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How do you calculate $P(X = n|G), P(X = n|G^C)$ by the Law of Total Probability, with extra conditioning?

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  1. Please see $P(U|G)$ and $P(U|G^C)$ below, beside my red line. Can you please expound these calculations?

  2. How was $s$ computed in both equations?

Example 2.4.5 (Unanimous agreement).

The article "Why too much evidence can be a bad thing" by Lisa Zyga [30] says:

Under ancient Jewish law, if a suspect on trial was unanimously found guilty by all judges, then the suspect was acquitted. This reasoning sounds counterintu- itive, but the legislators of the time had noticed that unanimous agreement often indicates the presence of systemic error in the judicial process.

There are n judges deciding a case. The suspect has prior probability p of being guilty. Each judge votes whether to convict or acquit the suspect. With probability s, a systemic error occurs (e.g., the defense is incompetent). If a systemic error occurs, then the judges unanimously vote to convict (i.e., all n judges vote to convict).

Whether a systemic error occurs is independent of whether the suspect is guilty. Given that a systemic error doesn't occur and that the suspect is guilty, each judge has probability c of voting to convict, independently. Given that a systemic error doesn't occur and that the suspect is not guilty, each judge has probability w of voting to convict, independently. Suppose that

$0 < p < 1; 0 < s < 1;$ and $0 < w < \frac12 < c < 1$.

(a) For this part only, suppose that exactly k out of n judges vote to convict, where k < n. Given this information, find the probability that the suspect is guilty.

(b) Now suppose that all n judges vote to convict. Given this information, find the probability that the suspect is guilty.

(c) Is the answer to (b), viewed as a function of n, an increasing function? Give a short, intuitive explanation in words.

Solution: (a) Since $k<n$, a systemic error didn't occur. We will implicitly condition on this in this part. Let $G$ be the event that the suspect is guilty and $X$ be the number of judges who vote to convict. Using Bayes' rule, LOTP, and the Binomial PMF, $$ P(G \mid X=k)=\frac{P(X=k \mid G) P(G)}{P(X=k)}=\frac{p c^{k}(1-c)^{n-k}}{p c^{k}(1-c)^{n-k}+(1-p) w^{k}(1-w)^{n-k}} $$ (b) Let $U$ be the event $X=n$ and $B$ be the event that a systemic error occurs. Then $$ P(G \mid U)=\frac{P(U \mid G) P(G)}{P(U)}=\frac{p P(U \mid G)}{p P(U \mid G)+(1-p) P\left(U \mid G^{c}\right)} $$ By LOTP with extra conditioning,

$$ \color{red}{\begin{aligned} P(U \mid G) &=P(U \mid G, B) P(B \mid G)+P\left(U \mid G, B^{c}\right) P\left(B^{c} \mid G\right)=s+(1-s) c^{n} \\ P\left(U \mid G^{c}\right) &=P\left(U \mid G^{c}, B\right) P\left(B \mid G^{c}\right)+P\left(U \mid G^{c}, B^{c}\right) P\left(B^{c} \mid G^{c}\right)=s+(1-s) w^{n} \end{aligned}} $$

Thus, $$ P(G \mid U)=\frac{p\left(s+(1-s) c^{n}\right)}{p\left(s+(1-s) c^{n}\right)+(1-p)\left(s+(1-s) w^{n}\right)} $$ (c) No, since a large value of $n$ yields a high chance of systemic error, and if a systemic error occurs then the judges' votes are uninformative about whether the suspect is guilty. The answer to (b) reverts to the prior probability $p$ as $n \rightarrow \infty$

Original image

Blitzstein, Introduction to Probability (2019 2 ed), Example 2.4.5, p 62.

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