total
Disease Present
Disease Absent
Total
There are four possible combinations of Disease and Test:
DandT ~ Disease Present and Screen Shows Positive
DandT* ~ Disease Present and Screen Shows Negative
DandT ~ Disease Absent and Screen Shows Positive
DandT* ~ Disease Absent and Screen Shows Negative
Pr{D|T} ~ Probability that a subject has the disease, given a positive test
Pr{D*|T} ~ Probability that a subject lacks the disease, given a positive test
Pr{D*|T*} ~ Probability that a subject lacks the disease, given a negative test
Pr{D|T*} ~ Probability that a subject has the disease, given a negative test
What is Pr{D|T}, the probability that disease is present given a positive screen?
Pr{D|T} =
Pr{DandT}/Pr{T} =
Pr{T|D}*Pr{D}/Pr{T} =
Pr{T|D}Pr{D}/(Pr{TandD} + Pr{TandD}) =
Pr{T|D}*Pr{D}/(Pr{T|D}Pr{D} + Pr{T|D}Pr{D})
That is,
Pr{D|T} = ( Pr{T|D}*Pr{D} ) / ( Pr{T|D}Pr{D} + Pr{T|D}Pr{D} )
So we need the prevalence of disease (Pr{D}), Pr{T|D} = 1 - Pr{T*|D}, where Pr{T*|D} is the false-negative rate, and Pr{T|D*}, the false positive rate.
Suppose that we are given that Pr{T|D*} = “False Positive” = .05, Pr{T*|D} = “False Negative” = .01 and Pr{D} = .001. Then Pr{T|D*} = “False Positive” = .05, so then Pr{T*|D*} = .95 and Pr{T*|D} = “False Negative” = .01, so then Pr{T|D} = .99
Plugging in what we know:
Pr{D|T} =(.99)(.001)/(.99(.001) + .05*(.999)) = .00099/(.00099+.04995) = .019 (<2%). So in this case, approximately 2% of the positive tests actually indicate disease, which leaves the other 98% with a false finding.
Let’s repeat the calculation, but with Pr{T|D*} = “False Positive” = .005, Pr{T*|D} = “False Negative” = .005 and Pr{D} = .001. Then Pr{T|D*} = “False Positive” = .05, so then Pr{T*|D*} = 1 - Pr{T|D*} = 1 - .005 = .995 and then Pr{T|D} = 1 - Pr{T*|D} = 1 - .005 = .995.
Pr{D|T} =(.995)(.001)/(.995(.001) + .005*(.999)) = .00099/(.00099+.04995) = .16611 (16%). So in this case, approximately 16% of the positive tests actually indicate disease, which leaves the other 84% with a false finding.
What is Pr{D*|T*}, the probability that disease is absent given a negative screen?
Pr{D*|T*} =
Pr{DandT}/Pr{T*} =
Pr{T*|D*}Pr{D}/Pr{T*} =
Pr{T*|D*}Pr{D}/(Pr{TandD} + Pr{TandD)) =
Pr{T|D*}Pr{D}/(Pr{T*|D*}Pr{D} + Pr{T*|D}*Pr{D})
That is,
Pr{D*|T*} = Pr{T*|D*}Pr{D}/(Pr{T*|D*}Pr{D} + Pr{T*|D}*Pr{D})
We need the prevalence of disease (Pr{D}), Pr{T*|D*} = 1 - Pr{T|D*}, where Pr{T|D*} is the false-positive rate, and Pr{T*|D}, the false negative rate. Recall that Pr{D*) = 1 - Pr{D}.
Suppose that we are given that Pr{T|D*} = “False Positive” = .05, Pr{T*|D} = “False Negative” = .01 and Pr{D} = .001. Then Pr{T|D*} = “False Positive” = .05, so then Pr{T*|D*} = .95 and Pr{T*|D} = “False Negative” = .01, so then Pr{T|D} = .99
Plugging in what we know:
Pr{D*|T*} = (.95)(.999)/( (.95)(.999) + (.01)*(.001)) = .9999+. So in this case, approximately 99.99% of the negative tests actually indicate absence of disease.
Let’s repeat the calculation, but with Pr{T|D*} = “False Positive” = .005, Pr{T*|D} = “False Negative” = .005 and Pr{D} = .001. Then Pr{T|D*} = “False Positive” = .05, so then Pr{T*|D*} = 1 - Pr{T|D*} = 1 - .005 = .995 and then Pr{T|D} = 1 - Pr{T*|D} = 1 - .005 = .995.
Pr{D*|T*} = .995*.999/(.995*.999 + .005*.001) = .9999+. So in this case, approximately 99.99% of the negative tests actually indicate absence of disease.
What this application suggests is that the likely problem with a diagnostic test is the proper interpretation of positive tests.