CS 188: Artificial Intelligence Spring 2007 Lecture 11: Probability 2/20/2007 Srini Narayanan – ICSI and UC Berkeley

Announcements  HW1 graded  Solutions to HW 2 posted Wednesday  HW 3 due Thursday 11:59 PM

Today  Probability  Random Variables  Joint and Conditional Distributions  Bayes Rule  Independence  You’ll need all this stuff for the next few weeks, so make sure you go over it!

What is this? Uncertainty

 Let action A t = leave for airport t minutes before flight  Will A t get me there on time?  Problems:  partial observability (road state, other drivers' plans, etc.)  noisy sensors (KCBS traffic reports)  uncertainty in action outcomes (flat tire, etc.)  immense complexity of modeling and predicting traffic  A purely logical approach either  Risks falsehood: “A 25 will get me there on time” or  Leads to conclusions that are too weak for decision making:  “A 25 will get me there on time if there's no accident on the bridge, and it doesn't rain, and my tires remain intact, etc., etc.''  A 1440 might reasonably be said to get me there on time but I'd have to stay overnight in the airport…

Probabilities  Probabilistic approach  Given the available evidence, A 25 will get me there on time with probability 0.04  P(A 25 | no reported accidents) = 0.04  Probabilities change with new evidence:  P(A 25 | no reported accidents, 5 a.m.) = 0.15  P(A 25 | no reported accidents, 5 a.m., raining) = 0.08  i.e., observing evidence causes beliefs to be updated

Probabilities Everywhere?  Not just for games of chance!  I’m snuffling: am I sick?  contains “FREE!”: is it spam?  Tooth hurts: have cavity?  Safe to cross street?  60 min enough to get to the airport?  Robot rotated wheel three times, how far did it advance?  Why can a random variable have uncertainty?  Inherently random process (dice, etc)  Insufficient or weak evidence  Unmodeled variables  Ignorance of underlying processes  The world’s just noisy!

Probabilistic Models  CSP/Prop Logic:  Variables with domains  Constraints: map from assignments to true/false  Ideally: only certain variables directly interact  Probabilistic models:  (Random) variables with domains  Joint distributions: map from assignments (or outcomes) to positive numbers  Normalized: sum to 1.0  Ideally: only certain variables are directly correlated ABP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3 ABP warmsunT warmrainF coldsunF coldrainT

Random Variables  A random variable is some aspect of the world about which we 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 in a CSP, each random variable has a domain  R in {true, false}  D in [0,  ]  L in possible locations

Distributions on Random Vars  A joint distribution over a set of random variables: is a map from assignments (or outcomes, or atomic events) to reals:  Size of distribution if n variables with domain sizes d?  Must obey:  For all but the smallest distributions, impractical to write out TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3

Examples  An event is a set E of assignments (or outcomes)  From a joint distribution, we can calculate the probability of any event  Probability that it’s warm AND sunny?  Probability that it’s warm?  Probability that it’s warm OR sunny? TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3

Marginalization  Marginalization (or summing out) is projecting a joint distribution to a sub-distribution over subset of variables TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3 TP warm0.5 cold0.5 SP sun0.6 rain0.4

Conditional Probabilities  A conditional probability is the probability of an event given another event (usually evidence) TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3

Conditional Probabilities  Conditional or posterior probabilities:  E.g., P(cavity | toothache) = 0.8  Given that toothache is all I know…  Notation for conditional distributions:  P(cavity | toothache) = a single number  P(Cavity, Toothache) = 2x2 table summing to 1  P(Cavity | Toothache) = Two 2-element vectors, each summing to 1  If we know more:  P(cavity | toothache, catch) = 0.9  P(cavity | toothache, cavity) = 1  Note: the less specific belief remains valid after more evidence arrives, but is not always useful  New evidence may be irrelevant, allowing simplification:  P(cavity | toothache, traffic) = P(cavity | toothache) = 0.8  This kind of inference, guided by domain knowledge, is crucial

Conditioning  Conditional probabilities are the ratio of two probabilities: TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3

Normalization Trick  A trick to get the whole conditional distribution at once:  Get the joint probabilities for each value of the query variable  Renormalize the resulting vector TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3 TP warm0.1 cold0.3 TP warm0.25 cold0.75 Select Normalize

