AIDS test

The AIDS test problem (Example 7 of Section ) is a very standard one. Let us solve it using the Bayes factor:

$$\frac{P(\mbox{HIV}\,\vert\,\mbox{Positive})} {P(\overline{\mbox{HIV}}\,\vert\,\mbox{Positive})} = \frac{P(\mbox{Positive}\,\vert\,\mbox{HIV})} {P(\mbox{Positive}\,... ...erline{\mbox{HIV}})} \cdot \frac{P_\circ(\mbox{HIV})}{P(\overline{\mbox{HIV}})}$$

$$= \frac{\approx 1}{0.002}\times \frac{0.1/60}{\approx 1} = 500 \times \frac{1}{600} = \frac{1}{1.2}$$

$$P( \,\vert\, ) = 45.5\%\,.$$

Writing Bayes' theorem in this way helps a lot in understanding what is going on. Stated in terms of signal to noise and selectivity (see problem 1 in Section ), we are in a situation in which the selectivity of the test is not enough for the noisy conditions. So in order to be practically sure that the patient declared `positive' is infected, with this performance of the analysis, one needs independent tests, unless the patient belongs to high-risk classes. For example, a double independent analysis on an average person would yield

$$P($$HIV$$\,\vert\,$$Positive$$_1\cap$$   Positive$$_2) = 99.76\%\,,$$

similar^8.22 to that obtained in the case where a physician had a `severe doubt' (i.e. $P_\circ($HIV$)\approx P_\circ(\overline{\mbox{HIV}}$) that the patient could be infected:

$$P($$HIV$$\,\vert\,$$Positive$$, P_\circ($$HIV$$)\approx 0.5) = 99.80\%\,.$$

We see then that, as discussed several times (see Section ), the conclusion obtained by arbitrary probability inversion is equivalent to assuming uniform priors.