Chapter 2 Statistical Rethinking Exercises
Easy
2E1
The expressions that corresponds to the statement “The probability of Rain on Monday” are: Pr(rain|Monday) and Pr(rain,Monday) / Pr(Monday)
2E2
The statement that corresponds to Pr(Monday|rain) is “The probability that it is Monday, given that it is raining.
2E3
Pr(Monday|rain) and Pr(rain|Monday) * Pr(Monday) / Pr(rain)
2E4
Given our limited knowledge and the results of the experiment, we believe that the true proportion of water in the globe is 0.7.
Medium
2M1
1:
#Grid approximation with 1000 points
prior <- rep(1, 1000)
p_grid <- seq(from=0, to=1, length.out = 1000)
likelihood <- dbinom(3, 3, prob = p_grid)
unstd.posterior <- prior * likelihood
posterior <- unstd.posterior / sum(unstd.posterior)plot(p_grid, posterior, type="b",
ylab="Posterior Probability",
xlab = "Probability of Water")
2:
#Grid approximation with 1000 points
prior <- rep(1, 1000)
p_grid <- seq(from=0, to=1, length.out = 1000)
likelihood <- dbinom(3, 4, prob = p_grid)
unstd.posterior <- prior * likelihood
posterior <- unstd.posterior / sum(unstd.posterior)plot(p_grid, posterior, type="b",
ylab="Posterior Probability",
xlab = "Probability of Water")
3:
#Grid approximation with 1000 points
prior <- rep(1, 1000)
p_grid <- seq(from=0, to=1, length.out = 1000)
likelihood <- dbinom(5, 7, prob = p_grid)
unstd.posterior <- prior * likelihood
posterior <- unstd.posterior / sum(unstd.posterior)plot(p_grid, posterior, type="b",
ylab="Posterior Probability",
xlab = "Probability of Water")
2M2
1:
prior <- ifelse(p_grid < 0.5, 0, 1)
p_grid <- seq(from=0, to=1, length.out = 1000)
likelihood <- dbinom(3, 3, prob = p_grid)
unstd.posterior <- prior * likelihood
posterior <- unstd.posterior / sum(unstd.posterior)plot(p_grid, posterior, type="b",
ylab="Posterior Probability",
xlab = "Probability of Water")
2:
prior <- ifelse(p_grid < 0.5, 0, 1)
p_grid <- seq(from=0, to=1, length.out = 1000)
likelihood <- dbinom(3, 4, prob = p_grid)
unstd.posterior <- prior * likelihood
posterior <- unstd.posterior / sum(unstd.posterior)plot(p_grid, posterior, type="b",
ylab="Posterior Probability",
xlab = "Probability of Water")
3:
prior <- ifelse(p_grid < 0.5, 0, 1)
p_grid <- seq(from=0, to=1, length.out = 1000)
likelihood <- dbinom(5, 7, prob = p_grid)
unstd.posterior <- prior * likelihood
posterior <- unstd.posterior / sum(unstd.posterior)plot(p_grid, posterior, type="b",
ylab="Posterior Probability",
xlab = "Probability of Water")
2M3
Pr(water|Earth) = 0.7 so Pr(land|Earth) = 0.3 Pr(land|Mars) = 1
Pr(Mars) = 0.5 Pr(Earth) = 0.5
We want Pr(Earth|land):
Pr(Earth|land) = Pr(land|Earth) * Pr(Earth) / Pr(land)
which expands to:
Pr(Earth|land) = Pr(land|Earth) * Pr(Earth) / (Pr(land|Earth) * Pr(Earth) + Pr(land|Mars) * Pr(Mars))
which ends up being:
Pr(Earth|land) = 0.3 * 0.5 /(0.3 * 0.5 + 1 * 0.5)
0.3 * 0.5 /(0.3 * 0.5 +1 * 0.5)## [1] 0.23076922M4
Card1: B/B Card2: B/W Card3: W/W
If we observed one black side up, there are 2 ways of producing the black side up if we pick the B/B card and one way if we pick the B/W card, so in total, 2 of 3 ways are consistent with the other side of the card being black. The answer is 2/3.
