A simple Bayesian network

Let us go back to our toy model of section 2 and let us complicate it just a little bit, adding the possibility of incorrect testimony (but we also simplify it using uniform priors, so that we can focus on the effect of the uncertain evidence). For example, imagine you do not see directly the color of the ball, but this is reported to you by a collaborator, who, however, might not tell you always the truth. We can model the possibility of a lie in following way: after each extraction he tosses a die and reports the true color only if the die gives a number smaller than 6. Using the formalism of Appendix I, we have
$\displaystyle P(E_T\,\vert\,E,I)$ $\displaystyle =$ $\displaystyle 5/6$ (52)
$\displaystyle P(E_T\,\vert\,\overline E,I)$ $\displaystyle =$ $\displaystyle 1/6\,,$ (53)
i.e.$\displaystyle \hspace{3cm}$   $\displaystyle \hspace{3cm}$  
$\displaystyle \lambda(I)$ $\displaystyle =$ $\displaystyle 1/5.$ (54)

The resulting belief network,64relative to five extractions and to the corresponding five reports is shown in figure 2, redrawn in a different way in figure 11.
Figure: Same belief network of figure 2. This representation shows the `monitors' giving the initial probabilities of all states of the variables. If you like to test your intuition, try the guess how all probabilities change when the information changes in the following order: a) witness 1 says white ($ E_1T=W$); b) witness 2 also reports white ($ E_2T=W$); c) then witness 3 claims, contrary to the previous two, to have observed black ($ E_3T=B$); c) finally we directly observe the result of the fourth extraction, resulting black ($ E_4=B$). The solutions are in figures 13 to 16.
In this diagram the nodes are represented by `monitors' that provide the probability of each state of the variable. The green bars mean that we are in condition of uncertainty with respect to all states of all variable. Let us describe the several nodes: Let us now see what happens if we observe white (red bar in figure 12).
Figure: Status of the network after the observation of a white ball.
All probabilities of the network have been updated (Hugin [12] has nicely done the job for us66). We recognize the 93% of box $ B_1$, that we already know. We also see that the increased belief on this box makes us more confident to observe white balls in the following extractions (after re-introduction).

More interesting is the case in which our inference is based on the reported color (figure 13).

Figure: Status of the network after the report of a white ball (compare with figure 12).
The fact that the witness could lie reduces, with respect to the previous case, our confidence on $ B_1$ and on white balls in future extractions. As an exercise on what we have learned in appendix H, we can evaluate the `effective' Bayes factor $ \tilde O_{B_1}(W_T,I)$ that takes into account the testimony. Applying Eq. (41) we get
$\displaystyle \tilde O_{B_1}(W_T,I)$ $\displaystyle =$ $\displaystyle \tilde O_{B_1}(W,I) \times
\frac{1 + \lambda(I) \cdot \left[
...\lambda(I) \cdot \left[
\frac{ \tilde O_{H}(W,I)}{P(W\,\vert\,H,I)} -1
\right]}$ (57)
  $\displaystyle =$ $\displaystyle 13\times \frac{5}{17} = 3.82\,,$ (58)

or $ \Delta $JL$ _{B_1}(W_T,I)=0.58$, about a factor of two smaller than $ \Delta $JL$ _{B_1}(W,I)$, that was 1.1 (this mean we need two pieces of evidence of this kind to recover the loss of information due to the testimony).

The network gives us also the probability that the witness has really told us the truth, i.e. $ P(W\,\vert\,W_T,I)$, that is different from $ P(W_T\,\vert\,W,I)$, the reason being that white was initially a bit more probable than black.

Let us see now what happens if we get two concording testimonies (figure 14).

Figure: Network of figure 13 updated by a second testimony in favor of white.
As expected, the probability of $ B_1$ increases and becomes closer to the case of a direct observation of white. As usual, also the probabilities of future white balls increase.

The most interesting thing that comes from the result of the network is how the probabilities that the two witness lie change. First we see that they are the same, about 95%, as expected for symmetry. But the surprise is that the probability the the first witness said the truth has increased, passing from 85% to 95%. We can justify the variation because, in qualitative agreement with intuition, if we have concordant witnesses, we tend to believe to each of them more than what we believed individually. Once again, the result is, perhaps after an initial surprise, in qualitative agreement with intuition. The important point is that intuition is unable to get quantitative estimates. Again, the message is that, once we agree on the basic assumption and we check, whenever it is possible, that the results are reasonable, it is better to rely on automatic computation of beliefs.

Let's go on with the experiment and suppose the third witness says black (figure 15).

Figure: Network of figure 14 updated by a third testimony in favor of black.
This last information reduces the probability of $ B_1$, but does not falsify this hypothesis, as if, instead, we had observed black. Obviously, it does also reduce the probability of white balls in the following extractions.

The other interesting feature concerns the probability that each witness has reported the truth. Our belief that the previous two witnesses really saw what they said is reduced to 83%. But, nevertheless we are more confident on the first two witnesses than on the third one, that we trust only at 76%, although the lie factor is the same for all of them. The result is again in agreement with intuition: if many witnesses state something and fewer say the opposite, we tend to believe the majority, if we initially consider all witnesses equally reliable. But a Bayesian network tells us also how much we have to believe the many more then the fewer.

Let us do, also in this case the exercise of calculating the effective Bayes factor, using however the first formula in footnote 63: the effective odds $ \tilde O_{H}(B_T,I)$ can be written as

$\displaystyle \tilde O_{B_1}(B_T,I)$ $\displaystyle =$ $\displaystyle \frac{1}{P(W\,\vert\,B_2)+P(B\,\vert\,B_2)/\lambda},$ (59)

i.e. $ {1}/{[1/13 + (12/13)/(1/5)]} = {13}/{61} = 0.213$, smaller then 1 because they provide an evidence against box $ B_1$ ( $ \Delta $JL$ =-0.67$). It is also easy to check that the resulting probability of 75.7% of $ B_1$ can be obtained summing up the three weights of evidence, two in favor of $ B_1$ and two against it: $ \Delta $JL$ _{B_1}(W_T,W_T,B_T,I)= 0.58+0.58-0.67=
0.49$, i.e. $ \tilde O_{B_1}(W_T,W_T,B_T,I)=10^{0.49}=3.1$, that gives a probability of $ B_1$ of 3.1/(1+3.1)=76%.

Finally, let us see what happens if we really see a black ball ($ E_4$ in figure 16).

Figure: Network of figure 15 updated by a direct observation of a black ball.
Only in this case we become certain that the box is of the kind $ B_2$, and the game is, to say, finished. But, nevertheless, we still remain in a state on uncertainty with respect to several things. The first one is the probability of a white ball in future extractions, that, from now becomes 1/13, i.e. 7.7%, and does not change any longer. But we also remain uncertain on whether the witnesses told us the truth, because what they said is not incompatible with the box composition. But, and again in qualitative agreement with the intuition, we trust much more whom told black (1.6% he lied) than the two who told white (70.6% they lied).

Another interesting way of analyzing the final network is to consider the probability of a black ball in the five extractions considered. The fourth is one, because we have seen it. The fifth is 92.3% () because we know the box composition. But in the first two extractions the probability is smaller than it (70.6%), while in the third is higher (98.4%). That is because in the two different cases we had an evidence respectively against and in favor of them.

Giulio D'Agostini 2010-09-30