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The theory of probability, particularly the causal Bayesian probability is quite intriguing, and sometimes gets challenging. I have one such problem at hand which makes sense to me, but not able to decode the mechanics of its sample space, or let's say the probability tree that defines the way I can look back to the cause. So, here it goes.

Before testing for COVID, Jemima has a prior probability of 0.45 of having COVID. If Jemima has COVID, there is a probability of 0.9 that the test would come positive and a probability of 0.1 that the test would come out negative. If Jemima does not have COVID, there is a probability of 0.14 that the test would come positive and a probability of 0.86 that the test would come out negative. Multiple tests provide independent information. Jemima takes two tests. Both tests come out negative. What is the posterior odds-ratio that Jemima has COVID?

So far, I have been able to figure out the obviously easy information that Jemima gets a negative result is

P(Negative) = $0.45 \times0.1+0.55\times0.86=0.518$

and hence, the Bayesian probability on a single trial that Jemima actually has COVID is

P(COVID = T | Negative) = $0.45 \times0.1\over 0.518 $$\approx 0.087$

But where does that lead me further, w.r.t. Jemima getting a negative test twice in a row, has stumped me honestly. I did try thinking as if i should take the square of the above probability of 0.518, for the joint probability of such an outcome, given independent events. But then, what to keep in the numerator to justify the objective? Thoughts??

Rupen Kohli
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  • Her test results are not independent events! They are both strongly dependent on whether or not she has COVID. Hence you must look at both cases (she has it or she doesn't). – lulu Nov 23 '21 at 18:24
  • To stress: saying that the test results yield independent information only means that an error on the first is independent of error on the second. This does not mean that the results are themselves independent (which, as I say, they clearly are not). – lulu Nov 23 '21 at 18:26
  • I am there with it, very good observation. – Rupen Kohli Nov 23 '21 at 18:41
  • So...what is your revised answer for the probability of getting two negatives? – lulu Nov 23 '21 at 18:41
  • So what ive done is that bunched together the two test outcomes together, within the total probability space. P(negative 1) = 0.450.10.1 = 0.0045 and P(Negative 2) = 0.550.860.86 = 0.40678. That gives total probability of 0.41128 and Hence, the causal probability is 0.0045/0.41128 = 0.01094... I think that makes sense now.... any correction there? – Rupen Kohli Nov 23 '21 at 18:56
  • Nope, that looks good! – lulu Nov 23 '21 at 18:58
  • Although test results aren't independent, you can certainly argue they are conditionally independent when you condition on the event that she has COVID. – Matthew H. Nov 23 '21 at 19:24
  • @MatthewPilling I am getting a good INCEPTION feeling there!! – Rupen Kohli Nov 23 '21 at 19:28

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This image is from my previous post Why is the disease frequency often included in examples on medical tests' reliability?. Please use its tree diagram and lowermost line as a reference for the following answer:

Given that two tests are independently negative, the probability that Jemima has COVID is $$\frac{0.45\times0.1\times0.1}{0.45\times0.1\times0.1+0.55\times0.86\times0.86}=1.0941\%.$$ Thus, the required odds is $\displaystyle\frac{1.0941}{100-1.0941}=0.01106$ $$=11:1000.$$

ryang
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