Empirical Independence

Calculate the independence of two events using both analytic and empirical techniques. I’m trying to assess whether the probability of having a meal classified as “dinner” depends on whether that meal includes “chicken” as its main dish.

The options for the main dish:

The options for the side dishes:

All possible combinations to create a meal:

dishes <- data.frame(
    main = c("chicken", "salmon", "pork", "chicken", "pancakes", "french toast"),
    side = c("salad", "salad", "green beans", "corn", "carrots", "bacon")
)

possible_meals <- dishes %>%
  cross_df() %>%
  mutate_if(is.factor,as.character)

possible_meals

# A tibble: 36 × 2
   main         side 
   <chr>        <chr>
 1 chicken      salad
 2 salmon       salad
 3 pork         salad
 4 chicken      salad
 5 pancakes     salad
 6 french toast salad
 7 chicken      salad
 8 salmon       salad
 9 pork         salad
10 chicken      salad
# … with 26 more rows

Event a will be, “the main course is chicken.” What is its probability?

# a = main course is chicken
# tally the number of meals that include chicken

sum(possible_meals$main == "chicken") / nrow(possible_meals)

[1] 0.3333333

So, p(a) = 0.333. Event b will be, “the meal is dinner.” What is its probability?

# b = the meal is dinner (rather than breakfast)
# tally the number of meals that are dinners rather than breakfast
# any meals with pancakes, french toast, or bacon are not dinner

# number of meal options for 'main" X number of meal options for 'side'
(
  sum(dishes$main != c('pancakes', 'french toast')) / nrow(dishes)
    *
  sum(dishes$side != "bacon") / nrow(dishes)
)

[1] 0.5555556

So, p(b) = 0.555. If a and b are independent, then p(b) should be the same as p(b | a). Does the probability of eating a meal classified as dinner depend on whether that meal includes chicken?

First, the analytic solution.

I need to find p(dinner & chicken) to solve. So tally the possible ways chicken can combine with other dishes to create a dinner platter.

tally_count <- 0
for(i in 1:nrow(possible_meals)){
  
  meal_df <- possible_meals[i,]
  
  contain_chicken <- meal_df$main == "chicken"
  no_bacon <- meal_df$side != "bacon"
  
  if(contain_chicken == T && no_bacon == T){tally_count <- tally_count + 1}
}

tally_count / nrow(possible_meals)

[1] 0.2777778

Cool, p(dinner & chicken) = 0.2777. Now I can calculate the conditional probability.

X does not equal p(b), so the two are dependent. How about the empirical solution?

# what is the empirical estimate of p(dinner | chicken)?
# to calculate, I need:
# p(dinner & chicken) / p(chicken)

sims <- 10000
df <- data.frame(
    chicken_and_dinner = c(rep(0, sims)),
    chicken = c(rep(0, sims))
    
)

for(j in 1:sims){
  
  eat_main <- sample(dishes$main, 1, replace = F)
  eat_side <- sample(dishes$side, 1, replace = F)
  
  chicken_and_dinner <- F
  
  if(eat_main == "chicken" && 
    (eat_side == "salad" | eat_side == "green beans" | eat_side == "corn" | eat_side == "carrots")){
    chicken_and_dinner <- T
    }
  
  
  chicken <- F
  if(eat_main == "chicken"){chicken <- T}
  
  single_run_result <- c(chicken_and_dinner, chicken)
  df[j, "chicken_and_dinner"] <- chicken_and_dinner
  df[j, "chicken"] <- chicken
  
}

tally_chicken_and_dinner <- sum(df$chicken_and_dinner == 1)
tally_chicken <- sum(df$chicken == 1)

prob_cd <- tally_chicken_and_dinner / sims
prob_c <- tally_chicken / sims

prob_cd / prob_c

[1] 0.8368014

Bo\(^2\)m =)