What do we learn by a second test?

Let us imagine that the same individual undergoes a second test and that the result is again Positive. How should we update our believes that this individual is infected, in the light of the second observation? The first idea would be to apply Bayes' rule in sequence, thus getting an overall Bayes' Factor of $(\pi_1/\pi_2)^2\approx 67$ that, multiplied by the initial odds of

, would give posterior odds of 7.4, or a probability of being infected of 88%, still far from a practical certainty. But the real question is if we can apply twice the same kind of test to the same person. It is easy to understand that the multiplication of the Bayes' factors assumes (stochastic) independence among them. In fact, according to probability theory we have to replace now Eq. (

) by

$\displaystyle \frac{ P(\mbox{Inf}\,\vert\,\mbox{Pos}^{(1)},\mbox{Pos}^{(2)})} { P(\mbox{NoInf}\,\vert\,\mbox{Pos}^{(1)},\mbox{Pos}^{(2)})}$

$\displaystyle =$

$\displaystyle \frac{P(\mbox{Pos}^{(1)},\mbox{Pos}^{(2)}\,\vert\,\mbox{Inf})} {P... ...(2)}\,\vert\,\mbox{NoInf})} \times \frac{P_0(\mbox{Inf})}{P_0(\mbox{NoInf})}\,,$

(20)

having indicated by Pos $^{(1)}$ and Pos $^{(2)}$ the two outcomes. Numerator and denominator of the Bayes' Factor are then

$\displaystyle P($ Pos $\displaystyle ^{(1)},$ Pos $\displaystyle ^{(2)}\,\vert\,$ Inf $\displaystyle )$	$\displaystyle =$	$\displaystyle P($ Pos $\displaystyle ^{(2)}\,\vert\,$ Pos $\displaystyle ^{(1)},$ Inf $\displaystyle )\cdot P($ Pos $\displaystyle ^{(1)}\,\vert\,$ Inf $\displaystyle )$
$\displaystyle P($ Pos $\displaystyle ^{(1)},$ Pos $\displaystyle ^{(2)}\,\vert\,$ NoInf $\displaystyle )$	$\displaystyle =$	$\displaystyle P($ Pos $\displaystyle ^{(2)}\,\vert\,$ Pos $\displaystyle ^{(1)},$ NoInf $\displaystyle )\cdot P($ Pos $\displaystyle ^{(1)}\,\vert\,$ NoInf $\displaystyle ) \,,$

which can be rewritten as

$\displaystyle P($ Pos $\displaystyle ^{(1)},$ Pos $\displaystyle ^{(2)}\,\vert\,$ Inf $\displaystyle )$	$\displaystyle =$	$\displaystyle P($ Pos $\displaystyle ^{(2)}\,\vert\,$ Inf $\displaystyle )\cdot P($ Pos $\displaystyle ^{(1)}\,\vert\,$ Inf $\displaystyle )$
$\displaystyle P($ Pos $\displaystyle ^{(1)},$ Pos $\displaystyle ^{(2)}\,\vert\,$ NoInf $\displaystyle )$	$\displaystyle =$	$\displaystyle P($ Pos $\displaystyle ^{(2)}\,\vert\,$ NoInf $\displaystyle )\cdot P($ Pos $\displaystyle ^{(1)}\,\vert\,$ NoInf $\displaystyle ) \,,$

and therefore we can factorize the two Bayes' factors, only if the two test results are independent. But this is far from being obvious. If the test response depends on something one has in the blood, different from the virus one is searching for, a second test of the same kind will most likely give the same result.