Taking into account the uncertainty on $\pi_1$ and $\pi_2$

As we have seen in Sec. , the way to take into account all possible values of $\pi_1$ and $\pi_2$, using the rules of probability theory, consists in evaluating the following integral

$$f(n_P\,\vert\,n_s,p_s) = \int_0^1\!\!\int_0^1\!f(n_P\,\vert\,n_s,p_s,\pi_1,\pi_2)\cdot f(\pi_1)\cdot f(\pi_2)\, \pi_1 \pi_2\,.$$ (48)

Before tacking the problem of how to evaluate this integral, a very important remark on how we are going to model the uncertainty about $\pi_1$ and $\pi_2$ is in order.

• When we write $f(n_P\,\vert\,n_s,p_s,\pi_1,\pi_2)$, we are assuming, trivially, the same exact values of $\pi_1$ and $\pi_2$ for all the tests performed on the $n_s$ individuals of the sample.
• If, instead, their value is uncertain, and we describe their uncertainty by $f(\pi_1)$ and $f(\pi_2)$, again it means that the same two numerical values influence the results of the $n_s$ tests. But we just do not know with certainty which are these values.
• In particular, associating to these two parameters the pdf's $f(\pi_1)$ and $f(\pi_2)$ does not mean that $\pi_1$ and $\pi_2$ fluctuate from one test to one other. The two pdf's only describe the uncertainty on their numerical values.
• It is however reasonable to think that, from how the `test devices' are built up, each item could perform slightly differently than the other, but we shall ignore these possible test-to-test fluctuations, although they could be taken into account just extending the model.

Going back to the practical issue of evaluating the integral, we use again Monte Carlo methods, employing e.g. the R script provided in Appendix B.2, for the case of $n_s=10000$ and $p_s=0.1$. The result, shown in the bottom plot of Fig. ,

Figure: Probabilistic prediction of the numbers of positives, based on a hypothetical test on 10000 individuals, exactly 1000 of them being infected. In the upper plot we use $\pi_1=0.978$ and $\pi_2=0.115$. In the lower plot we take into account their possible variability (see text). The over-imposed curve shows a Gaussian with average 2013 and standard deviation 200, values obtained by the approximated Eqs. () and (). (The top histogram is repeated, with enlarged horizontal scale, in Fig. .)

is quite impressive, compared to the top one, in which the precise values $\pi_1=0.978$ and $\pi_2=0.115$ were used. The mean of the distribution is unchanged, as more or less expected (see Sec. ), but its standard deviation, which quantifies the uncertainty of the prediction, increases by more than a factor six. We have then good reasons to expect a similar effect when we will be interested in the `reverse' problem, that is inferring the number of infectees in the sample from the resulting number of positives. Going into details, we see that the expected number of positives is essentially the same of Sec.  (the reduction from 2060 to 2013 is simply due to the new reference values for $\pi_1$ and $\pi_2$ we are using starting from Sec. ). But this number is now accounted by an uncertainty, which rises to about 10% of its value, when the uncertainties about $\pi_1$ and $\pi_2$ are also taken into account.