- Zhi Ying Chen (8.1)
- Michael Yampol (8.17) http://rpubs.com/myampol/MY606-HW8-15-Possums-Logistic-Regression
- Lewris Mota (8.16)
- Wei Zhou

May 1, 2019

- Zhi Ying Chen (8.1)
- Michael Yampol (8.17) http://rpubs.com/myampol/MY606-HW8-15-Possums-Logistic-Regression
- Lewris Mota (8.16)
- Wei Zhou

Kruschke’s videos are an excelent introduction to Bayesian Analysis https://www.youtube.com/watch?v=YyohWpjl6KU!

Doing Bayesian Data Analysis, Second Edition: A Tutorial with R, JAGS, and Stan

*The Theory That Would Not Die: How Bayes’ Rule Cracked the Enigma Code, Hunted Down Russian Submarines, and Emerged Triumphant from Two Centuries of Controversy* by Sharon Bertsch McGrayne

\[ P\left( A|B \right) =\frac { P\left( B|A \right) P\left( A \right) }{ P\left( B|A \right) P\left( A \right) +P\left( B|{ A }^{ ` } \right) P\left( { A }^{ ` } \right) } \]

Consider the following data from a cancer test:

- 1% of women have breast cancer (and therefore 99% do not).
- 80% of mammograms detect breast cancer when it is there (and therefore 20% miss it).
- 9.6% of mammograms detect breast cancer when it’s not there (and therefore 90.4% correctly return a negative result).

Cancer (1%) | No Cancer (99%) | |
---|---|---|

Test postive | 80% | 9.6% |

Test negative | 20% | 90.4% |

Now suppose you get a positive test result. What are the chances you have cancer? 80%? 99%? 1%?

- Ok, we got a positive result. It means we’re somewhere in the top row of our table. Let’s not assume anything — it could be a true positive or a false positive.
- The chances of a true positive = chance you have cancer * chance test caught it = 1% * 80% = .008
- The chances of a false positive = chance you don’t have cancer * chance test caught it anyway = 99% * 9.6% = 0.09504

Cancer (1%) | No Cancer (99%) | |
---|---|---|

Test postive | True +: 1% * 80% | False +: 99% * 9.6% |

Test negative | False -: 1% * 20% | True -: 99% * 90.4% |

\[ Probability = \frac{desired\quad event}{all\quad possibilities} \]

The chance of getting a real, positive result is .008. The chance of getting any type of positive result is the chance of a true positive plus the chance of a false positive (.008 + 0.09504 = .10304).

**So, our chance of cancer is .008/.10304 = 0.0776, or about 7.8%.**

It all comes down to the chance of a true positive result divided by the chance of any positive result. We can simplify the equation to:

\[ P\left( A|B \right) =\frac { P\left( B|A \right) P\left( A \right) }{ P\left( B \right) } \]

- Catch them all, count them. Not practical (or even possible)!
- We can sample some fish.

Our strategy:

- Catch some fish.
- Mark them.
- Return the fish to the pond. Let them get mixed up (i.e. wait a while).
- Catch some more fish.
- Count how many are marked.

For example, we initially caught 20 fish, marked them, returned them to the pond. We then caught another 20 fish and 5 of them were marked (i.e they were caught the first time).

Adopted from Rasmath Bååth useR! 2015 workshop: http://www.sumsar.net/files/academia/user_2015_tutorial_bayesian_data_analysis_short_version.pdf

Step 1: Define Prior Distribution. Draw a lot of random samples from the “prior” probability distribution on the parameters.

n_draw <- 100000 n_fish <- sample(20:250, n_draw, replace = TRUE) head(n_fish, n=10)

## [1] 209 130 107 81 81 166 81 128 186 98

hist(n_fish, main="Prior Distribution")

Step 2: Plug in each draw into the generative model which generates “fake” data.

pick_fish <- function(n_fish) { # The generative model fish <- rep(0:1, c(n_fish - 20, 20)) sum(sample(fish, 20)) } n_marked <- rep(NA, n_draw) for(i in 1:n_draw) { n_marked[i] <- pick_fish(n_fish[i]) } head(n_marked, n=10)

## [1] 1 3 3 2 2 2 4 2 2 7

Step 3: Keep only those parameter values that generated the data that was actually observed (in this case, 5).

post_fish <- n_fish[n_marked == 5] hist(post_fish, main='Posterior Distribution') abline(v=median(post_fish), col='red') abline(v=quantile(post_fish, probs=c(.25, .75)), col='green')

An “expert” believes there are around 200 fish in the pond. Insteand of a uniform distribution, we can use a binomial distribution to define our “prior” distribution.

n_fish <- rnbinom(n_draw, mu = 200 - 20, size = 4) + 20 hist(n_fish, main='Prior Distribution')

n_marked <- rep(NA, n_draw) for(i in 1:n_draw) { n_marked[i] <- pick_fish(n_fish[i]) } post_fish <- n_fish[n_marked == 5] hist(post_fish, main='Posterior Distribution') abline(v=median(post_fish), col='red') abline(v=quantile(post_fish, probs=c(.25, .75)), col='green')

Consider a pool table of length one. An 8-ball is thrown such that the likelihood of its stopping point is uniform across the entire table (i.e. the table is perfectly level). The location of the 8-ball is recorded, but not known to the observer. Subsequent balls are thrown one at a time and all that is reported is whether the ball stopped to the left or right of the 8-ball. Given only this information, what is the position of the 8-ball? How does the estimate change as more balls are thrown and recorded?

shiny_demo('BayesBilliards', package='DATA606')

See also: http://www.bryer.org/post/2016-02-21-bayes_billiards_shiny/