In richelbilderbeek/statistical_rethinking: These are my solutions to the exercises of 'Statistical Rethinking' by Richard McElreath

Chapter 2

2E1

To help my thoughts, I define:

• Probablity of rain = 1/3
• Probability of Monday = 1/7

p_rain <- 1/3
p_monday <- 1/7

(1) Yes: the probability of rain is independent of which day it is; knowing the day does not give you any new information (2) Yes: the is simply the notation of the verbal sentence (3) No: the chance that it is Monday is independent of which weather it is. The probability of Monday is something else than the probability of rain. Note that their values are the same in countries with a chance of rain of 1/7 (4) Yes: the chance of rain and Monday are 1/21 (it happens once per 21 days on average). Dividing this by the chance it is Monday, results in the chance it rains. You could read the notation as 'the chance of rain and Monday per chance it is Monday'

p_rain <- p_rain # 1
p_rain_on_monday <- p_rain #2
p_monday_on_rainy_day <- p_monday # 3
p_rain_and_monday_per_monday <- p_rain * p_monday / p_monday # 4

Now the answers

answer_1 <- p_rain_on_monday == p_rain
answer_2 <- p_rain_on_monday == p_rain_on_monday
answer_3 <- p_rain_on_monday == p_monday_on_rainy_day
answer_4 <- p_rain_on_monday == p_rain_and_monday_per_monday

Answers:

1. r answer_1
2. r answer_2
3. r answer_3
4. r answer_4

2E2

1. No, that would be Pr(rain, Monday)
2. No, that would be Pr(rain| Monday)
3. Yes, that would be Pr(Monday| rain)
4. No, that would be Pr(Monday, rain)

2E3

1. Yes, this is the notation of the verbal sentence
2. No, this is 'the possibility of rain (given it is Monday)'
3. No, this is 'the possibility of the combination of rain and Monday'
4. Yes, this is 'the possibility of the combination of rain and Monday per probability of rain', which simplifies to the probability of Monday
5. No, this is 'the possibility of the combination of Monday and rain per probability of Monday', which simplifies to the probability of rain

2E4

If the probability of water is 70%, this means that 70% of the Earth is covered with water.

2M1

This is the number of probabilities we will investigate:

n_probabilities <- 20
testit::assert(n_probabilities >= 1)

Our experiment involves throwing n_throws times and observing water n_water times.

n_throws <- 9
n_water <- 6
testit::assert(n_water <= n_throws)

Here we create a range of n_probabilities probabilities from 0 to 1.

p_grid <- seq(from = 0, to = 1, length.out = n_probabilities)
plot(p_grid, main = "The p_grid values")

Our prior knowledge is that we assume all probabilities are equally likely:

prior <- rep(1, n_probabilities)
plot(prior, main = "Prior", type = "b")

Using the binomial distribution, we can calculatethe likelihood of having n_water times water out of n_throws throws, for our range of probabilities:

likelihood <- dbinom(n_water, size = n_throws, prob = p_grid)
plot(likelihood, main = "Likelihood", type = "b")

We can get our marginal posterior:

posterior <- likelihood * prior / sum(likelihood * prior)
#unstd_posterior <- likelihood * prior / sum(likelihood * prior)
#posterior <- unstd_posterioir / sum(unstd_posterior)
plot(p_grid, posterior, main = "Posterior", type = "b")

richelbilderbeek/statistical_rethinking documentation built on May 27, 2019, 8:01 a.m.