Solutions to exercises for Class 1
+Supporting code
+Below is code from Chapter 2 of the book that you can use to do the +exercises. Not all the code you require is here though. Make sure to not +rush it and take time to play around with each function. :))
+# Grid Approximation
+
+# define grid
+p_grid = seq(from = 0, to = 1, length.out=20)
+
+# define prior
+prior = rep(1, 20)
+
+# compute likelihood at each value in grid
+likelihood = dbinom(6, size = 9, prob = p_grid)
+
+# compute product of likelihood and prior
+unstd.posterior = likelihood * prior
+
+# standardize the posterior so it sums to 1
+posterior = unstd.posterior / sum(unstd.posterior)
+
+# plot the distribution
+plot(p_grid, posterior, type = "b",
+ xlab = "probability of water", ylab = "posterior_probability")
+mtext("20 points")
Exercises
+Medium
+2M1.
+# define grid
+p_grid = seq(from = 0, to = 1, length.out=20)
+
+# define prior
+prior = rep(1, 20)
+
+# compute likelihood at each value in grid
+likelihood = dbinom(5, size = 7, prob = p_grid)
+# change to dbinom(3, size = 3
+# change to dbinom(3, size = 4
+# change to dbinom(5, size = 7
+
+# compute product of likelihood and prior
+unstd.posterior = likelihood * prior
+
+# standardize the posterior so it sums to 1
+posterior = unstd.posterior / sum(unstd.posterior)
+
+# plot the distribution
+plot(p_grid, posterior, type = "b",
+ xlab = "probability of water", ylab = "posterior_probability")
+mtext("20 points")
2M2.
+# define grid
+p_grid = seq(from = 0, to = 1, length.out=20)
+
+# step prior, which codes our assumption that all probabilities before 0.5 are impossible
+prior = ifelse(p_grid < 0.5, 0 , 1)
+
+# compute likelihood at each value in grid
+likelihood = dbinom(5, size = 7, prob = p_grid)
+# change to dbinom(3, size = 3
+# change to dbinom(3, size = 4
+# change to dbinom(5, size = 7
+
+# compute product of likelihood and prior
+unstd.posterior = likelihood * prior
+
+# standardize the posterior so it sums to 1
+posterior = unstd.posterior / sum(unstd.posterior)
+
+# plot the distribution
+plot(p_grid, posterior, type = "b",
+ xlab = "probability of water", ylab = "posterior_probability")
+mtext("20 points")
2M3.
+# Pr(Earth|Land) = Pr(Land|Earth) * Pr(Earth) / Pr(Land)
+
+answer = 0.3 * 0.5 / ((1 + 0.3) / 2)
+
+answer## [1] 0.23076922M4.
+Using the table similar to one on page 25.
+B - black, W - white
+| Cards | +Ways to produce observation B | +
|---|---|
| BB | +2 | +
| BW | +1 | +
| WW | +0 | +
Total ways to produce observation B: 3
+Since BB card can produce B in two different ways (via each +side) out of total possible ways, the probability that the called pulled +out was indeed BB is 2/3 or 0.66.
+2M5.
+Here we just add another BB card.
+B - black, W - white
+| Cards | +Ways to produce observation B | +
|---|---|
| BB | +2 | +
| BB | +2 | +
| BW | +1 | +
| WW | +0 | +
Total ways to produce observation B: 5
+Now that we have two BB cards, we simply sum the number ways the can +produce observation B and divide it by the total number of +possible observations of B. So 4/5 or +0.8.
+2M6.
+Here we add another column to our table - our prior.
+B - black, W - white
+| Cards | +Ways to produce observation B | +Prior | +Updated count | +
|---|---|---|---|
| BB | +2 | +1 | +2 | +
| BW | +1 | +2 | +2 | +
| WW | +0 | +3 | +0 | +
Total ways to produce observation B: 4
+So both BW and BB cards are equally likely to +produce observation B.
+2M7.
+This task is best solved by drawing up the ‘garden of forking paths’ +from the beginning of the chapter. Essentially, we count how many ways +each starting card could have produced observation +W on the second draw.
+Let’s start with BB.
+library(dagitty)
+
+# Create a dagitty object
+garden <- dagitty("dag {
+ BB -> B1
+ BB -> B2
+ B1 -> BW
+ B1 -> WW
+ B2 -> BW
+ B2 -> WW
+}")
+
+# Plot it
+plot(garden)## Plot coordinates for graph not supplied! Generating coordinates, see ?coordinates for how to set your own.
As we can see above, if our first card was BB, there are 6 ways the +second draw could have been W - 3 for one side of BB (B1) and 3 +for the other side of BB (B2).
+Now let’s assume the first card drawn was BW and count the ways for +the second draw to be W.
+# Create our garden
+garden <- dagitty("dag {
+ BW -> B
+ B -> WW
+}")
+
+# Plot it
+plot(garden)## Plot coordinates for graph not supplied! Generating coordinates, see ?coordinates for how to set your own.
If the first card was BW, then the only way to get W on the +second draw is by pulling the WW card out - either of it’s side.
+Now we table our findings.
+| Cards | +Ways to produce observation W on second draw | +
|---|---|
| BB | +6 | +
| BW | +2 | +
| WW | +0 | +
Total ways to produce observation W on second draw: 8
+Therefore, to get the probability that the first card drawn was BB is +6/8 or 0.75.
+