Question

a) How many pieces of evidence must the prosecutor gather before the jury decides to convict the accused? Use Bayes theorem and compute P(G|E1, ··· Ej ) for all j.

b) The defense is putting a lot of preassure in dismissing the case. The prosecutor has only time to discover 1 piece of evidence. Clearly with the standard of presumption of innocence and guilty beyond any reasonable doubt it would result in failure to convict. What is the minimal prior probability P(G) = p that the jury must have, before seeing any evidence in order to convict after seeing E1?

All the information is written above. Please answer both the parts.

Answer 1

a) From the evidence probabilities and the exchangability, by Bayes' Theorem, we get that

$P(G|E_1,\dots,E_j) = \frac{P(E_1,\dots,E_j|G)P(G)}{P(E_1,\dots,E_j|G)P(G)+P(E_1,\dots,E_j|G^c)P(G^c)} \\ = \frac{P(G)\prod_{i=1}^j P(E_i|G)}{P(G)\prod_{i=1}^j P(E_i|G)+P(G^c)\prod_{i=1}^j P(E_i|G^c)} \\ = \frac{0.05\times(0.9)^j}{0.05\times(0.9)^j+0.95\times(0.1)^j} = \frac{9^j}{9^j+19}$

Now, if prosecutor should gather n many evidences such that the jury convicts the accused, we get that

$P(G|E_1,\dots,E_n) \ge 0.95\\ \Rightarrow \frac{9^n}{9^n+19} \ge 0.95 \\ \Rightarrow 1+19\cdot9^{-n} \le \frac{20}{19} \\ \Rightarrow 19\cdot9^{-n} \le \frac1{19} \\ \Rightarrow 9^{n} \ge 361 \Rightarrow n \ge \log_9361 \approx 2.68$

Hence, the prosecutor must gather at least 3 pieces of evidence.

b) In this case, the probaility of being guilty after seeing 1 piece of evidence would be:

$P(G|E_1) = \frac{P(E_1|G)P(G)}{P(E_1|G)P(G)+P(E_1|G^c)P(G^c)} \\ = \frac{0.9p}{0.9p+0.1(1-p)} = \frac{0.9p}{0.8p+0.1} = \frac{9p}{8p+1}$

In order to prove guilty after 1 evidence, we get that

PG|E1) 0.95 9p >0.95 8p+1 9p 7.6p +0.95 >1.4p > 0.95 19 p +AI 00

This should be the minimum prior probability.