Expected number of positives and its standard uncertainty

In Sec.  we have considered the numbers of positives and negatives that we expect to observe, analyzing 10000 individuals, using our initial parameters ($p_s=0.10$, $\pi_1=0.98$, $\pi_2=0.12$) but without taking into account the unavoidable `statistical fluctuation'. We do it now, using the probabilistic graphical model shown in Fig. ,

Figure: Graphical model in which the number of positives could come from infected or not infected individuals. Arrows with dashed lines stand for a deterministic link, being $n_{P}$ simply equal to the sum of $n_{P_I}$ and $n_{P_{NI}}$.

obtained by doubling the basic one of Fig. , one branch for the infectees and a second for the others. Then the numbers of positives resulting from the two contributions are added up. Note in Fig.  the dashed arrows from the nodes $n_{P_I}$ and $n_{P_{NI}}$ to the node $n_{P}$: they indicate a deterministic link,²⁸being $n_{P} = n_{P_I} + n_{P_{NI}}$.

The probability distribution of $n_P$ is with good approximation Gaussian, due to the well known large numbers behavior of the binomial distribution (and, moreover, to the properties of the sum of `random variables'). On the other hand, the expected value and the standard deviation of $n_P$ can been calculated exactly, using the properties of expected values and variances, thus getting (summarizing for sake of space with the symbol $I$, staying for all available information, the conditions on which the various quantities depend):

$$(n_P\,\vert\,I) = (n_{P_I}\,\vert\,I) + (n_{P_{NI}}\,\vert\,I)$$

$$= \pi_1\cdot n_I + \pi_2\cdot n_{NI}$$

$$= \pi_1\cdot p_s\cdot n_s + \pi_2\cdot (1-p_s)\cdot n_s$$ (43)

$$\sigma^2(n_P\,\vert\,I) = \sigma^2(n_{P_I}) + \sigma^2(n_{P_I})$$

$$= \pi_1\cdot (1-\pi_1)\cdot p_s\cdot n_s + \pi_2\cdot (1-\pi_2)\cdot (1-p_s)\cdot n_s$$ (44)

$$\sigma(n_P\,\vert\,I) = \sqrt{ \pi_1\cdot (1-\pi_1)\cdot p_s\cdot n_s + \pi_2\cdot (1-\pi_2)\cdot (1-p_s)\cdot n_s}\,,$$ (45)

with $n_s$ the sample size. Expected value and standard deviation of the fraction of the number of individuals tagged as positive ( $f_P=n_P/n_s$) are then

$$(f_P\,\vert\,I) = \frac{\mbox{E}(n_P\,\vert\,I)}{n_s} = \pi_1\cdot p_s + \pi_2\cdot (1-p_s)$$ (46)

$$\sigma(f_P\,\vert\,I) = \frac{ \sigma(n_P\,\vert\,I)}{n_s} = \sqrt{\frac{\,\pi_1\cdot (1-\pi_1)\cdot p_s + \pi_2\cdot (1-\pi_2)\cdot (1-p_s)\,}{\,n_s}}. \ \ \ \ \$$ (47)

For example, making use of our reference numbers ($n_s=10000$, $\pi_1=0.978$ and $\pi_2=0.115$) we get for some values of $p_s$ (expected value $\pm$ standard uncertainty):

$$\left.n_P\right\vert _{(n_s=10000,\,\pi_1= 0.978,\,\pi_2= 0.115,\,{\footnotesize\mbox{\boldmath$ p_s=0.0$}})} = 1150\pm 32 \ \ \ \longrightarrow \, f_P=0.1150\pm0.0032$$

$$\left.n_P\right\vert _{(n_s=10000,\,\pi_1= 0.978,\,\pi_2= 0.115,\,{\footnotesize\mbox{\boldmath$ p_s=0.1$}})} = 2013\pm 31 \ \ \ \longrightarrow \, f_P=0.2013\pm0.0031$$

$$\left.n_P\right\vert _{(n_s=10000,\,\pi_1= 0.978,\,\pi_2= 0.115,\,{\footnotesize\mbox{\boldmath$ p_s=0.2$}})} = 2876\pm 29 \ \ \ \longrightarrow \, f_P=0.2876\pm0.0029$$

$$\left.n_P\right\vert _{(n_s=10000,\,\pi_1= 0.978,\,\pi_2= 0.115,\,{\footnotesize\mbox{\boldmath$ p_s=0.5$}})} = 5465\pm 25 \ \ \ \longrightarrow \, f_P=0.5465\pm0.0025\,.$$

From this numbers we can get an idea about the precision we could get on $p_s$, if $\pi_1$ and $\pi_2$ were perfectly known, although their values are rather far from what one would ideally desire. For example, since under the hypotheses $p_s=0.1$ and $p_s=0$ (and similar numbers are obtained varying $p_s$ from $0.1$ to $0.2$) the expected difference of positives is $\Delta_{n_P}=863\pm 45$, it follows that, varying $p_s$ by $\pm 0.01$ the expected number of positives would vary by $\approx \pm\, (86\pm 4.5)$. This means that, roughly speaking, it could be possible to estimate $p_s$ with an uncertainty of $\pm 0.01$ or better.

Before taking into account the effects due to the uncertainties of $\pi_1$ and $\pi_2$, let us also see how the quality of the measurement depends on the sample size. In order to do this, we fix this time $p_s$ to our arbitrary value of $0.1$ and vary the sample size by about half order of magnitude (that is $\approx 10^{k/2}$, with $k=6, 7, \ldots, 10$), reporting in this case directly the expected fraction of positives:

$$\left.f_P\right\vert _{({\footnotesize\mbox{\boldmath$n_s=1000$}},\,\pi_1= 0.978,\,\pi_2= 0.115,\,p_s=0.1)} = 0.2013\pm 0.0097$$

$$\left.f_P\right\vert _{({\footnotesize\mbox{\boldmath$n_s=3000$}},\,\pi_1= 0.978,\,\pi_2= 0.115,\,p_s=0.1)} = 0.2013\pm 0.0056$$

$$\left.f_P\right\vert _{({\footnotesize\mbox{\boldmath$n_s=10000$}},\,\pi_1= 0.978,\,\pi_2= 0.115,\,p_s=0.1)} = 0.2013\pm 0.0031$$

$$\left.f_P\right\vert _{({\footnotesize\mbox{\boldmath$n_s=30000$}},\,\pi_1= 0.978,\,\pi_2= 0.115,\,p_s=0.1)} = 0.2013\pm 0.0018$$

$$\left.f_P\right\vert _{({\footnotesize\mbox{\boldmath$n_s=100000$}},\,\pi_1= 0.978,\,\pi_2= 0.115,\, p_s=0.1)} = 0.2013\pm 0.0010\,.$$

As we can see, if we knew perfectly $\pi_1$ and $\pi_2$, already a sample of a few thousands individuals would allow us to predict the fraction of tagged positives with a relative uncertainty of a few percent. However there are other effects to be taken into account:

• there is uncertainty about $\pi_1$ and $\pi_2$;
• the proportion of infectees in the sample is different from that in the population (that is, in general $p_s$ differs from $p$);
• the inference from the observed numbers of positives to $p_s$, and then to $p$, has to be done using sound probabilistic inferential methods.