Assignment #1

Questions

2E1. Which of the expressions below correspond to the statement: the probability of rain on Monday? (1) Pr(rain) (2) Pr(rain|Monday) (3) Pr(Monday|rain) (4) Pr(rain, Monday)/ Pr(Monday)

Answer:(2)(4)

2E2. Which of the following statements corresponds to the expression: Pr(Monday|rain)? (1) The probability of rain on Monday. (2) The probability of rain, given that it is Monday. (3) The probability that it is Monday, given that it is raining. (4) The probability that it is Monday and that it is raining.

Answer:(3)

2E3. Which of the expressions below correspond to the statement: the probability that it is Monday, given that it is raining? (1) Pr(Monday|rain) (2) Pr(rain|Monday) (3) Pr(rain|Monday) Pr(Monday) (4) Pr(rain|Monday) Pr(Monday)/ Pr(rain) (5) Pr(Monday|rain) Pr(rain)/ Pr(Monday)

Answer:(1)(4)

2E4. The Bayesian statistician Bruno de Finetti (1906–1985) began his 1973 book on probability theory with the declaration: “PROBABILITY DOES NOT EXIST.” The capitals appeared in the original, so I imagine de Finetti wanted us to shout this statement. What he meant is that probability is a device for describing uncertainty from the perspective of an observer with limited knowledge; it has no objective reality. Discuss the globe tossing example from the chapter, in light of this statement. What does it mean to say “the probability of water is 0.7”?

Answer:According to what we know the estimate of the water value is 0.7 ```

2M1. Recall the globe tossing model from the chapter. Compute and plot the grid approximate posterior distribution for each of the following sets of observations. In each case, assume a uniform prior for p. (1) W, W, W (2) W, W, W, L (3) L, W, W, L, W, W, W

(1)

## [1] 1

p_grid <- seq(from = 0, to = 1, length.out = 20)
prior <- rep(1, 20)
likelihood <- dbinom(3, size = 3, prob = p_grid)
unstd.posterior <- likelihood * prior
posterior <- unstd.posterior / sum(unstd.posterior)
plot(p_grid, posterior, type = "b", 
  xlab = "water probability", ylab = "posterior probability")

(2)

## [1] 2

p_grid <- seq(from = 0, to = 1, length.out = 20)
prior <- rep(1, 20)
likelihood <- dbinom(3, size = 4, prob = p_grid)
unstd.posterior <- likelihood * prior
posterior <- unstd.posterior / sum(unstd.posterior)
plot(p_grid, posterior, type = "b", 
  xlab = "water probability", ylab = "posterior probability")

(3)

## [1] 3

p_grid <- seq(from = 0, to = 1, length.out = 20)
prior <- rep(1, 20)
likelihood <- dbinom(5, size = 7, prob = p_grid)
unstd.posterior <- likelihood * prior
posterior <- unstd.posterior / sum(unstd.posterior)
plot(p_grid, posterior, type = "b", 
  xlab = "water probability", ylab = "posterior probability")

2M2. Now assume a prior for p that is equal to zero when p < 0.5 and is a positive constant when p ≥ 0.5. Again compute and plot the grid approximate posterior distribution for each of the sets of observations in the problem just above.

(1)

## [1] 1

p_grid <- seq(from = 0, to = 1, length.out = 20)
prior <- ifelse(p_grid < 0.5, 0, 1)
likelihood <- dbinom(3, size = 3, prob = p_grid)
unstd.posterior <- likelihood * prior
posterior <- unstd.posterior / sum(unstd.posterior)
plot(p_grid, posterior, type = "b", 
  xlab = "water probability", ylab = "posterior probability")

(2)

## [1] 2

p_grid <- seq(from = 0, to = 1, length.out = 20)
prior <- ifelse(p_grid < 0.5, 0, 1)
likelihood <- dbinom(3, size = 4, prob = p_grid)
unstd.posterior <- likelihood * prior
posterior <- unstd.posterior / sum(unstd.posterior)
plot(p_grid, posterior, type = "b", 
  xlab = "water probability", ylab = "posterior probability")

p_grid <- seq(from = 0, to = 1, length.out = 20)
prior <- ifelse(p_grid < 0.5, 0, 1)
likelihood <- dbinom(5, size = 7, prob = p_grid)
unstd.posterior <- likelihood * prior
posterior <- unstd.posterior / sum(unstd.posterior)
plot(p_grid, posterior, type = "b", 
  xlab = "water probability", ylab = "posterior probability")

2M3. Suppose there are two globes, one for Earth and one for Mars. The Earth globe is 70% covered in water. The Mars globe is 100% land. Further suppose that one of these globes—you don’t know which—was tossed in the air and produced a “land” observation. Assume that each globe was equally likely to be tossed. Show that the posterior probability that the globe was the Earth, conditional on seeing “land” (Pr(Earth|land)), is 0.23.


**2M4.** Suppose you have a deck with only three cards. Each card has two sides, and each side is either
black or white. One card has two black sides. The second card has one black and one white side. The
third card has two white sides. Now suppose all three cards are placed in a bag and shuffled. Someone
reaches into the bag and pulls out a card and places it flat on a table. A black side is shown facing up,
but you don’t know the color of the side facing down. Show that the probability that the other side is
also black is 2/3. Use the counting method (Section 2 of the chapter) to approach this problem. This
means counting up the ways that each card could produce the observed data (a black side facing up
on the table).

Answer:
BB two black sides 2
BW one black and one white side 1
WW two white sides 0
Pr(BB)=BB/(BB+BW+WW)=2/(2+1+0)=2/3

2M5. Now suppose there are four cards: B/B, B/W, W/W, and another B/B. Again suppose a card is drawn from the bag and a black side appears face up. Again calculate the probability that the other side is black.

Answer: Pr(BB)=BB/(BB+BW+WW)=(2+2)/(2+2+1+0)=4/5


**2M6.** Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white
sides. As a result, it’s less likely that a card with black sides is pulled from the bag. So again assume
there are three cards: B/B, B/W, and W/W. After experimenting a number of times, you conclude that
for every way to pull the B/B card from the bag, there are 2 ways to pull the B/W card and 3 ways to
pull the W/W card. Again suppose that a card is pulled and a black side appears face up. Show that
the probability the other side is black is now 0.5. Use the counting method, as before.


Answer:             
BB 2*1=2
BW 1*2+2          
WW 0*3=0
Pr(BB)=BB/(BB+BW+WW)=2/(2+2+0)=1/2

2M7. Assume again the original card problem, with a single card showing a black side face up. Before looking at the other side, we draw another card from the bag and lay it face up on the table. The face that is shown on the new card is white. Show that the probability that the first card, the one showing a black side, has black on its other side is now 0.75. Use the counting method, if you can. Hint: Treat this like the sequence of globe tosses, counting all the ways to see each observation, for each possible first card.

Answer: BB 3+3=6 BW 2 WW 0 Pr(BB)=BB/(BB+BW+WW)=6/(6+2+0)=3/4


**2H1.** Suppose there are two species of panda bear. Both are equally common in the wild and live
in the same places. They look exactly alike and eat the same food, and there is yet no genetic assay
capable of telling them apart. They differ however in their family sizes. Species A gives birth to twins
10% of the time, otherwise birthing a single infant. Species B births twins 20% of the time, otherwise
birthing singleton infants. Assume these numbers are known with certainty, from many years of field
research.
Now suppose you are managing a captive panda breeding program. You have a new female panda
of unknown species, and she has just given birth to twins. What is the probability that her next birth
will also be twins?

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Assignment #1

Yongting Tan

2020-07-28

Chapter 2 - Large Worlds and Small Worlds

Questions