bayesian_statistics
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bayesian_statistics [2019/10/03 00:38] floyd |
bayesian_statistics [2019/10/03 02:40] floyd |
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| $$P(A|B)P(B)=P(A \cap B)\mbox{.}$$ | $$P(A|B)P(B)=P(A \cap B)\mbox{.}$$ | ||
| (Note to draw a Venn diagram here to illustrate.) | (Note to draw a Venn diagram here to illustrate.) | ||
| - | For example, you go to a pet store looking for a small mammal with white fur. There is a box and you are told that one out of six animals in the box have white fur and that there are both eight mice and ten hamsters in the box for a total of 18 individuals. 2 of the mice have white fur and 1 of the hamsters have white fur. You reach in and can feel that you have grabbed a mouse but can't see it yet. Given that you have a mouse (finish later) | + | For example, you go to a pet store looking for a small mammal with white fur. There is a box and you are told that one out of six animals in the box have white fur and that there are both eight mice and ten hamsters in the box for a total of 18 individuals. 2 of the mice have white fur and 1 of the hamsters have white fur. Before you reach in the box you know the probability of grabbing an individual with white fur is $3/18 = 0.1\bar{6}$ |
| + | $$P(\mbox{white}|\mbox{mouse})=\frac{\mbox{white}\cap\mbox{mouse}}{P(\mbox{mouse})}=\frac{2/ | ||
| + | and indeed 1/4 of the mice have white fur (2 out of 8). This is an awkward way to calculate the probability, | ||
| - | Once we accept the relationships between the probabilities | + | Once we accept the relationships between the probabilities we can easily rearranged the system to the classical Bayesian equation |
| $$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 · Last modified: 2019/10/03 18:21 by floyd
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