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AIDS test

The AIDS test problem (Example 7 of Section [*]) is a very standard one. Let us solve it using the Bayes factor:
$\displaystyle \frac{P(\mbox{HIV}\,\vert\,\mbox{Positive})}
{P(\overline{\mbox{HIV}}\,\vert\,\mbox{Positive})}$ $\displaystyle =$ $\displaystyle \frac{P(\mbox{Positive}\,\vert\,\mbox{HIV})}
{P(\mbox{Positive}\,...
...erline{\mbox{HIV}})}
\cdot \frac{P_\circ(\mbox{HIV})}{P(\overline{\mbox{HIV}})}$  
  $\displaystyle =$ $\displaystyle \frac{\approx 1}{0.002}\times \frac{0.1/60}{\approx 1}
= 500 \times \frac{1}{600} = \frac{1}{1.2}$  
$\displaystyle P($HIV$\displaystyle \,\vert\,$Positive$\displaystyle )$ $\displaystyle =$ $\displaystyle 45.5\%\,.$  

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

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

similar8.22 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:

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

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


next up previous contents
Next: Gold/silver ring problem Up: Solution to some problems Previous: Solution to some problems   Contents
Giulio D'Agostini 2003-05-15