Mudanças entre as edições de "Bayes Theorem - Teorema de Bayes"
Calsaverini (disc | contribs) m (New page: The **Bayes' theorem** is a result of probability theory that states that: math P(A|B) = \frac{P(B|A) P(A)}{P(B)} math where P(A...) |
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| − | The | + | The [http://en.wikipedia.org/wiki/Bayes%27_theorem Bayes' theorem] is a result of probability theory that states that: |
| − | + | <math>P(A|B) = \frac{P(B|A) P(A)}{P(B)}</math> | |
| − | P(A|B) = \frac{P(B|A) P(A)}{P(B)} | + | |
| − | + | ||
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. | 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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By the Bayes' theorem we have: | By the Bayes' theorem we have: | ||
| − | + | <math> | |
P(p | n_1,n_2) = \frac{P(n_1,n_2| p) P(p) }{P(n1,n2)} | P(p | n_1,n_2) = \frac{P(n_1,n_2| p) P(p) }{P(n1,n2)} | ||
| − | + | </math> | |
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: | 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: | ||
Edição das 19h29min de 25 de julho de 2007
The Bayes' theorem is a result of probability theory that states that:
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.
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?
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?
By the Bayes' theorem we have:
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: math \int_0^1 P(p| n_1, n_2) dp = 1 math
Then we have: math P(p | n_1,n_2) = \frac{P(n_1,n_2| p) }{ \int_0^1 P(n_1, n_2|p) dp} math
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: math P(n_1,n_2| p) \propto p^{n_1} (1-p)^{n_2} math
Noting that: math \int p^{n_1} (1-p)^{n_2} dp = \frac{n_1!n_2!}{(n_1+n_2+1)!} math
we finally get: math 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} math
The mean probability of 1 happening is: math \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)!} math So: math \left\langle p \right\rangle = \frac{n_1+1}{n_1+n_2+2} math
The mean quadratic probability is: math \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)!} math
So: math \left\langle p^2 \right\rangle = \frac{(n_1+1)(n_1+2)}{(n_1+n_2+2)(n_1+n_2+3)} math
And the variance is: math \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} math math \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)} math
