Probabilistic Calculus to the Rescue

1. Suppose we know the likelihood of each of the (propositional) worlds (aka Joint Probability distribution) Then we can use standard rules of probability to compute the likelihood of all queries (as I will remind you) So, Joint Probability Distribution is all that you ever need! In the case of Pearl example, we just need the joint probability distribution over B,E,A,J,M (32 numbers) --In general 2n separate numbers (which should add up to 1) If Joint Distribution is sufficient for reasoning, what is domain knowledge supposed to help us with? --Answer: Indirectly by helping us specify the joint probability distribution with fewer than 2n numbers ---The local relations between propositions can be seen as “constraining” the form the joint probability distribution can take! Burglary => Alarm Earth-Quake => Alarm Alarm => John-calls Alarm => Mary-calls Probabilistic Calculus to the Rescue Only 10 (instead of 32) numbers to specify!

2. If in addition, each proposition is equally likely to be true or false, Then the joint probability distribution can be specified without giving any numbers! All worlds are equally probable! If there are n props, each world will be 1/2n probable Probability of any propositional conjunction with m (< n) propositions will be 1/2m If there are no relations between the propositions (i.e., they can take values independently of each other) Then the joint probability distribution can be specified in terms of probabilities of each proposition being true Just n numbers instead of 2n Easy Special Cases Is this a good world to live in?

3. Will we always need 2n numbers? • If every pair of variables is independent of each other, then • P(x1,x2…xn)= P(xn)* P(xn-1)*…P(x1) • Need just n numbers! • But if our world is that simple, it would also be very uninteresting & uncontrollable(nothing is correlated with anything else!) • We need 2n numbers if every subset of our n-variables are correlated together • P(x1,x2…xn)= P(xn|x1…xn-1)* P(xn-1|x1…xn-2)*…P(x1) • But that is too pessimistic an assumption on the world • If our world is so interconnected we would’ve been dead long back…  A more realistic middle ground is that interactions between variables are contained to regions. --e.g. the “school variables” and the “home variables” interact only loosely (are independent for most practical purposes) -- Will wind up needing O(2k) numbers (k << n)

4. Takes O(2n) for most natural queries of type P(D|Evidence) NEEDS O(2n) probabilities as input Probabilities are of type P(wk)—where wk is a world Directly using Joint Distribution Can take much less than O(2n) time for most natural queries of type P(D|Evidence) STILL NEEDS O(2n) probabilities as input Probabilities are of type P(X1..Xn|Y) Directly using Bayes rule Can take much less than O(2n) time for most natural queries of type P(D|Evidence) Can get by with anywhere between O(n) and O(2n) probabilities depending on the conditional independences that hold. Probabilities are of type P(X1..Xn|Y) Using Bayes rule With bayes nets

5. Prob. Prop logic: The Game plan • We will review elementary “discrete variable” probability • We will recall that joint probability distribution is all we need to answer any probabilistic query over a set of discrete variables. • We will recognize that the hardest part here is not the cost of inference (which is really only O(2n) –no worse than the (deterministic) prop logic • Actually it is Co-#P-complete (instead of Co-NP-Complete) (and the former is believed to be harder than the latter) • The real problem is assessing probabilities. • You could need as many as 2n numbers (if all variables are dependent on all other variables); or just n numbers if each variable is independent of all other variables. Generally, you are likely to need somewhere between these two extremes. • The challenge is to • Recognize the “conditional independences” between the variables, and exploit them to get by with as few input probabilities as possible and • Use the assessed probabilities to compute the probabilities of the user queries efficiently.

6. Propositional Probabilistic Logic

8. If B=>A then P(A|B) = ? P(B|~A) = ? P(B|A) = ?

9. CONDITIONAL PROBABLITIES Non-monotonicity w.r.t. evidence– P(A|B) can be either higher, lower or equal to P(A)

10. P(A|B=T;C=False) P(A|B=T) P(A) Most useful probabilistic reasoning involves computing posterior distributions Probability Variable values Important: Computing posterior distribution is inference; not learning

11. If you know the full joint, You can answer ANY query

12. & Marginalization

13. P(CA & TA) = P(CA) = P(TA) = P(CA V TA) = P(CA|~TA) =

14. P(CA & TA) = 0.04 P(CA) = 0.04+0.06 = 0.1 (marginalizing over TA) P(TA) = 0.04+0.01= 0.05 P(CA V TA) = P(CA) + P(TA) – P(CA&TA) = 0.1+0.05-0.04 = 0.11 P(CA|~TA) = P(CA&~TA)/P(~TA) = 0.06/(0.06+.89) = .06/.95=.0631 Think of this as analogous to entailment by truth-table enumeration!

