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A person does two independent tests to check for an infection, both are positive. How high is the probability of an infection, if sensitivity $P(T|I)$ and specificity $P(\neg T| \neg I)$ of both tests as well as the prevalence $P(I)$ are known?

My approach: \begin{align} P(I|T_1, T_2) &= \frac{P(T_1, T_2|I)P(I)}{P(T_1, T_2)}\\ &= \frac{P(T_1|I)P(T_2|I)P(I)}{P(T_1)P(T_2)}\\ &= \frac{P(T_1|I)P(T_2|I)P(I)}{\big[P(T_1|I)P(I)+P(T_1|\neg I)P(\neg I)\big]\cdot \big[P(T_2|I)P(I)+P(T_2|\neg I)P(\neg I)\big]}\\ &= \frac{P(T_1|I)P(T_2|I)P(I)}{\big[P(T_1|I)P(I)+[1-P(\neg T_1|\neg I)][1-P(I)]\big]\cdot \big[P(T_2|I)P(I)+[1-P(\neg T_2|\neg I)][1-P(I)]\big]}\\ \end{align}

However, if I input numbers I often get $P(I|T_1, T_2) > 1$, which should not be possible (e.g. with $P(T|I) = P(\neg T| \neg I) = 0.95$ and $P(I) = 0.1$ the result is $P(I|T_1, T_2)\approx 4.6$).

Where is the mistake?

Alduno
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    I can't follow your computation of the denominator . "Independence" does not mean what you think. It does not, for instance, mean that the two tests report True independently of each other, since the results are (one hopes) strongly dependent on the true state of the world. In the numerator you've already done half the work. Now you just need the parallel expression for the probability that they both report True for a non-infected patient. – lulu Feb 21 '22 at 15:12

1 Answers1

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Based on @lulu's comment, this should be the solution:

\begin{align} P(I|T_1, T_2) &= \frac{P(T_1, T_2|I)P(I)}{P(T_1, T_2)}\\ &= \frac{P(T_1|I)P(T_2|I)P(I)}{P(T_1, T_2|I)P(I)+P(T_1, T_2|\neg I)P(\neg I)}\\ &= \frac{P(T_1|I)P(T_2|I)P(I)}{P(T_1|I)P(T_2|I)P(I)+P(T_1|\neg I)P(T_2|\neg I)P(\neg I)}\\ &= \frac{P(T_1|I)P(T_2|I)P(I)}{P(T_1|I)P(T_2|I)P(I)+[1-P(\neg T_1|\neg I)][1-P(\neg T_2|\neg I)][1-P(I)]} \end{align}

The problem was that $T_1$ and $T_2$ are not independent if they are not conditioned on $I$ or $\neg I$.

Alduno
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