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CS 511a: Artificial Intelligence Spring 2013 Lecture 13: Probability/ State Space Uncertainty 03/18/2013 Robert Pless Class directly adapted from Kilian.

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Presentation on theme: "CS 511a: Artificial Intelligence Spring 2013 Lecture 13: Probability/ State Space Uncertainty 03/18/2013 Robert Pless Class directly adapted from Kilian."— Presentation transcript:

1 CS 511a: Artificial Intelligence Spring 2013 Lecture 13: Probability/ State Space Uncertainty 03/18/2013 Robert Pless Class directly adapted from Kilian Weinberger, with multiple slides adapted from Dan Klein and Deva Ramanan

2 Announcements  Upcoming  Project 3 out now(-ish)  Due Monday, April 1, 11:59pm + 1 minute  Midterm graded, to be returned on Wednesday. 2

3 Course Status:  An MDP:  set of states s  S  set of actions a  A  transition function T(s,a,s’)  Prob that a from s leads to s’  reward function R(s, a, s’)  Sometimes just R(s) or R(s’)  start state (or distribution)  one or many terminal states  Solution is a policy dictating action to take in each state.  Reinforcement learning: MDPs where we don’t know the transition or reward functions  A search problem:  set of states s  S  set of actions a  A  start state  goal test  cost function  Solution is a sequence of actions (a plan) which transforms the start state to a goal state

4 Uncertainty  General situation:  Evidence: Agent knows certain things about the state of the world (e.g., sensor readings or symptoms)  Hidden variables: Agent needs to reason about other aspects (e.g. where an object is or what disease is present, or how sensor is bad.)  Model: Agent knows something about how the known variables relate to the unknown variables  Probabilistic reasoning gives us a framework for managing our beliefs and knowledge 4

5 Inference in Ghostbusters  A ghost is in the grid somewhere  Sensor readings tell how close a square is to the ghost  On the ghost: red  1 or 2 away: orange  3 or 4 away: yellow  5+ away: green P(red | 3)P(orange | 3)P(yellow | 3)P(green | 3) 0.050.150.50.3  Sensors are noisy, but we know P(Color | Distance) [Demo]

6 Today  Goal:  Modeling and using distributions over LARGE numbers of random variables  Probability  Random Variables  Joint and Marginal Distributions  Conditional Distribution  Product Rule, Chain Rule, Bayes’ Rule  Inference  Independence  You’ll need all this stuff A LOT for the next few weeks, so make sure you go over it now! 6

7 Random Variables  A random variable is some aspect of the world about which we (may) have uncertainty  R = Is it raining?  D = How long will it take to drive to work?  L = Where am I?  We denote random variables with capital letters  Like variables in a CSP, random variables have domains  R in {true, false} (sometimes write as {+r,  r})  D in [0,  )  L in possible locations, maybe {(0,0), (0,1), …} 7

8 Probabilistic Models 8

9 Probability Distributions  Unobserved random variables have distributions  A distribution is a TABLE of probabilities of values  A probability (lower case value) is a single number  Must have: 9 TP warm0.5 cold0.5 WP sun0.6 rain0.1 fog0.2999999999999 meteor0.0000000000001

10 Joint Distributions  A joint distribution over a set of random variables: specifies a real number for each assignment (or outcome):  Size of distribution if n variables with domain sizes d?  Must obey:  Joint Distribution Table list probabilities of all combinations. For all but the smallest distributions, this is impractical to write out TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 10

11 Probabilistic Models  A probabilistic model is a joint distribution over a set of random variables  Probabilistic models:  (Random) variables with domains Assignments are called outcomes  Joint distributions: say whether assignments (outcomes) are likely  Normalized: sum to 1.0  Ideally: only certain variables directly interact  Constraint satisfaction probs:  Variables with domains  Constraints: state whether assignments are possible  Ideally: only certain variables directly interact TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 TWP hotsunT hotrainF coldsunF coldrainT Distribution over T,W Constraint over T,W

