I have trouble understanding this exercise from A First Course in Probability by Sheldon Ross:

Evidence concerning the guilt or innocence of a defendant in a criminal investigation can be summarized by the value of an exponential random variable $X$ whose mean $μ$ depends on whether the defendant is guilty. If innocent, $μ = 1$; if guilty, $μ = 2$. The deciding judge will rule the defendant guilty if $X > c$ for some suitably chosen value of $c$.

(a) If the judge wants to be 95 percent certain that an innocent man will not be convicted, what should be the value of $c$?

This is the answer from the book:

Let $X_i$ be an exponential random variable with mean $i$, $i = 1, 2$.

The value $c$ should be such that $P\{X_1 > c\} = .05$. Therefore, $$e^{−c} = .05 = \frac{1}{20}$$ or $c = \log(20) = 2.996$.

But why should we compute conditional probability here instead of the probability of intersection of events, namely the probability that the defendant is innocent and the judge will rule him guilty? If we want to be sure that an innocent person won't be convicted, isn't it more natural to also consider the probability that he is innocent, instead of assuming it from the start?

  • $\begingroup$ That is just based on the interpretation of "the judge wants to be 95 percent certain that an innocent man will not be convicted." Interpreting this as a conditional probability, given the person is innocent, is what the book wants you to do. Interpreting this as an absolute probability would require you to know the a-priori probability of being innocent or guilty, and that is not given in the problem. If we do not know this a-priori probability, the extreme case is when $P[innocent]=1$ and in this case your solution is the same as the book. $\endgroup$ – Michael Sep 11 '16 at 17:53
  • $\begingroup$ Thank you. Yes, I understand what the author meant, but I want to know if my interpretation is good as well (better, worse?). I just want to know if the problem is in me or is this question really as ambiguous as I think. $\endgroup$ – bg5 Sep 11 '16 at 19:20
  • $\begingroup$ Your interpretation is valid. The problem could have been interpreted either way, but only one way gives you all the info you need. $\endgroup$ – Michael Sep 11 '16 at 19:23

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