hugh farey
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- Joined
- Apr 20, 2013
- Messages
- 506
You're telling me. You seem to know a lot more about Bayesian statistics than I do, so I hope you can clarify a few more things.Hugh,
- A is the hypothesis that the shroud is authentic. ~A is the complement of A -- IOW, the shroud is not authentic.
- The other info we have about the shroud has been taken into account as estimates for P(A) and P(~A).
- Unfortunately, this stuff is difficult to get one's brain around if one hasn't had much practice at it (the category that Jay puts me in).
Presumably we start with the premise that we don't know whether the Shroud is either authentic or not. If we knew, then P(A) and P(~A) would be 1 and 0, or 0 and 1, whichever we knew.
Now, in your calculation, P(A) is the probability that the Shroud is authentic, given all we already know about it, but excluding the information about the presence of limestone - is that right? And then P(A|x) is the probability that the Shroud is authentic, given all we already know about it, and including the information about the presence of limestone.
Now if I may try to spread out your equation a bit - and please correct me if I get it wrong, but on the right hand of the equation there seem to be 9 opening parentheses, but only 7 closing parentheses. Surely this cannot be correct? The calculation seems to depend on the number and positioning of these parentheses, so I should be grateful if you could check them. Will that be OK?
Is this correct?
P(A|x) = (L(x|A)*P(A) / ((L(x|A)*P(A)+L(x|~A)*P(~A))
Removing the the dangling parentheses:
P(A|x) = L(x|A)*P(A) / ( L(x|A)*P(A) + L(x|~A)*P(~A) )
Unfortunately, I can't see the rationale behind your multiplying and dividing. I've been doing a bit of experimenting with coins to try to work out what we need. Would you be so kind as to check my working?
The probability of three fair coins falling all heads is 1/8. However, if we are told that one of the coins has a 75% chance of being two-headed, then the probability of the three coins falling all heads is (1/8 * 2) * 3/4 = 3/16. I multiplied the original probability by 2 as, by separating one of the coins, the probability of the other two both falling heads is doubled.
That seems to work. It also works with two coins. The probability of two fair coins falling all heads is 1/4. However, if we are told that one of the coins has a 75% chance of being two-headed, then the probability of the two coins falling all heads is (1/4 * 2) * 3/4 = 3/8.
In more general terms, the probability of x coins all falling heads is 1/2^x, assuming ignorance of the chance of any coin being double-headed. If one of them has a 75% chance of being double headed, the chance becomes:
(1/2^x) * 2 * 3/4
Now let's use your figures.
You give P(A) as 0.1, and therefore P(~A) as 0.9, which is generously in favour of non-authenticity, and you give L(x|A) as 0.95 and L(x|~A) as 0.5, which is debatable.
So, the probability of the Shroud being authentic, given a 50/100 chance of limestone, is 10/100. However, if we are told that there is an 95/100 chance of limestone being present, then the probability of the Shroud being authentic is 10/100 * 2 * 95/100 = 1900/10000 = 19/100.
If this is correct, then why do we have to divide by anything?
Many thanks to anybody who has got this far!