12 Events  An event is a set E of outcomes  From a joint distribution, we can calculate the probability of any event  Probability that it’s hot AND sunny?  Probability that it’s hot?  Probability that it’s hot OR sunny?  Typically, the events we care about are partial assignments, like P(T=hot) TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 12

13 Marginal Distributions  Marginal distributions are sub-tables which eliminate variables  Marginalization (summing out): Combine collapsed rows by adding TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 TP hot0.5 cold0.5 WP sun0.6 rain0.4 13

14 Conditional Probabilities  A simple relation between joint and conditional probabilities  In fact, this is taken as the definition of a conditional probability TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 14

15 Conditional Distributions  Conditional distributions are probability distributions over some variables given fixed values of others TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 WP sun0.8 rain0.2 WP sun0.4 rain0.6 Conditional Distributions Joint Distribution 15

16 Normalization Trick  A trick to get a whole conditional distribution at once:  Select the joint probabilities matching the evidence  Normalize the selection (scale values so table sums to one)  Compute (T | W = rain)  Why does this work? Sum of selection is P(evidence)! (here, P(r)) TWP hotsun0.4 hotrain0.1 coldsun0.2 coldrain0.3 TWP hotrain0.1 coldrain0.3 TP hot0.25 cold0.75 Select Normalize 16

17 The Product Rule  Sometimes have conditional distributions but want the joint  Example with new weather condition, “dryness”. WP sun0.8 rain0.2 DWP wetsun0.1 drysun0.9 wetrain0.7 dryrain0.3 DWP wetsun0.08 drysun0.72 wetrain0.14 dryrain0.06 17

18 Inference 18

19 Probabilistic Inference  Probabilistic inference: compute a desired probability from other known probabilities (e.g. conditional from joint)  We generally compute conditional probabilities  P(on time | no reported accidents) = 0.90  These represent the agent’s beliefs given the evidence  Probabilities change with new evidence:  P(on time | no accidents, 5 a.m.) = 0.95  P(on time | no accidents, 5 a.m., raining) = 0.80  Observing new evidence causes beliefs to be updated 19

20 Inference by Enumeration  P(W=sun)?  P(W=sun | S=winter)?  P(W=sun | S=winter, T=hot)? STWP summerhotsun0.30 summerhotrain0.05 summercoldsun0.10 summercoldrain0.05 winterhotsun0.10 winterhotrain0.05 wintercoldsun0.15 wintercoldrain0.20 20

21 Inference by Enumeration  General case:  Evidence variables:  Query* variable:  Hidden variables:  We want:  First, select the entries consistent with the evidence  Second, sum out H to get joint of Query and evidence:  Finally, normalize the remaining entries to conditionalize  Obvious problems:  Worst-case time complexity O(d n )  Space complexity O(d n ) to store the joint distribution All variables * Works fine with multiple query variables, too

22 Monty Hall Problem 22  Car behind one of three doors  Contestant picks door #1  Host (Monty) shows that behind door #2 is no car  Should contest switch to door #3 or stay at #1?  Random Variables:  C(=car), M(=Monty’s pick)  What is the query, evidence?  What is P(C)?  What are the CPTs P(M|C=c)  How do you solve for P(C|evidence)?

23 Probability Tables 23 Assume (without loss of generality) candidate picks door 1. CP door1 door2 door3 M|C=1P door1 door2 door3 M|C=2P door1 door2 door3 M|C=3P door1 door2 door3

24 Probability Tables 24 MCP door1 door2door1 door3door1 door2 door3door2 door1door3 door2door3 Assume (without loss of generality) candidate picks door 1. CP door11/3 door21/3 door31/3 M|C=1P door10 door20.5 door30.5 M|C=2P door10 door20 door31 M|C=3P door10 door21 door30

25 Probability Tables 25 MCP door1 0 door2door11/6 door3door11/6 door1door20 0 door3door21/3 door1door30 door2door31/3 door3 0 Assume (without loss of generality) candidate picks door 1. CP door11/3 door21/3 door31/3 M|C=1P door10 door20.5 door30.5 M|C=2P door10 door20 door31 M|C=3P door10 door21 door30

