1.2 Conditional probabilities and bayes' rule

If $A$ and $B$ are two separate but possibly dependent random events, then:

• Probability of $A$ and $B$ occurring together = $(,(A, B))$
• The conditional probability of $A$ , given that $B$ occurs = $(B, A)$
• The conditional probability of $B$ , given that $A$ occurs = $(A, B)$

• $(A)$ is the prior probability of the model $A$ (in the absence of any evidence);
• $(B)$ is the probability of the evidence $B$ ;
• $(A, B)$ is the likelihood that the evidence $B$ was produced, given that the model was $A$ ;
• $(B, A)$ is the posterior probability of the model being $A$ , given that the evidence is $B$ .

The following example illustrates the concepts of Bayesian model selection.

Loaded dice

Problem:

Given a tub containing 100 six-sided dice, in which one die is known to be loaded towards the six to a specified extent,derive an expression for the probability that, after a given set of throws, an arbitrarily chosen die is the loaded one?Assume the other 99 dice are all fair (not loaded in any way). The loaded die is known to have the following pmf: $$p_L(1)=0.05$$ $$\{p_L(2), \dots , p_L(5)\}=0.15$$ $$p_L(6)=0.35$$ Here derive a good strategy for finding the loaded die from the tub.

Solution:

The pmfs of the fair dice may be assumed to be: $$\forall i, i=\{1, \dots , 6\}\colon p_F(i)=\frac{1}{6}$$ Let each die have one of two states, $S=L$ if it is loaded and $S=F$ if it is fair. These are our two possible models for the random process and they have underlying pmfs given by $\{p_L(1), \dots , p_L(6)\}$ and $\{p_F(1), \dots , p_F(6)\}$ respectively.

After $N$ throws of the chosen die, let the sequence of throws be ${\Theta }_N=\{{\theta }_1, \dots , {\theta }_N\}$ , where each ${\theta }_i\in \{1, \dots , 6\}$ . This is our evidence .

We shall now calculate the probability that this die is the loaded one. We therefore wish to find the posterior $({\Theta }_N, S=L)$ .

We cannot evaluate this directly, but we can evaluate the likelihoods , $(S=L, {\Theta }_N)$ and $(S=F, {\Theta }_N)$ , since we know the expected pmfs in each case. We also know the prior probabilities $(S=L)$ and $(S=F)$ before we have carried out any throws, and these are $\{0.01, 0.99\}$ since only one die in the tub of 100 is loaded. Hence we can use Bayes' rule:

We may calculate this iteratively by noting that

The choice of these thresholds for $({\Theta }_N, S=L)$ is a function of the desired tradeoff between speed of searching versus the probability of failure to findthe loaded die, either by moving on to the next die even when the current one is loaded, or by selecting a fair dieas the loaded one.

The lower threshold, $p_1=10^{-4}$ , is the more critical, because it affects how long we spend before discarding each fair die. The probability ofcorrectly detecting all the fair dice before the loaded die is reached is $(1-p_1)^n\approx 1-n{p}_1$ , where $n\approx 50$ is the expected number of fair dice tested before the loaded one is found. So the failure probability due toincorrectly assuming the loaded die to be fair is approximately $n{p}_1\approx 0.005$ .

The upper threshold, $p_2=0.995$ , is much less critical on search speed, since the loaded result only occurs once, so it is a good idea to set it very close to unity. The failureprobability caused by selecting a fair die to be the loaded one is just $1-p_2=0.005$ . Hence the $\mathrm{overall failure probability }=0.005+0.005=0.01$

One solution to this problem is to compute only the ratio of likelihoods, as in . A more generally useful solution is to computelog(likelihoods) instead. The product operations in the expressions for the likelihoods then become sums oflogarithms. Even the calculation of likelihood ratios such as $R_N$ and comparison with appropriate thresholds can be done in the log domain. After this, it is OK to return tothe linear domain if necessary since $R_N$ should be a reasonable value as it is the ratio of very small quantities.