Mudanças entre as edições de "Bayes Theorem - Teorema de Bayes"
Calsaverini (disc | contribs) |
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By the Bayes' theorem we have: | 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)} | P(p | n_1,n_2) = \frac{P(n_1,n_2| p) P(p) }{P(n1,n2)} | ||
| Linha 17: | Linha 18: | ||
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: | ||
| + | |||
<math> | <math> | ||
\int_0^1 P(p| n_1, n_2) dp = 1 | \int_0^1 P(p| n_1, n_2) dp = 1 | ||
| Linha 22: | Linha 24: | ||
Then we have: | Then we have: | ||
| + | |||
<math> | <math> | ||
P(p | n_1,n_2) = \frac{P(n_1,n_2| p) }{ \int_0^1 P(n_1, n_2|p) dp} | P(p | n_1,n_2) = \frac{P(n_1,n_2| p) }{ \int_0^1 P(n_1, n_2|p) dp} | ||
</math> | </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: | 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> | <math> | ||
P(n_1,n_2| p) \propto p^{n_1} (1-p)^{n_2} | P(n_1,n_2| p) \propto p^{n_1} (1-p)^{n_2} | ||
| Linha 31: | Linha 36: | ||
Noting that: | Noting that: | ||
| + | |||
<math> | <math> | ||
\int p^{n_1} (1-p)^{n_2} dp = \frac{n_1!n_2!}{(n_1+n_2+1)!} | \int p^{n_1} (1-p)^{n_2} dp = \frac{n_1!n_2!}{(n_1+n_2+1)!} | ||
</math> | </math> | ||
we finally get: | we finally get: | ||
| + | |||
<math> | <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} | 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> | </math> | ||
The mean probability of 1 happening is: | The mean probability of 1 happening is: | ||
| + | |||
<math> | <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)!} | \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> | </math> | ||
So: | So: | ||
| + | |||
<math> | <math> | ||
\left\langle p \right\rangle = \frac{n_1+1}{n_1+n_2+2} | \left\langle p \right\rangle = \frac{n_1+1}{n_1+n_2+2} | ||
| Linha 48: | Linha 57: | ||
The average quadratic probability is: | The average quadratic probability is: | ||
| + | |||
<math> | <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)!} | \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)!} | ||
| Linha 53: | Linha 63: | ||
So: | So: | ||
| + | |||
<math> | <math> | ||
\left\langle p^2 \right\rangle = \frac{(n_1+1)(n_1+2)}{(n_1+n_2+2)(n_1+n_2+3)} | \left\langle p^2 \right\rangle = \frac{(n_1+1)(n_1+2)}{(n_1+n_2+2)(n_1+n_2+3)} | ||
| Linha 58: | Linha 69: | ||
And the variance is: | And the variance is: | ||
| + | |||
<math> | <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} | \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> | ||
| + | |||
<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)} | \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> | </math> | ||
Edição das 01h47min de 4 de agosto 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:
Then we have:
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:
Noting that:
we finally get:
The mean probability of 1 happening is:
So:
The average quadratic probability is:
So:
And the variance is:
