第 65 章 贝叶斯假设检验

library(tidyverse)
library(tidybayes)
library(rstan)

rstan_options(auto_write = TRUE)
options(mc.cores = parallel::detectCores())
theme_set(bayesplot::theme_default())

65.1 人们会给爱情片打高分?

这是一个关于电影评分的数据。我们想看下爱情片与动作片的平均得分是否存在显著不同?

movies <- read_rds(here::here("demo_data", "movies.rds"))
movies
## # A tibble: 100 × 5
##   title                 year rating genre genre_numeric
##   <chr>                <int>  <dbl> <fct>         <dbl>
## 1 One of Our Spies Is…  1966    5.6 Acti…             1
## 2 Black Pirate, The     1926    7.5 Acti…             1
## 3 Caged Heat 3000       1995    2.9 Acti…             1
## 4 G.P.U.                1942    8.8 Acti…             1
## 5 Goal Club             2001    9.2 Acti…             1
## 6 Dead or Alive: Hanz…  1999    6.9 Acti…             1
## # … with 94 more rows

65.1.1 可视化探索

看下两种题材电影评分的分布

movies %>%
  ggplot(aes(x = genre, y = rating, color = genre)) +
  geom_boxplot() +
  geom_jitter() +
  scale_x_discrete(
    expand = expansion(mult = c(0.5, 0.5))
  ) +
  theme(legend.position = "none") 

65.1.2 计算均值差

统计两种题材电影评分的均值

group_diffs <- movies %>% 
  group_by(genre) %>% 
  summarize(avg_rating = mean(rating, na.rm = TRUE)) %>% 
  mutate(diff_means = avg_rating - lag(avg_rating))

group_diffs
## # A tibble: 2 × 3
##   genre   avg_rating diff_means
##   <fct>        <dbl>      <dbl>
## 1 Action        5.55      NA   
## 2 Romance       6.22       0.67

65.1.3 t检验

传统的t检验

t.test(
  rating ~ genre,
  data = movies,
  var.equal = FALSE
) 
## 
##  Welch Two Sample t-test
## 
## data:  rating by genre
## t = -2.3, df = 84, p-value = 0.03
## alternative hypothesis: true difference in means between group Action and group Romance is not equal to 0
## 95 percent confidence interval:
##  -1.2571 -0.0829
## sample estimates:
##  mean in group Action mean in group Romance 
##                  5.55                  6.22

65.2 stan 代码

65.2.1 normal分布

先假定rating评分,服从正态分布,同时不同的电影题材分组考虑

\[ \begin{aligned} \textrm{rating} & \sim \textrm{normal}(\mu_{\textrm{genre}}, \, \sigma _{\textrm{genre}}) \\ \mu &\sim \textrm{normal}(\textrm{mean_rating}, \, 2) \\ \sigma &\sim \textrm{cauchy}(0, \, 1) \end{aligned} \]

stan_program <- '
data {
  int<lower=1> N;                            
  int<lower=2> n_groups;                     
  vector[N] y;                               
  int<lower=1, upper=n_groups> group_id[N];  
}
transformed data {
  real mean_y;
  mean_y = mean(y); 
}
parameters {
  vector[2] mu;                    
  vector<lower=0>[2] sigma;        
}
model {
  mu ~ normal(mean_y, 2);
  sigma ~ cauchy(0, 1);

  for (n in 1:N){
    y[n] ~ normal(mu[group_id[n]], sigma[group_id[n]]);
  }
}

generated quantities {
  real mu_diff;
  mu_diff = mu[2] - mu[1];
}

'

stan_data <- movies %>% 
  select(genre, rating, genre_numeric) %>% 
  tidybayes::compose_data(
    N        = nrow(.), 
    n_groups = n_distinct(genre), 
    group_id = genre_numeric, 
    y        = rating
  )

stan_best_normal <- stan(model_code = stan_program, data = stan_data)
stan_best_normal
## Inference for Stan model: anon_model.
## 4 chains, each with iter=2000; warmup=1000; thin=1; 
## post-warmup draws per chain=1000, total post-warmup draws=4000.
## 
##            mean se_mean   sd   2.5%    25%    50%
## mu[1]      5.55    0.00 0.25   5.06   5.39   5.55
## mu[2]      6.22    0.00 0.16   5.90   6.11   6.22
## sigma[1]   1.77    0.00 0.19   1.45   1.64   1.76
## sigma[2]   1.15    0.00 0.12   0.95   1.06   1.14
## mu_diff    0.67    0.00 0.30   0.09   0.46   0.66
## lp__     -86.87    0.03 1.47 -90.58 -87.59 -86.55
##             75%  97.5% n_eff Rhat
## mu[1]      5.72   6.05  3946    1
## mu[2]      6.32   6.55  3707    1
## sigma[1]   1.88   2.19  3885    1
## sigma[2]   1.22   1.40  4169    1
## mu_diff    0.86   1.25  3771    1
## lp__     -85.79 -85.09  1840    1
## 
## Samples were drawn using NUTS(diag_e) at Sun May  8 09:52:31 2022.
## For each parameter, n_eff is a crude measure of effective sample size,
## and Rhat is the potential scale reduction factor on split chains (at 
## convergence, Rhat=1).
stan_best_normal %>% 
  tidybayes::spread_draws(mu_diff) %>%
  ggplot(aes(x = mu_diff)) +
  tidybayes::stat_halfeye() +
  geom_vline(xintercept = 0)
stan_best_normal %>% 
  tidybayes::spread_draws(mu_diff) %>%
  
    ggplot(aes(x = mu_diff)) +
  stat_eye(side = "right", 
           fill = "skyblue",
             point_interval = mode_hdi, 
             .width = c(0.5, 0.89),
             interval_colour = "red", 
             point_colour = "red",
             width = 15.5, 
             height = 0.1
             ) +
  geom_vline(xintercept = c(-0.1, 0.1), linetype = "dashed", size = 1) +

  coord_cartesian(xlim = c(-1, 2)) +
    labs(x = "mu_diff", y = NULL)

65.2.2 等效检验

我们一般会采用实用等效区间 region of practical equivalence ROPE。实用等效区间,就是我们感兴趣值附近的一个区间,比如这里的均值差。频率学中的零假设是看均值差是否为0,贝叶斯则是看均值差有多少落入0附近的区间。具体方法就是,先算出后验分布的高密度区间,然后看这个高密度区间落在[-0.1, 0.1]的比例.

lower <- -0.1*sd(movies$rating)
upper <-  0.1*sd(movies$rating)

stan_best_normal %>% 
  tidybayes::spread_draws(mu_diff) %>%
  filter(
    mu_diff > ggdist::mean_hdi(mu_diff, .width = c(0.89))$ymin,
    mu_diff < ggdist::mean_hdi(mu_diff, .width = c(0.89))$ymax
  ) %>%
  summarise(
    percentage_in_rope = mean(between(mu_diff, lower, upper))
  )
## # A tibble: 1 × 1
##   percentage_in_rope
##                <dbl>
## 1                  0

在做假设检验的时候,我们内心是期望,后验概率的高密度区间落在实际等效区间的比例越小越小,如果小于2.5%,我们就可以拒绝零假设了;如果大于97.5%,我们就接受零假设。

stan_best_normal %>% 
  tidybayes::spread_draws(mu_diff) %>%
  pull(mu_diff) %>% 
  bayestestR::rope(x,
    range = c(-0.1, 0.1)*sd(movies$rating),
    ci = 0.89,
    ci_method = "HDI"
  )
## # Proportion of samples inside the ROPE [-0.15, 0.15]:
## 
## inside ROPE
## -----------
## 0.00 %

65.2.3 Student-t 分布

标准正态分布是t分布的极限分布

for (nu in c(1, seq(5, 50, by = 10))) {
 p <- tibble(x = seq(-5, 5, by=0.1)) %>% 
    ggplot(aes(x)) + 
    stat_function(fun = dnorm, color = 'gray') + 
    stat_function(fun = dt, args = list(df = nu), color = 'blue') +
    theme_classic() + 
    ylab("Density") + 
    xlab('Value') + 
    ggtitle(paste("df =", nu))
 
  print(p)
}

假定rating评分服从student-t分布,

\[ \begin{aligned} \textrm{rating} & \sim \textrm{student_t}(\nu, \,\mu_{\textrm{genre}}, \, \sigma _{\textrm{genre}}) \\ \mu &\sim \textrm{normal}(\textrm{mean_rating}, \, 2) \\ \sigma &\sim \textrm{cauchy}(0, \, 1) \\ \nu &\sim \textrm{exponential}(1.0/29) \end{aligned} \]

stan_program <- '
data {
  int<lower=1> N;                            
  int<lower=2> n_groups;                     
  vector[N] y;                               
  int<lower=1, upper=n_groups> group_id[N];  
}
transformed data {
  real mean_y;
  mean_y = mean(y); 
}
parameters {
  vector[2] mu;                    
  vector<lower=0>[2] sigma;        
  real<lower=0, upper=100> nu;     
}
model {
  mu ~ normal(mean_y, 2);
  sigma ~ cauchy(0, 1);
  nu ~ exponential(1.0/29);

  for (n in 1:N){
    y[n] ~ student_t(nu, mu[group_id[n]], sigma[group_id[n]]);
  }
}

generated quantities {
  real mu_diff;
  mu_diff = mu[2] - mu[1];
}

'

stan_data <- movies %>% 
  select(genre, rating, genre_numeric) %>% 
  tidybayes::compose_data(
    N        = nrow(.), 
    n_groups = n_distinct(genre), 
    group_id = genre_numeric, 
    y        = rating
  )

stan_best_student <- stan(model_code = stan_program, data = stan_data)
stan_best_student
## Inference for Stan model: anon_model.
## 4 chains, each with iter=2000; warmup=1000; thin=1; 
## post-warmup draws per chain=1000, total post-warmup draws=4000.
## 
##             mean se_mean    sd    2.5%     25%     50%
## mu[1]       5.54    0.00  0.25    5.06    5.36    5.54
## mu[2]       6.24    0.00  0.16    5.92    6.12    6.23
## sigma[1]    1.71    0.00  0.19    1.37    1.58    1.69
## sigma[2]    1.05    0.00  0.13    0.81    0.96    1.04
## nu         27.13    0.41 19.82    5.27   12.11   20.88
## mu_diff     0.70    0.00  0.30    0.11    0.49    0.70
## lp__     -119.52    0.04  1.60 -123.55 -120.37 -119.21
##              75%   97.5% n_eff Rhat
## mu[1]       5.71    6.03  4778    1
## mu[2]       6.35    6.55  4020    1
## sigma[1]    1.82    2.12  4126    1
## sigma[2]    1.13    1.31  3223    1
## nu         36.49   78.45  2313    1
## mu_diff     0.90    1.27  4327    1
## lp__     -118.34 -117.38  1797    1
## 
## Samples were drawn using NUTS(diag_e) at Sun May  8 09:54:10 2022.
## For each parameter, n_eff is a crude measure of effective sample size,
## and Rhat is the potential scale reduction factor on split chains (at 
## convergence, Rhat=1).
stan_best_student %>% 
  tidybayes::spread_draws(mu_diff) %>%
  ggplot(aes(x = mu_diff)) +
  tidybayes::stat_halfeye() +
  geom_vline(xintercept = 0)
stan_best_student %>%
  as.data.frame() %>% 
  head()
##   mu[1] mu[2] sigma[1] sigma[2]     nu mu_diff   lp__
## 1 5.586 6.099    1.922   1.0341 10.863  0.5128 -118.7
## 2 5.262 6.450    2.232   1.2170 13.952  1.1879 -122.9
## 3 5.166 6.334    1.665   0.8879 13.429  1.1676 -118.9
## 4 5.306 5.983    1.757   1.1454  7.552  0.6778 -120.7
## 5 5.525 6.312    1.716   0.9283 63.953  0.7872 -119.9
## 6 5.410 6.351    1.712   0.8915 70.707  0.9407 -121.5
stan_best_student %>%
  as.data.frame() %>%
  ggplot(aes(x = `mu[1]`)) +
  geom_density()
stan_best_student %>%
  tidybayes::gather_draws(mu[i], sigma[i]) %>%
  tidybayes::mean_hdi(.width = 0.89)
## # A tibble: 4 × 8
##       i .variable .value .lower .upper .width .point
##   <int> <chr>      <dbl>  <dbl>  <dbl>  <dbl> <chr> 
## 1     1 mu          5.54  5.14    5.94   0.89 mean  
## 2     1 sigma       1.71  1.41    2.01   0.89 mean  
## 3     2 mu          6.24  5.98    6.49   0.89 mean  
## 4     2 sigma       1.05  0.833   1.24   0.89 mean  
## # … with 1 more variable: .interval <chr>

65.3 小结

65.4 作业

  • 将上一章线性模型的stan代码应用到电影评分数据中

\[ \begin{aligned} \textrm{rating} & \sim \textrm{Normal}(\mu, \, \sigma) \\ \mu & = \alpha + \beta \, \textrm{genre} \\ \alpha &\sim \textrm{Normal}(0, \, 5) \\ \beta &\sim \textrm{Normal}(0, \, 1) \\ \sigma &\sim \textrm{Exponential}(1) \\ \end{aligned} \]

stan_program <- '
data {
  int<lower=1> N;            
  vector[N] y;        
  vector[N] x;           
}
parameters {
  real<lower=0> sigma;    
  real alpha;                     
  real beta;                     
}

model {
  y ~ normal(alpha + beta * x, sigma);  
  
  alpha ~ normal(0, 5);
  beta ~ normal(0, 1);
  sigma ~ exponential(1);  
}

'

stan_data <- list(
    N   = nrow(movies), 
    x   = as.numeric(movies$genre),  
    y   = movies$rating
  )

stan_linear <- stan(model_code = stan_program, data = stan_data)
stan_linear
## Inference for Stan model: anon_model.
## 4 chains, each with iter=2000; warmup=1000; thin=1; 
## post-warmup draws per chain=1000, total post-warmup draws=4000.
## 
##         mean se_mean   sd   2.5%    25%    50%    75%
## sigma   1.49    0.00 0.10   1.30   1.41   1.48   1.55
## alpha   4.92    0.01 0.46   4.07   4.60   4.91   5.23
## beta    0.64    0.01 0.29   0.06   0.45   0.64   0.84
## lp__  -91.23    0.03 1.24 -94.31 -91.80 -90.92 -90.33
##        97.5% n_eff Rhat
## sigma   1.70  1954    1
## alpha   5.83  1530    1
## beta    1.19  1543    1
## lp__  -89.80  1275    1
## 
## Samples were drawn using NUTS(diag_e) at Sun May  8 09:55:40 2022.
## For each parameter, n_eff is a crude measure of effective sample size,
## and Rhat is the potential scale reduction factor on split chains (at 
## convergence, Rhat=1).
stan_linear %>% 
  tidybayes::spread_draws(beta) %>%
  ggplot(aes(x = beta)) +
  tidybayes::stat_halfeye() +
  geom_vline(xintercept = 0)
stan_linear %>%
  tidybayes::gather_draws(beta) %>%
  tidybayes::mean_hdi(.width = 0.89)
## # A tibble: 1 × 7
##   .variable .value .lower .upper .width .point
##   <chr>      <dbl>  <dbl>  <dbl>  <dbl> <chr> 
## 1 beta       0.641  0.149   1.06   0.89 mean  
## # … with 1 more variable: .interval <chr>