Let T be an event whose occurrence is to be investigated and the probability with which a witness(W) tells the truth is $p$. Let E be the event of witness testifying that the event T has occurred, then how can we express $p$ in term of T and E?

In my view expressing $p$ as $Pr[T | E]$ should be the correct way of expressing this probability but I cannot find any fault if we express $p$ as $Pr[E | T]$. Some insight into it would be useful.


I'm going to disagree with the other answer here.

The two values you gave in your question are not correct. The first one is closer, but it is the probability that a witness tells the truthgiven the event occurred. The problem doesn't tell you to assume the event occurred. So we need to consider both truthful circumstances.

  • The event occurred and the witness said it did

  • The event did not occur and the witness said it did not.

The presence of the word "and" in these circumstances tells us we want to use the joint distribution, instead of the conditional. I don't want to write the answer here until I know this is not an assignment, so I hope this helps.

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  • $\begingroup$ Derp, missed the probability for the conditions. Otherwise this is just repetitive now. $\endgroup$ – Eric Czech Sep 12 '15 at 13:13
  • $\begingroup$ I still think the joint distribution is cleaner. And you said both his answers were correct when they're both incorrect. I think is issue had to do with confusing conditionals and joints, as well as forgetting about the possibility of the two "not"s. $\endgroup$ – jlimahaverford Sep 12 '15 at 13:16

You're right, both ways work but I don't think that's what the problem is getting at. Or at least both ways work as long as the problem is really as narrow as having an event space with two binary, non-mutually-exclusive events (would be a stupid problem otherwise though, without more info).

I think the point of the problem is really just to think about the probability of being truthful when the event doesn't occur. That's the classic "bayes" logic example, 2 binary variables, 4 outcomes.

The four possibilities in this case are:

  1. $ T \wedge E $
  2. $ T \wedge \neg E $
  3. $ \neg T \wedge E $
  4. $ \neg T \wedge \neg E $

where "$\neg$" indicates that the event did not occur.

Regardless of which variable you condition on first, the probability of the witness being truthful is the same and you could express that truthfulness as:

$$p = P(E|T)P(T) + P(\neg E|\neg T)P(\neg T)$$ OR $$p = P(T|E)P(E) + P(\neg T|\neg E)P(\neg E).$$

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  • $\begingroup$ Adding conditionals conditioned on different things is almost never the way to go. $\endgroup$ – jlimahaverford Sep 12 '15 at 12:57
  • $\begingroup$ How does that make any sense? I assume you're somehow trying to argue that working with joint distributions is more intuitive but to say using conditional probability is "almost never the way to go" is ridiculous. $\endgroup$ – Eric Czech Sep 12 '15 at 13:21
  • $\begingroup$ Please reread the comments. It said conditionals Conditioned On Different Things (events). I wrote this before you edited your post adding a factor to each term. In the future when you edit an answer please note that it is being edited otherwise previous comments are no longer in context. $\endgroup$ – jlimahaverford Sep 12 '15 at 13:27

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