Combining probabilities / results

[pinball1970]

TL;DR Summary

Six lateral flow tests and counting all negative. I started thinking if I took one test right after another how do I combine those probabilities correctly?

I had a LFT test this week which gives a 75% accuracy result.Negative.I also wanted to stick in some prior knowledge if I can, 20% of the UK currently has Covid say.Vanadium 50 illustrated this to me a while back so I hope I don’t mess it up

		20	False pos
No Covid in UK	80%	60	Neg
Covid in UK	20%	5	False Neg
		15	Pos

So if I am negative I am either positive with 5% probability or negative with 60% probability is this right?

If I had another test which is also negative how much confidence can I have now?

I am not sure how to combine the results, the “and” “or” x and add rule does not work I end up with probabilities over one

 

[PeroK]

There are three factors: the prevalence of the disease in the population ($d$); the sensitivity ($p$) of the test (which is the probability of a postive test given you have the virus); and, the specificity ($q$) of the test (which is the probability of a negative test given you do not have the virus).

The simplest way to analyse this is to use a probability tree:

$dp$ - Virus and postive test
$d(1-p)$ - Virus and negative test (false negative)

$(1-d)q$ - No virus and negative test
$(1-d)(1-q)$ - No virus and positive test (false positive)

If you test negative, then the probablity you do not have the virus is:$$P_0 = \frac{(1-d)q}{(1-d)q + d(1-p)}$$
I had to post this to get the Latex to render. Notes about second test to follow.

 

[PeroK]

Doing a second test (if the first is negative) has the following effect:

$d(1-p)p$ - Virus, negative first test, positive second test.
$d(1-p)(1-p)$ - Virus and two negative tests

$(1-d)q^2$ - No virus and two negative tests
$(1-d)q(1-q)$ - No virus, negative first test and positive second test

If you have two negative tests, then the probablity you do not have the virus is:$$P_1 = \frac{(1-d)q^2}{(1-d)q^2 + d(1-p)^2}$$
Note that assuming that the test is sufficently sensitive, then $(1-p)^2$ is very small compared to $q^2$ and $P_1 \approx 1$.

(This assumes that the test results are independent and there is nothing specific about an individual that would imply repeated false negatives.)

 

[pinball1970]

Doing a second test (if the first is negative) has the following effect:

$d(1-p)p$ - Virus, negative first test, positive second test.
$d(1-p)(1-p)$ - Virus and two negative tests

$(1-d)q^2$ - No virus and two negative tests
$(1-d)q(1-q)$ - No virus, negative first test and positive second test

If you have two negative tests, then the probablity you do not have the virus is:$$P_1 = \frac{(1-d)q^2}{(1-d)q^2 + d(1-p)^2}$$
Note that assuming that the test is sufficently sensitive, then $(1-p)^2$ is very small compared to $q^2$ and $P_1 \approx 1$.

(This assumes that the test results are independent and there is nothing specific about an individual that would imply repeated false negatives.)

I will try this and feedback thanks

 

[pbuk]

(This assumes that the test results are independent and there is nothing specific about an individual that would imply repeated false negatives.)

And therein lies the problem.

Assuming you are swabbing correctly then a negative test simply shows that the test has not detected antigens in your saliva/nasal mucus. It is perfectly possible that you have COVID-19 but do not currently have antigens in your saliva/nasal mucus*; you gain no information about this situation by doing another lateral flow test.

 

[PeroK]

And therein lies the problem.

Assuming you are swabbing correctly then a negative test simply shows that the test has not detected antigens in your saliva/nasal mucus. It is perfectly possible that you have COVID-19 but do not currently have antigens in your saliva/nasal mucus; you gain no information about this situation by doing another lateral flow test.

In general, the independence of tests on the same subject is another factor. That's a clinical question!

 

[pinball1970]

And therein lies the problem.

Assuming you are swabbing correctly then a negative test simply shows that the test has not detected antigens in your saliva/nasal mucus. It is perfectly possible that you have COVID-19 but do not currently have antigens in your saliva/nasal mucus*; you gain no information about this situation by doing another lateral flow test.

I put this in maths rather medical as I am genuinely interested in probability side of it.
All the numbers I have got so far make sense, If I increase/decrease the sensitivity for instance.

I am assuming “all else being equal” for the medical side.

I am not 100% on how the tree is built.

 

[PeroK]

I am not 100% on how the tree is built.

For a simple case like this, the tree has an initial branch into $d$ (has the virus) and $1-d$ (does not have the virus). Then, each of these branches according to negative and positive tests.

You then look at the end of each branch and decide whether it's in your sample space - in this case a negative test - or not. Then you pick out the cases you are interested in (does not have the disease) from this reduced sample space This gives you the probability of not having the disease given a negative test.

This is a good alternative to using the algebraic version of Bayes' theorem, as the theorem is emedded in the tree structure.

In the case where you have two negative tests, you simply extend the tree by branching each of the first negative tests again.

One of the great advantages is that if we complicate the situation with more options and more levels, then using the raw Bayes' theorem becomes algebraically unworkable compared to the tree representation.