SR2 Chapter 3 Hard

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SR2 Chapter 3 Hard

Here’s my solutions to the hard exercises in chapter 3 of McElreath’s Statistical Rethinking, 2nd edition.

\(\DeclareMathOperator{\dbinomial}{Binomial} \DeclareMathOperator{\dbernoulli}{Bernoulli} \DeclareMathOperator{\dpoisson}{Poisson} \DeclareMathOperator{\dnormal}{Normal} \DeclareMathOperator{\dt}{t} \DeclareMathOperator{\dcauchy}{Cauchy} \DeclareMathOperator{\dexponential}{Exp} \DeclareMathOperator{\duniform}{Uniform} \DeclareMathOperator{\dgamma}{Gamma} \DeclareMathOperator{\dinvpamma}{Invpamma} \DeclareMathOperator{\invlogit}{InvLogit} \DeclareMathOperator{\logit}{Logit} \DeclareMathOperator{\ddirichlet}{Dirichlet} \DeclareMathOperator{\dbeta}{Beta}\)

Let’s first put the data into a tibble for easier manipulation later.

data(homeworkch3)

df <- tibble(birth1 = birth1, birth2 = birth2) %>% 
  mutate(birth = row_number())
The first few rows of the data.
birth1 birth2 birth
1 0 1
0 1 2
0 0 3
0 1 4
1 0 5
1 1 6

3H1

Let’s check we have the correct total cound and the correct number of boys.

h1_counts <- df %>% 
  gather(order, gender, -birth) %>% 
  summarise(boys = sum(gender), births = n())

Now we can grid approximate the posterior as before.

granularity <- 1000

h1_grid <- tibble(p = seq(0, 1, length.out = granularity)) %>% 
  mutate(prior = 1)

h1_posterior <- h1_grid %>% 
  mutate(
    likelihood = dbinom(h1_counts$boys, h1_counts$births, p),
    posterior = prior * likelihood,
    posterior = posterior / sum(posterior)
  )

The maximum a posteriori (MAP) value is the value of p that maximises the posterior.

h1_map <- h1_posterior %>% 
  slice(which.max(posterior)) %>% 
  pull(p)

h1_map
[1] 0.5545546
Solution 3H1: posterior probability of giving birth to a boy.
Solution 3H1: posterior probability of giving birth to a boy.

3H2

We draw samples with weight equalt to the posterior. We then apply the HPDI function to these samples, each time with a different width.

h2_samples <- h1_posterior %>% 
  sample_n(10000, replace = TRUE, weight = posterior) %>% 
  pull(p)

h2_hpdi <- h2_samples %>% 
  crossing(prob = c(0.5, 0.89, 0.97)) %>% 
  group_by(prob) %>% 
  group_map(HPDI) 

h2_hpdi
[[1]]
     |0.5      0.5| 
0.4574575 0.5735736 

[[2]]
    |0.89     0.89| 
0.4534535 0.6606607 

[[3]]
    |0.97     0.97| 
0.4294294 0.6616617 

3H3

The posterior predictive samples are possible observations according to our posterior.

h3_posterior_predictive <- rbinom(10000, 200, h2_samples)
Solution 3H3: the posterior predictive distribution for 200 births
Solution 3H3: the posterior predictive distribution for 200 births

The number of observed births is very close to the MAP of the posterior predictive distribution, suggesting we have a decent fit.

3H4

Our data are from birth pairs and so far we didn’t make any distinction between the first and second births. To test this assumption, we can perform a posterior predictive check as in 3H3, but this time for first births.

h4_posterior_predictive <- rbinom(10000, 100, h2_samples)
Solution 3H4: the posterior predictive distribution for 100 births
Solution 3H4: the posterior predictive distribution for 100 births

The fit doesn’t look quite as good for first births as it did for all births together. It also doesn’t look bad since there is still a fair bit of probability mass around the observed number of first birth boys.

3H5

As the final posterior predictive check, let’s check the number of boys born after a girl.

h5_counts <- df %>% 
  filter(birth1 == 0) %>% 
  summarise(boys = sum(birth2), births = n())

h5_posterior_predictive <- rbinom(10000, h5_counts$births, h2_samples)
Solution 3H5: the posterior predictive distribution for 100 births
Solution 3H5: the posterior predictive distribution for 100 births

The fit here looks bad, since the observed number of boys is higher than the bulk of the model’s expectations.


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