Mudanças entre as edições de "Bayes Theorem - Teorema de Bayes"

De Stoa
Ir para: navegação, pesquisa
Linha 1: Linha 1:
−
<math>a=b</math>
+
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>
 +
 
 +
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:
 +
<math>
 +
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:
 +
<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 average 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>

Edição das 01h45min de 4 de agosto de 2007

The Bayes' theorem is a result of probability theory that states that:

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.

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: P(p | n_1,n_2) = \frac{P(n_1,n_2| p) P(p) }{P(n1,n2)}

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: \int_0^1 P(p| n_1, n_2) dp = 1

Then we have: P(p | n_1,n_2) = \frac{P(n_1,n_2| p) }{ \int_0^1 P(n_1, n_2|p) dp} 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: P(n_1,n_2| p) \propto p^{n_1} (1-p)^{n_2}

Noting that: \int p^{n_1} (1-p)^{n_2} dp = \frac{n_1!n_2!}{(n_1+n_2+1)!} we finally get: 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} The mean probability of 1 happening is: \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)!} So: \left\langle p \right\rangle = \frac{n_1+1}{n_1+n_2+2}

The average quadratic probability is: \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)!}

So: \left\langle p^2 \right\rangle = \frac{(n_1+1)(n_1+2)}{(n_1+n_2+2)(n_1+n_2+3)}

And the variance is: \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} \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)}

Disponível em "http://wiki.stoa.usp.br/index.php?title=Bayes_Theorem_-_Teorema_de_Bayes&oldid=3513"
Ferramentas pessoais
Espaços nominais

Variantes
Ações
Navegação
Imprimir/exportar
Ferramentas