26 Probability Tables 26 MCP door1 0 door2door11/6 door3door11/6 door1door20 0 door3door21/3 door1door30 door2door31/3 door3 0 Assume (without loss of generality) candidate picks door 1. CP door11/3 door21/3 door31/3 C|M=door2P door1 door2 door3 M|C=1P door10 door20.5 door30.5 M|C=2P door10 door20 door31 M|C=3P door10 door21 door30

27 Probability Tables 27 M|C=1P door10 door20.5 door30.5 M|C=2P door10 door20 door31 M|C=3P door10 door21 door30 MCP door1 0 door2door11/6 door3door11/6 door1door20 0 door3door21/3 door1door30 door2door31/3 door3 0 Assume (without loss of generality) candidate picks door 1. CP door11/3 door21/3 door31/3 C|M=door2P door11/6 door20 door31/3 (un-normalized!!)

28 Probability Tables 28 M|C=1P door10 door20.5 door30.5 M|C=2P door10 door20 door31 M|C=3P door10 door21 door30 MCP door1 0 door2door11/6 door3door11/6 door1door20 0 door3door21/3 door1door30 door2door31/3 door3 0 Assume (without loss of generality) candidate picks door 1. CP door11/3 door21/3 door31/3 C|M=door2P door11/3 door20 door32/3

29 Humans don’t get it… 29

30 Humans don’t get it… 30

31 Pigeons … 31 SourceSource: Discover

32 Pigeons do! 32 SourceSource: Discover

33 Inference by Enumeration  General case:  Evidence variables:  Query* variable:  Hidden variables:  We want:  First, select the entries consistent with the evidence  Second, sum out H to get joint of Query and evidence:  Finally, normalize the remaining entries to conditionalize  Obvious problems:  Worst-case time complexity O(d n )  Space complexity O(d n ) to store the joint distribution All variables

34 The Chain Rule  More generally, can always write any joint distribution as an incremental product of conditional distributions  Why is this always true? 34 Parse this carefully

35 Bayes’ Rule  Two ways to factor a joint distribution over two variables:  Dividing, we get:  Why is this at all helpful?  Lets us build one conditional from its reverse  Often one conditional is tricky but the other one is simple  Foundation of many systems we’ll see later (e.g. Automated Speech Recognition, Machine Translation)  In the running for most important AI equation! That’s my rule! 35

36 Inference with Bayes’ Rule  Example: Diagnostic probability from causal probability:  Example:  m is meningitis, s is stiff neck  Note: posterior probability of meningitis still very small  Note: you should still get stiff necks checked out! Why? Example givens 36

37 Probability Tables 37 M|C=1P door10 door20.5 door30.5 M|C=2P door10 door20 door31 M|C=3P door10 door21 door30 Assume (without loss of generality) candidate picks door 1. CP door11/3 door21/3 door31/3 C|M=door2P door11/3 door20 door32/3

38 Ghostbusters, Revisited  Let’s say we have two distributions:  Prior distribution over ghost location: P(G)  Let’s say this is uniform  Sensor reading model: P(R | G)  Given: we know what our sensors do  R = reading color measured at (1,1)  E.g. P(R = yellow | G=(1,1)) = 0.1  We can calculate the posterior distribution P(G|r) over ghost locations given a reading using Bayes’ rule: 38

39 Summary  Probabilistic model:  All we need is joint probability table (JPT)  Can perform inference by enumeration  From JPT we can obtain CPT and marginals  (Can obtain JPT from CPT and marginals)  Bayes Rule can perform inference directly from CPT and marginals  Next several lectures:  How to avoid storing the entire JPT 39

40 Why might these tables get so big?  Markov property: p(pixel | rest of image) = p(pixel | neighborhood) ?

41 Example application domain: image repairp Synthesizing a pixel


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