bayesian_statistics
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bayesian_statistics [2019/10/03 02:34] floyd |
bayesian_statistics [2019/10/03 02:40] floyd |
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| and indeed 1/4 of the mice have white fur (2 out of 8). This is an awkward way to calculate the probability, | and indeed 1/4 of the mice have white fur (2 out of 8). This is an awkward way to calculate the probability, | ||
| - | Once we accept the relationships between the probabilities | + | Once we accept the relationships between the probabilities we can easily rearranged the system to the classical Bayesian equation |
| $$P(M|D) P(D) = P(D|M) P(M)\mbox{, | $$P(M|D) P(D) = P(D|M) P(M)\mbox{, | ||
| $$P(M|D) = \frac{P(D|M) P(M)}{P(D)}\mbox{.}$$ | $$P(M|D) = \frac{P(D|M) P(M)}{P(D)}\mbox{.}$$ | ||
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| Let's bring this into our example. | Let's bring this into our example. | ||
| $$P(M_1|D) = \frac{P(D|M_1) P(M_1)}{P(D)}\mbox{.}$$ | $$P(M_1|D) = \frac{P(D|M_1) P(M_1)}{P(D)}\mbox{.}$$ | ||
| - | We saw above that $P(D) = P(M_1) P(D|M_1) + P(M_2) P(D|M_2)\mbox{, | + | We saw above that $P(D) = P(M_1) P(D|M_1) + P(M_2) P(D|M_2)\mbox{, |
| $$P(M_1|D) = \frac{P(D|M_1) P(M_1)}{ P(D|M_1) P(M_1) + P(D|M_2) P(M_2)}$$ | $$P(M_1|D) = \frac{P(D|M_1) P(M_1)}{ P(D|M_1) P(M_1) + P(D|M_2) P(M_2)}$$ | ||
| The equation is really a fraction out of the total of all possibilities (of the data under all models). | The equation is really a fraction out of the total of all possibilities (of the data under all models). | ||
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| $$P(M_1|D) = \frac{0.1 \times 0.05}{ 0.1 \times 0.05 + 1 \times 0.95} = \frac{0.005}{0.955} \approx 0.00524$$ | $$P(M_1|D) = \frac{0.1 \times 0.05}{ 0.1 \times 0.05 + 1 \times 0.95} = \frac{0.005}{0.955} \approx 0.00524$$ | ||
| - | This $P(M_1|D) | + | This $P(M_1|D) |
| What if we observed a second +/+ offspring? The probability of the data under the first hypothesis is now $P(D|M_1) = 0.1^2=0.01$ and $P(D|M_2) = 1^2=1$. | What if we observed a second +/+ offspring? The probability of the data under the first hypothesis is now $P(D|M_1) = 0.1^2=0.01$ and $P(D|M_2) = 1^2=1$. | ||
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| Let's say that we didn't see a heterozygous offspring; how many observations are enough to be reasonably confident that the male parent was a +/+ homozygote? Bayes factor is a proposed way to address this. It is a ratio of the two probabilities of the data given the models (here $M_2$ is in the numerator because it has more support from the data). | Let's say that we didn't see a heterozygous offspring; how many observations are enough to be reasonably confident that the male parent was a +/+ homozygote? Bayes factor is a proposed way to address this. It is a ratio of the two probabilities of the data given the models (here $M_2$ is in the numerator because it has more support from the data). | ||
| - | $$K = \frac{P(D|M_2)}{P(D|M_1} $$ | + | $$K = \frac{P(D|M_2)}{P(D|M_1)} $$ |
| So with the observation of a single +/+ offspring $K_1 = \frac{1}{0.1} = 10$. | So with the observation of a single +/+ offspring $K_1 = \frac{1}{0.1} = 10$. | ||
| Two +/+ offspring gives $K_2 = \frac{1}{0.01} = 100$. | Two +/+ offspring gives $K_2 = \frac{1}{0.01} = 100$. | ||
bayesian_statistics.txt ยท Last modified: 2019/10/03 18:21 by floyd
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