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Linha 1: |
| − | The [http://en.wikipedia.org/wiki/Bayes%27_theorem Bayes' theorem] is a result of probability theory that states that:
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| − | <math>P(A|B) = \frac{P(B|A) P(A)}{P(B)}</math>
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| − | <math>a^b</math>
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| − | where P(A|B) is the conditional probability of event A given event B, P(A) is the probability of event A, and so on.
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| − | Let's wonder about the following problem: given a histogram of the frequencies measured of some event, what is the probability that the real distribution is a given function?
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| − | Let's simplify by assuming an experiment with two possible outcomes (1 and 2). Let's assume that in a statistical survey, event 1 is find n1 times and event 2 n2 times. What is the probability that the real probability of event 1 is p?
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| − | By the Bayes' theorem we have:
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| | <math> | | <math> |
| − | P(p | n_1,n_2) = \frac{P(n_1,n_2| p) P(p) }{P(n1,n2)} | + | \begin{align} |
| | + | P(A|B) & = \frac{P(B | A)\, P(A)}{P(B)} \\ |
| | + | & \propto L(A | B)\, P(A) |
| | + | \end{align} |
| | </math> | | </math> |
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| − | As we do not now the (unconditional) probability of the distribution being p, let's make the assumption that any distribution is as good as any other, that is, let's assume that P(p) is a constant. As P(n1,n2) does not depend on p, and we must satisfy:
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| − | [[math]]
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| − | \int_0^1 P(p| n_1, n_2) dp = 1
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| − | [[math]]
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| − | Then we have:
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| − | [[math]]
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| − | P(p | n_1,n_2) = \frac{P(n_1,n_2| p) }{ \int_0^1 P(n_1, n_2|p) dp}
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| − | [[math]]
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| − | The probability that 1 happens n1 times and 2 happens n2 times given that the probability of 1 is p is given by the binomial distribution:
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| − | [[math]]
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| − | P(n_1,n_2| p) \propto p^{n_1} (1-p)^{n_2}
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| − | [[math]]
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| − | Noting that:
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| − | [[math]]
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| − | \int p^{n_1} (1-p)^{n_2} dp = \frac{n_1!n_2!}{(n_1+n_2+1)!}
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| − | [[math]]
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| − | we finally get:
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| − | [[math]]
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| − | P(p | n_1,n_2) = \frac{p^{n_1} (1-p)^{n_2}}{ \int_0^1 p^{n_1} (1-p)^{n_2} dp} = \frac{(n_1+n_2+1)!}{n_1!n_2!}p^{n_1} (1-p)^{n_2}
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| − | [[math]]
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| − | The mean probability of 1 happening is:
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| − | [[math]]
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| − | \left\langle p \right\rangle = \int P(p | n_1,n_2) p dp = \frac{(n_1+n_2+1)!}{n_1!n_2!} \int p^{n_1+1} (1-p)^{n_2} dp =\frac{(n_1+n_2+1)!}{n_1!n_2!}\frac{(n_1+1)!n_2!}{(n_1+n_2+2)!}
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| − | [[math]]
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| − | So:
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| − | [[math]]
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| − | \left\langle p \right\rangle = \frac{n_1+1}{n_1+n_2+2}
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| − | [[math]]
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| − | The mean quadratic probability is:
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| − | [[math]]
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| − | \left\langle p^2 \right\rangle = \int P(p | n_1,n_2) p^2 dp = \frac{(n_1+n_2+1)!}{n_1!n_2!} \int p^{n_1+2} (1-p)^{n_2} dp =\frac{(n_1+n_2+1)!}{n_1!n_2!}\frac{(n_1+2)!n_2!}{(n_1+n_2+3)!}
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| − | [[math]]
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| − | So:
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| − | [[math]]
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| − | \left\langle p^2 \right\rangle = \frac{(n_1+1)(n_1+2)}{(n_1+n_2+2)(n_1+n_2+3)}
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| − | [[math]]
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| − | And the variance is:
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| − | [[math]]
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| − | \sigma = \left\langle p^2 \right\rangle - \left\langle p \right\rangle^2 = \frac{(n_1+1)(n_1+2)}{(n_1+n_2+2)(n_1+n_2+3)} - \frac{(n_1+1)^2}{(n_1+n_2+2)^2}
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| − | [[math]]
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| − | [[math]]
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| − | \sigma = \frac{(n_1+1)(n_1+2)(n_1+n_2+2) - (n_1+1)^2(n_1+n_2+3)}{(n_1+n_2+2)^2(n_1+n_2+3)}
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| − | [[math]]
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