a) Your initial belief is that a defendant in a court case is guilty with probability 0.5. A witness comes forward claiming he saw the defendant committed the crime. You know the witness is not totally reliable and tells the truth with probability p. Calculate the posterior probability that the defendant is guilty, based on the witness’s evidence.

b) A second witness, equally unreliable, comes forward and claims she saw the defendant committed the crime. Assuming the witnesses are not colluding, what is your posterior probability of guilt?

c) In total, $n$ equally unreliable witnesses claim that they saw the defendant committed the crime. If there is no collusion among them, what is your posterior probability of guilt?

d) Compare the answers to a), b) and c). How do you explain this result?

So far I know I have to make a probability tree have and I two marginal probabilities which are $P(\mathrm{Guilty})=0.5$ and $P(\mathrm{NotGuilty})=0.5$. How to extend the tree branches from these two marginal probabilities?

  • $\begingroup$ While you can ask these kind of questions (routine bookwork questions) here, they are treated somewhat differently. Please add the self-study tag, and read its tag wiki, altering your question if necessary. $\endgroup$ – Glen_b Sep 28 '14 at 6:32
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    $\begingroup$ WHAT ABOUT b? can someone please post the answer for part b aswell. Thank you. $\endgroup$ – user56661 Sep 29 '14 at 19:47
  • $\begingroup$ @Hassan is right: the really interesting aspect of this question concerns how multiple unreliable witnesses can change the posterior probability (and in which direction they change it). For many people this result may be unintuitive. $\endgroup$ – whuber Oct 31 '14 at 14:32

Instead of a tree, use a contingency table (which is the same thing, but lays out the calculations in a more convenient form). Instead of probabilities, perform the calculations in terms of odds.

Because the problem eventually concerns multiple witnesses, let's address the case where after $n-1$ witnesses have come forward our belief in guilt has probability $q$, say (and therefore our belief in lack of guilt is $1-q$). Because the next witness will split each row into two parts in the proportion $p:(1-p)$, the table must divide up like this:

               Witness:     Guilty Not guilty  Total
                        ---------- ---------- |-----
Prior belief is guilty:         pq     (1-p)q |    q
        ... not guilty: (1-p)(1-q)     p(1-q) |  1-q

The updated belief in guilt when the witness testifies to guilt will be found in the proportions of the "Guilty" column appearing in the first row. (The second column becomes irrelevant.) Since the total in the "Guilty" column is $pq + (1-p)(1-q)$, this proportion equals

$$\frac{pq}{pq + (1-p)(1-q)}.$$

A neater way to express this is in terms of the odds, assuming neither $p$ nor $q$ equals $1$. (When either does equal $1$, the odds become infinite but the calculations in terms of probabilities are easy.) The odds are found by dividing the belief in guilt (first row) given the witness is testifying to guilt (left column) by the belief in not guilty (second row) within the same column. The final odds are therefore

$$\frac{pq}{(1-p)(1-q)} = \frac{p}{1-p} \frac{q}{1-q}.$$

We see that the initial odds of $q/(1-q)$ have been updated to odds of $p/(1-p)\times q/(1-q)$. In other words,

The posterior odds of guilt are the prior odds of guilt $q/(1-q)$ multiplied by the witness's odds of guilt $p(1-p)$.

This updating will happen each time a witness comes forward. (That explains how to extend the probability tree, if one really wanted to: it consists of a sequence of identical steps, each leading to the same calculations.) Therefore, after $n$ witnesses appear, the initial odds $q_0 = (1/2)/(1 - 1/2) = 1$ have been multiplied by $\left(p/(1-p)\right)^n$. That is, with $n$ witnesses the odds of our belief in guilt should be updated to

$$\text{Odds(Guilty)} = \left(\frac{p}{1-p}\right)^n \frac{q_0}{1-q_0} = \left(\frac{p}{1-p}\right)^n.$$

This forms a geometric sequence with initial value $q_0/(1-q_0)=1$ and common ratio $\lambda=p/(1-p)$. When $\lambda\gt 1$ (equivalently, $p\gt 1/2$), which occurs when the witnesses are somewhat reliable, the sequence increases without bound, showing that the probability of guilt is growing large. When $\lambda\lt 1$ (i.e., $p\lt 1/2$), which occurs when the witnesses are more likely to lie than not, the sequence converges to zero, showing that the probability of guilt is becoming small: a pack of lying bastards should convince you the defendant is innocent! When $\lambda=1$ (which corresponds to witnesses who are doing no better than randomly guessing), the odds remain the same as always.

  • $\begingroup$ Although I hope it's obvious, it may be worth pointing out that this approach of multiplying the odds provides an immediate generalization to the case where the witnesses may have varying degrees of reliability, and thereby is useful for analyses in which the witnesses are drawn randomly and independently from some distribution of reliabilities. $\endgroup$ – whuber Nov 3 '14 at 23:14

P(guilty) = 0.5 P(witness says guilty | guilty) = p (Witness is telling the truth) P(witness says guilty | not guilty) = 1- p (Witness is telling a lie)

P(guilty and witness says guilty) = 0.5p P(not guilty and witness says guilty) = 0.5(1-p) P(witness says guilty) = 0.5[p + (1-p)] = 0.5

by bayes' rule P(guilty | witness says guilty)= P(guilty and witness says guilty)/P(witness says guilty) = 0.5p/0.5 = p

Source: http://slaystats.com/adms2320assignment.html

  • $\begingroup$ Okay makes sense. Is this answering only part a? $\endgroup$ – Jafri Sep 28 '14 at 18:52
  • $\begingroup$ Isn't part b asking the same thing? $\endgroup$ – Jafri Sep 28 '14 at 19:07

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