The Product Rule  Sometimes joint P(X,Y) is easy to get  Sometimes easier to get conditional P(X|Y)  Example: P(sun, dry)? RP sun0.8 rain0.2 DSP wetsun0.1 drysun0.9 wetrain0.7 dryrain0.3 DSP wetsun0.08 drysun0.72 wetrain0.14 dryrain0.06

Lewis Carroll's Sack Problem  Sack contains a red or blue token, 50/50  We add a red token  If we draw a red token, what’s the chance of drawing a second red token?  Variables:  F={r,b} is the original token  D={r,b} is the first token we draw  Query: P(F=r|D=r) FP r0.5 b FDP rr rb br bb FDP rr1.0 rb0.0 br0.5 bb

Lewis Carroll's Sack Problem  Now we have P(F,D)  Want P(F=r|D=r) FDP rr0.5 rb0.0 br0.25 bb

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

More Bayes’ Rule  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?

Inference by Enumeration  P(sun)?  P(sun | winter)?  P(sun | winter, warm)? STRP summerwarmsun0.30 summerwarmrain0.05 summercoldsun0.10 summercoldrain0.05 winterwarmsun0.10 winterwarmrain0.05 wintercoldsun0.15 wintercoldrain0.20

Inference by Enumeration  General case:  Evidence variables:  Query variables:  Hidden variables:  We want:  First, select the entries consistent with the evidence  Second, sum out H:  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

Independence  Two variables are independent if:  This says that their joint distribution factors into a product two simpler distributions  Independence is a modeling assumption  Empirical joint distributions: at best “close” to independent  What could we assume for {Weather, Traffic, Cavity}?  How many parameters in the joint model?  How many parameters in the independent model?  Independence is like something from CSPs: what?

Example: Independence  N fair, independent coin flips: H0.5 T H T H T

Example: Independence?  Arbitrary joint distributions can be poorly modeled by independent factors TSP warmsun0.4 warmrain0.1 coldsun0.2 coldrain0.3 TSP warmsun0.3 warmrain0.2 coldsun0.3 coldrain0.2 TP warm0.5 cold0.5 SP sun0.6 rain0.4

Conditional Independence  P(Toothache,Cavity,Catch) has 2 3 = 8 entries (7 independent entries)  If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:  P(catch | toothache, cavity) = P(catch | cavity)  The same independence holds if I don’t have a cavity:  P(catch | toothache,  cavity) = P(catch|  cavity)  Catch is conditionally independent of Toothache given Cavity:  P(Catch | Toothache, Cavity) = P(Catch | Cavity)  Equivalent statements:  P(Toothache | Catch, Cavity) = P(Toothache | Cavity)  P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)

Conditional Independence  Unconditional (absolute) independence is very rare (why?)  Conditional independence is our most basic and robust form of knowledge about uncertain environments:  What about this domain:  Traffic  Umbrella  Raining  What about fire, smoke, alarm?

The Chain Rule II  Can always factor any joint distribution as an incremental product of conditional distributions  Why?  This actually claims nothing…  What are the sizes of the tables we supply?

The Chain Rule III  Trivial decomposition:  With conditional independence:  Conditional independence is our most basic and robust form of knowledge about uncertain environments  Graphical models (next class) will help us work with independence

The Chain Rule IV  Write out full joint distribution using chain rule:  P(Toothache, Catch, Cavity) = P(Toothache | Catch, Cavity) P(Catch, Cavity) = P(Toothache | Catch, Cavity) P(Catch | Cavity) P(Cavity) = P(Toothache | Cavity) P(Catch | Cavity) P(Cavity) Cav TCat P(Cavity) P(Toothache | Cavity)P(Catch | Cavity) Graphical model notation: Each variable is a node The parents of a node are the other variables which the decomposed joint conditions on MUCH more on this to come!

Combining Evidence P(cavity | toothache, catch) =  P(toothache, catch | cavity) P(cavity) =  P(toothache | cavity) P(catch | cavity) P(cavity)  This is an example of a naive Bayes model: C E1E1 EnEn E2E2