# Posterior Probability

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 for part a I have this:

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

P(guilty | witness says guilty)= P(guilty and witness says guilty)/P(witness says guilty) = 0.5p/0.5 = p

Is part B asking the same thing?

• use tex typesetting please – Seyhmus Güngören Sep 29 '14 at 0:54
• Duplicate of this question asked (and answered) a day earlier on stats.SE. – Dilip Sarwate Sep 29 '14 at 2:33

Parts (a), (b), and (c) are all the same thing, but with different numbers of equally unreliable witnesses all claiming the defendant is guilty.

Let $G$ be the event of the defendant being guilty.

Let $W_n$ be the event of $n$ witnesses all claiming guilt.

So what you have so far for part (a) is:

\begin{align} \\ \mathsf P(G) & = 0.5 \\ \mathsf P(G^c) & = 0.5 \\ \mathsf P(W_1 \mid G) &= p \\ \mathsf P(W_1 \mid G^c) & = 1-p \\\therefore \mathsf P(G\mid W_1) &= \frac{\mathsf P(G)\mathsf P(W_1 \mid G)}{\mathsf P(G)\mathsf P(W_1 \mid G)+\mathsf P(G^c)\mathsf P(W_1 \mid G^c )} \\ & = \frac{0.5 p}{0.5 p + 0.5(1-p)} \\ & = p & \color{green}{\checkmark} \end{align}

Extending this, for part (b) and (c) you want: \begin{align} \mathsf P(G\mid W_2) &= \frac{\mathsf P(G)\mathsf P(W_2 \mid G)}{\mathsf P(G)\mathsf P(W_2 \mid G)+\mathsf P(G^c)\mathsf P(W_2 \mid G^c )} \\[2ex] \mathsf P(G\mid W_n) &= \frac{\mathsf P(G)\mathsf P(W_n \mid G)}{\mathsf P(G)\mathsf P(W_n \mid G)+\mathsf P(G^c)\mathsf P(W_n \mid G^c )} \end{align}

Now when two or more witnesses claim the defendant is guilty, they are either all reliable or all unreliable, depending on the defendent's guilt.

So, the probability that all $n$ witnesses claim guilt, when the defendant is guilty, is the probability of all witnesses being reliable: $$\mathsf P(W_n\mid G) = \underline{\qquad}$$

Likewise probability that all $n$ witnesses claim guilt, when the defendant is innocent, is the probability that they are all unreliable: $$\mathsf P(W_n\mid G^c) = \underline{\qquad}$$

Fill in the blanks, as functions of $p$ and $n$.

Hint: you already have them for when $n=1$.

• Alright so would P(W2|G)=p as well? That is one thing I'm not understanding – user179498 Sep 29 '14 at 2:38
• I'm also trying to figure out how the tree diagram would look. Any ideas? – user179498 Sep 29 '14 at 2:40
• No, @user179498 . If the defendant is guilty, then for two witnesses to both say that then they must both be reliable. When you flip two equally biased coins, with a bias towards heads of $p$, what is the probability that they both come up heads? – Graham Kemp Sep 29 '14 at 2:49
• If there is a bias towards heads, the probability would be heads comes almost every time – user179498 Sep 29 '14 at 3:06
• So the probability of it landing on heads is greater than p=0.5 – user179498 Sep 29 '14 at 3:16