Reasoning with uncertainty

Reasoning tasks involving uncertainty

Causes and effects
Kinds of reasoning
- Diagnostic inference (a form of abduction): reasoning from effects to causes
  Given a stalled robot, infer the probability that its motor needs to be rebuilt.
- Causal inference: reasoning from causes to effects
  Given that a robot's motor needs to be rebuilt, infer the probability that it will stall.
- Intercausal inference: reasoning between causes of a common effect
  Given a stalled robot, infer the change in the probability that the motor needs to be rebuilt if you know in addition that its battery needs to be replaced.
- Making decisions based on probabilities, costs, and utilities associated with different values for variables
  Given a probability that rebuilding the motor will fix the problem, a cost of rebuilding the motor, and a utility of having the robot working again, decide whether to rebuild the motor.

Basic concepts

Joint probability distribution: probability for each variable assignment (each atomic event)
Probabilities of conjunctions and disjunctions
Statistical dependence
Statistical independence
Prior and conditional probability

For independent variables,
The chain rule, relating joint and conditional probability
P(A ∧ ... ∧ Z) = P(A|B ∧ ... ∧ Z) P(B|C ∧ ... ∧ Z) ... P(Y|Z) P(Z)
Conditional independence
Conditional independence of S₁ and S₂, given D,

Not conditionally independent:
Where probabilities come from: experiments vs. beliefs, inconsistencies in beliefs
Bayes' rule
Avoiding estimation of prior probability of effects (symptoms)
Combining evidence with Bayes' rule: Bayesian updating

For "symptoms" which are unconditionally independent and conditionally independent given the "disease",

An example of Bayesian reasoning

You would like to know the chance that it will rain within the next day on a particular planet. You can see that there are green clouds in the sky but no purple clouds. You also know that the chance of rain in general during this season is .1, that when it's raining there have been green clouds in the sky during the past day 60% of the time and purple clouds in the sky 30% of the time, and that when it isn't raining there have been green clouds in the sky during the last day 20% of the time and purple clouds 50% of the time. Calculate the probability that it will rain, making the usual simplifying assumptions. Also explain what these assumptions are.

Answer

Bayesian belief networks

Network (directed acyclic graph) of nodes representing predicates and links dependencies (conditional probabilities) between them
Constraint built into the network: the probability of a node depends only on the probability of its immediate parents. This means that any two nodes that A and B that are connected via a bridging node C are conditionally independent given C.
A probability table specifying for each node how it depends on its parents (where these probabilities come from is still an issue)

Based on a set of observations about some of the nodes in the network, we want to infer the probabilities of other nodes. Consider the case where we observe earth shaking (es), no news report (¬nr), and equipment audible (ea). We want to find the probabiliity that an earthquake happened (eq), given these observations:

P(eq | es ∧ ¬nr ∧ ea)

We will consider both cases where eq is T and where it is F, but because the values of the observed predicates also depend on construction (c), we will have to consider all four subcases:
P(eq ∧ c | es ∧ ¬nr ∧ ea)      (1)
P(eq ∧ ¬c | es ∧ ¬nr ∧ ea)      (2)
P(¬eq ∧ c | es ∧ ¬nr ∧ ea)      (3)
P(¬eq ∧ ¬c | es ∧ ¬nr ∧ ea)      (4)
Once we have these, we will add the first two for our answer.
Using Bayes' Rule, we can rewrite each of these. (1) becomes
P(eq ∧ c | es ∧ ¬nr ∧ ea) =
P(es ∧ ¬nr ∧ ea | eq ∧ c) P(eq ∧ c) / P(es ∧ ¬nr ∧ ea) (5)
Similarly for (2), (3), and (4). We'll ignore the denominator (we can't calculate it anyway) because it's the same for all four cases, and we'll scale them so that they add up to 1.0 after we calculate them.
Because of the statistical independence of the observation nodes, of eq from c, of nr from c, and of ea from eq, the numerator of (5) is

P(es | eq ∧ c) P(¬nr | eq) P(ea | c) P(eq) P(c) (6)
ea doesn't depend directly on c, but rather depends on it via en. Considering both possible values of en and of c, we have
P(ea | c) = P(ea ∧ en | c) + P(ea ∧ ¬en | c) (7)
P(ea | ¬c) = P(ea ∧ en | ¬c) + P(ea ∧ ¬en | ¬c) (8)
Since ea and c are conditionally independent given en, we have (using the chain rule)
P(ea ∧ en | c) = P(ea | en ∧ c) P(en | c) = P(ea | en) P(en | c) (9)
And similarly for other values of en and c.
So for all four combinations of c and en, we have the following (italics are used for predicates whose values vary):

c en P(ea | en) P(en | c) = P(ea ∧ en | c) P(ea | c)

T T 0.7 × 0.9 = 0.63 0.64

T F 0.1 × 0.1 = 0.01

F T 0.7 × 0.05 = 0.035 0.13

F F 0.1 × 0.95 = 0.095

c	en	P(ea \| en) P(en \| c) = P(ea ∧ en \| c)	P(ea \| c)
T	T	0.7 × 0.9 = 0.63	0.64
T	F	0.1 × 0.1 = 0.01
F	T	0.7 × 0.05 = 0.035	0.13
F	F	0.1 × 0.95 = 0.095