2M5
With an extra B/B card:
If we observe one black side up, there are 4 ways of producing the black side up (2 for each B/B card) and 1 for picking the B/W card. The total probability that the other side is black is thus: 4/5
2M6
Card1: B/B Card2: B/W Card3: W/W
Black sides are heavier than cards with white sides so it is less likely that a card with black sides is pulled from the bag.
There are still 2 ways for B/B to produce a black side up, 1 way for B/W and zero ways for W/W. But according to the prior information, there is 1 way to get the B/B card, 2 ways to get the B/W card and 3 ways to get the W/W card so:
The total ways for the B/B card to produce a black side up are 2 * 1 and the total ways for the B/W card to produce a black side up are 1 * 2.
So, there are 4 ways total to see a black side up and 2 of these come from the B/B card so:
2/4## [1] 0.52M7
Card1: B/B Card2: B/W Card3: W/W
The B/B card has 2 ways to produce an observation with the black side up. This leaves the B/W and the W/W card with the job of producing the next observation. The B/W has only 1 way to produce a white side up while the W/W has 2 ways. So there are 3 ways to get the second card to show the white side up. Assuming the first card is B/B, there are 6 ways to see the BW sequence.
If however the B/W card is drawn first, there is 1 way for it to show the black side up. This leaves the other two cards to show the white side up. The B/B can’t show white while W/W has 2 ways to show white up so that is 2 ways to see the sequence Black and White when the first card drawn is B/W.
The probability is thus: 8 total ways of producing the sequence BW with 6 ways with B/B being drawn first.
6/8## [1] 0.752H1
We are trying to find the probability that the second birth is twins, knowing that the first birth was twins.
Pr(twin2|twin1) = Pr(twin1 & twin2) / Pr(twins)
Pr(twins) = 1/2 * (0.1) + 1/2 * (0.2) for species A and B respectively
Pr of a Species A female having two twins is: 1/2 * (0.1 * 0.1)
Pr of a Species B female having two twins is: 1/2 * (0.2 * 0.2)
Thus:
Pr_twins <- 0.5 * 0.1 + 0.5 * 0.2
Pr_twin1and2 <- 0.5 * (0.1 * 0.1) + 0.5 * (0.2 * 0.2)
Final_pr <- Pr_twin1and2 / Pr_twins
Final_pr## [1] 0.16666672H2
We are now trying to find Pr(SpeciesA|Twins)
According to Bayes’ Theorem this equals:
Pr(Twins|SpeciesA) * Pr(SpeciesA) / Pr(Twins)
The result thus is:
(0.1 * 0.5) / 0.15## [1] 0.33333332H3
Now this is an interesting problem concerning Bayesian Updating.
From the previous problem we know the posterior probability of being Species A is 1/3. This will be the new prior for this problem
We are thus trying to find Pr(SpeciesA|Singleton) = Pr(Singleton|SpeciesA) * Pr(SpeciesA) / Pr(Singleton)
This ends up being:
SpeciesA <- 1/3
Singleton <- 1/3 * (0.9) + 2/3 * (0.8)
Final_Pr <- (0.9 * SpeciesA) / Singleton
Final_Pr## [1] 0.362H4
Pr(+|SpeciesA) = 0.8 Pr(+|SpeciesB) = 0.65
Without previous information
Pr(SpeciesA|+) = Pr(+|SpeciesA) * Pr(SpeciesA) / Pr(+) (which expands to Pr(+|SpeciesA) * Pr(SpeciesA) + Pr(+|SpeciesB) * Pr(SpeciesB))
Pr(+) = 0.8 * 0.5 + 0.65 * 0.5 Pr(SpeciesA) = 0.5
positive_prob <- 0.8 * 0.5 + 0.65 * 0.5
SpeciesA <- 0.5
Final_Pr <- (0.8 * 0.5) / positive_prob
Final_Pr## [1] 0.5517241With previous information
Pr(SpeciesA) = 0.36
positive_prob <- 0.36 * 0.8 + (1-0.36) * 0.65
SpeciesA <- 0.36
Final_Pr <- (0.8 * 0.36) / positive_prob
Final_Pr## [1] 0.4090909