Consider the following.
A test T has a sensitivity p and a selectivity q, for a disease that has a prevalence in the population of r.
If Bob gets a positive test for the disease D, then the probability they have the disease, according to Bayes theorem is:
P(D | T) = p.r / ( p.r + q.(1-r) )
call this r', and let me use this as the updated prior. If Bob gets a second test, which is also positive, then I can repeat the calculation:
P(D | T T) = p.r' / ( p.r' + q.(1-r') ) and then substitute r' with the first equation:
P( D | T T) = p.(p.r / ( p.r.+ q.(1-r) ) ) / ( p.(p.r / ( p.r.+ q.(1-r) ) ) + q.( 1 - (p.r / ( p.r.+ q.(1-r) ) ) ) )
which after rearranging becomes:
P( D | T T ) = p.p.r / ( p.p.r + q.q.(1-r) )
...which is the same as if I had put the two test results into the original equation for P(D | T), without modifying the initial prior.
Is there a reason for this? I can't help but feel if I had come from some other direction, this conclusion would be obvious.
Any enlightenment welcome.
A test T has a sensitivity p and a selectivity q, for a disease that has a prevalence in the population of r.
If Bob gets a positive test for the disease D, then the probability they have the disease, according to Bayes theorem is:
P(D | T) = p.r / ( p.r + q.(1-r) )
call this r', and let me use this as the updated prior. If Bob gets a second test, which is also positive, then I can repeat the calculation:
P(D | T T) = p.r' / ( p.r' + q.(1-r') ) and then substitute r' with the first equation:
P( D | T T) = p.(p.r / ( p.r.+ q.(1-r) ) ) / ( p.(p.r / ( p.r.+ q.(1-r) ) ) + q.( 1 - (p.r / ( p.r.+ q.(1-r) ) ) ) )
which after rearranging becomes:
P( D | T T ) = p.p.r / ( p.p.r + q.q.(1-r) )
...which is the same as if I had put the two test results into the original equation for P(D | T), without modifying the initial prior.
Is there a reason for this? I can't help but feel if I had come from some other direction, this conclusion would be obvious.
Any enlightenment welcome.