Now we're ready to calculate the probabilities for different combinations of eq and c. The last column accounts for the missing denominator by normalizing the probabilities so that they sum to 1.0

eq	c	P(es \| eq ∧ c) P(¬nr \| eq) P(ea \| c) P(eq) P(c)	P(eq ∧ c \| es ∧ ¬nr ∧ ea)
T	T	0.99 × 0.1 × 0.64 × 0.01 × 0.1 = .000063	0.0035
T	F	0.95 × 0.1 × 0.13 × 0.01 × 0.9 = 0.00011	0.0061
F	T	0.2 × 0.95 × 0.64 × 0.99 × 0.1 = 0.012	0.67
F	F	0.05 × 0.95 × 0.13 × 0.99 × 0.9 = 0.0055	0.31

Combining the first two rows on the right gives the cases where eq is T, so the probability we want is 0.0096.

Inference in Bayesian networks
- Another example (R & N, p. 494)
- A Bayesian network contains evidence variables (at the bottom) representing aspects of the situation that we can observe and hidden variables that we can't observe.
- Any hidden variable can be a query variable when we want to know that probability of that variable having a particular value given values for some evidence variables. For example,
  P(b|m ∧ ¬j)
- By the definition of conditional probability
  P(b|m ∧ ¬j) = P(b ∧ m ∧ ¬j) / P(m ∧ ¬j)
- We don't have any direct way of calculating the denominator of this expression, but if we consider both the true and false cases for the query variable, they have the same demoninator and sum to 1:
  P(b|m ∧ ¬j) + (P¬b|m ∧ ¬j) =
  P(b ∧ m ∧ ¬j) / P(m ∧ ¬j) + P(¬b ∧ m ∧ ¬j) / P(m ∧ ¬j) = 1
  That is, the denominator is the sum of the joint probabilities:
  P(m ∧ ¬j) = P(b ∧ m ∧ ¬j) + P(¬b ∧ m ∧ ¬j)
- So now the problem of calculating the original conditional probability reduces to that of calculating two joint probabilities. Each of those makes reference to some, but not all, of the variables in the network.
  For variables that are missing, we have to consider all combinations of values:
  P(b ∧ m ∧ ¬j) =
  P(b ∧ e ∧ a ∧ m ∧ ¬j) +
  P(b ∧ e ∧ ¬a ∧ m ∧ ¬j) +
  P(b ∧ ¬e ∧ a ∧ m ∧ ¬j) +
  P(b ∧ ¬e ∧ ¬a ∧ m ∧ ¬j)
- If one of the missing variables is an evidence variable or a hidden variable that's not an ancestor of an evidence variable that is included, then it can be ignored.
- We can calculate any joint probability in the network using the conditional and prior probabilities that are available. Using the chain rule, the first joint probability above becomes
  P(b ∧ e ∧ a ∧ m ∧ ¬j) =
  P(¬j|m ∧ a &and e &and b) P(m|a ∧ e ∧ b) P(a|e ∧ b) P(e|b) P(b)
  Because of the independence constraints that are built into the network, this product is equal to
  P(¬j|a) P(m|a) P(a|e ∧ b) P(e) P(b)
  That is, any joint probability for a set of variable values is just the product of the conditional and prior probabilities for those values.
A tutorial on Bayesian belief networks with a worked-out example
An applet illustrating Bayesian belief networks

Decision making

Possible outcomes: their probabilities and utilities
Expected utility of an option =
(prob of outcome 1) X (utility of outcome 1) + (prob of outcome 2) X (utility of outcome 2) ...
Choice among actions made on the basis of maximum expected utility
Decision tree
- At outcome nodes, take weighted average of utilities of outcomes, using probabilities as weights
- At decision nodes, compute expected utility for the outcome of each action and subtract the cost of that action. The expected utility at the decision node is the maximum of the utilities of each action

Human reasoning under uncertainty and decision making

Some tests
1. Which are more frequent, words beginning with the letter 'R' or words in which 'R' is the third letter?
2. What's the result of the calculation: 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8? (Average response: 512) What's the result of the calculation: 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1? (Average response: 2,250) (Right answer: 40,320).
3. You are playing roulette and observe the sequence: Red Red Red Red Red. What color should you bet on next?
4. Bill is 34 years old. He is intelligent, but unimaginative, compulsive, and generally lifeless. In school, he was strong in math but weak in social studies and humanities. How likely would you rate (on a scale of 1 to 7) each of the following propositions?
  1. Bill is an accountant.
  2. Bill is an accountant who plays jazz for a hobby.
  3. Bill is a physician who plays poker for a hobby.
  4. Bill is an architect.
  5. Bill climbs mountains for a hobby.
  6. Bill plays jazz for a hobby.
5. Which of the following lotteries would you prefer?
  A: 80% chance at $4000
  B: 100% chance at $3000
6. Which of the following lotteries would you prefer?
  C: 20% chance at $4000
  D: 25% chance at $3000
Some events are perceived as so unique that past history does not seem relevant to the evaluation of their likelihood. Instead we create scenarios. The plausibility of the scenarios that come to mind, or the difficulty of producing them, serve as clues about the likelihood of the event.
People are risk-averse with high probability events but are willing to take more risks with unlikely payoffs.