Project 3.1: Probability
DUE DATE EXTENDED TO 10/25
Due 10/25 at 11:59pm: Submit your assignment in a .pdf or a .txt file. You may submit one assignment per group, as always,
though we recommend everyone work through this assignment on their own before meeting with a partner.
Question 1 (2 points).
Consider the following 3-sided dice with the given side values. Assume the dice are all fair
and all rolls are independent.
A: 2, 2, 5
B: 1, 4, 4
C: 3, 3, 3
- What is the expected value of each die?
- Consider the indicator function better(X,Y) which has value 1 if X>Y and
value -1 if X<Y. What are the expected values of better(A, B), better(B, C), better(C, A)? Why
are these sometimes called non-transitive dice?
Question 2 (2 points). Assume that a joint distribution
over two variables, X = {x, ¬x}and Y = {y, ¬y} is known to have the marginal distributions
P(x) = P(¬x) = P(y) = P(¬y). Give joint distributions satisfying these marginals for
each of these conditions:
- X and Y are independent
- Observing Y=y increases the belief in X=x, i.e. P(x | y) > P(x)
- Observing Y=y decreases the belief in X=x, i.e. P(x | y) < P(x)
Question 3 (2 points). On a day when an assignment
is due (A=a), the newsgroup tends to be busy (B=b), and the computer lab tends
to be full (C=c). Consider the following conditional probability tables
for the domain, where A = {a, ¬a}, B = {b, ¬b}, C = {c, ¬c}.
| P(A) |
P(B|A) |
P(C|A) |
|
|
|
B |
A |
P |
| b |
a |
0.90 |
| ¬b |
a |
0.10 |
| b |
¬a |
0.40 |
| ¬b |
¬a |
0.60 |
|
|
C |
A |
P |
| c |
a |
0.70 |
| ¬c |
a |
0.30 |
| c |
¬a |
0.50 |
| ¬c |
¬a |
0.50 |
|
- Construct the joint distribution out of these conditional probabilities
tables assuming B and C are independent given A.
- What is the marginal distribution P(B,C)? Are these two variables
absolutely independent in this model? Justify your answer using the
actual probabilities, not your intuitions.
- What is the posterior distribution over A given that B=b, P(A |
B=b)? What is the posterior distribution over A given that C=c, P(A |
C=c)? What about P(A | B=b, C=c)? Explain the pattern among
these posteriors and why it holds.
Question 4 (2 points). Sometimes, there is traffic
(cars) on the freeway (C=c). This could either be because of a ball game (B=b) or
because of an accident (A=a). Consider the following joint probability
table for the domain, where A = {a, ¬a}, B = {b, ¬b}, C = {c, ¬c}.
| P(A, B, C) |
|
A |
B |
C |
P |
| a |
b |
c |
0.018 |
| a |
b |
¬c |
0.002 |
| a |
¬b |
c |
0.126 |
| a |
¬b |
¬c |
0.054 |
| ¬a |
b |
c |
0.064 |
| ¬a |
b |
¬c |
0.016 |
| ¬a |
¬b |
c |
0.072 |
| ¬a |
¬b |
¬c |
0.648 |
|
- What is the distribution P(A,B)? Are A and B independent in this
model given no evidence? Justify your answer using the actual
probabilities, not your intuitions.
- What is the marginal distribution over A given no evidence?
- How does this change if we observe that C=c; what is the posterior
distribution P(A | C=c)? Does this change intuitively make
sense? Why or why not?
- What is the conditional distribution over A if we then learn there is a
ball game, P(A | B=b, C=c)? Does it make sense that observing B should
cause this update to A (called explaining-away)? Why or why not?
Question 5 (2 points). Often we need to carry out
reasoning over some pair of variables X, Y conditioned on the value of other
variable E.
- Using the definitions of conditional probabilities, prove the
conditionalized version of the product rule: P(x, y | e) = P(x | y, e) P(y |
e)
- Prove the conditionalized version of Bayes' rule: P(y | x, e) = P(x | y,
e) P(y | e) / P(x | e)
Question 6 (2 points). Suppose we wish to calculate
P(C=c | A=a, B=b).
- If we have no conditional independence information, which of the following
sets of tables are sufficient to calculate P(C=c | A=a, B=b)?
- P(A, B), P(C), P(A | C), P(B | C)
- P(A, B), P(C), P(A, B | C)
- P(A, B, C)
- P(C), P(A| C), P(B | C)
- P(C | A, B), P(A)
- Which are sufficient if we know that A and B are conditionally independent
given C?