16. Problem: --Need too many numbers… --The needed numbers are harder to assess You can avoid assessing P(E=e) if you assess P(~Y|E=e) since it must add up to 1

17. Happy Spring Break!

18. 3/5 Homework 2 due today Mid-term on Tuesday after Springbreak Will cover everything upto (but not including) probabilistic reasoning

19. Mid-term Syllabus

20. Reviewing Last Class • What is the difficult part of reasoning with uncertainty? • Complexity of reasoning? • Assessing Probabilities? • If Joint distribution is enough for answering all queries, what exactly is the role of domain knowledge? • What, if any, is the difference between probability and statistics? • Statistics  Model-finding (learning) • Probability Model-using (inference) • Given your midterm marks, I am interested in finding the model of the generative process that generated those marks. I will go ahead and use the assumption that each of your performance is independent of others in the class (clearly bogus since I taught you all), and another huge bias that the individual distributions are all Gaussian. Then I just need to find the mean and standard deviation for the data to give the full distribution. • Given the distribution, now I can compute random queries the probability of more than 10 people getting more than 50 marks on the test!

21. Joint distribution requires us to assess probabilities of type P(x1,~x2,x3,….~xn) This means we have to look at all entities in the world and see which fraction of them have x1,~x2,x3….~xm true Difficult experiment to setup.. Conditional probabilities of type P(A|B) are relatively easier to assess You just need to look at the set of entities having B true, and look at the fraction of them that also have A true Eventually, they too can get baroque P(x1,~x2,…xm|y1..yn) Doc, Doc, I have flu. Can you tell if I have a runny nose? Relative ease/utility of Assessing various types of probabilities • Among the conditional probabilities, causal probabilities of the form P(effect|cause) are better to assess than diagnostic probabilities of the form P(cause|effect) • Causal probabilities tend to me more stable compared to diagnostic probabilities • (for example, a text book in dentistry can publish P(TA|Cavity) and hope that it will hold in a variety of places. In contrast, P(Cavity|TA) may depend on other fortuitous factors—e.g. in areas where people tend to eat a lot of icecream, many tooth aches may be prevalent, and few of them may be actually due to cavities.

22. Generalized bayes rule P(A|B,e) = P(B|A,e) P(A|e) P(B|e) Get by with easier to assess numbers Think of this as analogous to inference rules (like modus-ponens) A be Anthrax; Rn be Runny Nose P(A|Rn) = P(Rn|A) P(A)/ P(Rn)

23. Can we avoid assessing P(S)? P(M|S) = P(S|M) P(M)/P(S) P(~M|S) = P(S|~M) P(~M)/P(S) ---------------------------------------------------------------- 1 = 1/P(S) [ P(S|M) P(M) + P(S|~M) P(~M) ] So, if we assess P(S|~M), then we don’t need to assess P(S) “Normalization”

24. Is P(S|~M) any easier to assess than P(~S)? • P(S|M) is clearly easy to assess (just look at the fraction of meningitis patients that have stiff neck • P(S) seems hard to assess—you need to ask random people whether they have stiff neck or not • P(S|~M) seems just as hard to assess… • And in general there seems to be no good argument that it is always easier to assess than P(S) • In fact they are related in a quite straightforward way • P(S) =P(S|M)*P(M) + P(S|~M)*P(~M) • (To see this, note that P(S)= P(S&M)+P(S&~M) and then use product rule) • The real reason we assess P(S|~M) is that often we need the posterior distribution rather than just the single probability • For boolean variables, you can get the distribution given one value • But for multi-valued variables, we need to assess P(D=di|S) for all values di of the variable D. To do this, we need P(S|D=di) type probabilities anyway…

25. Need to know this! If n evidence variables, We will need 2n probabilities! What happens if there are multiple symptoms…? Conditional independence To the rescue Suppose P(TA,Catch|cavity) = P(TA|Cavity)*P(Catch|Cavity) Patient walked in and complained of toothache You assess P(Cavity|Toothache) Now you try to probe the patients mouth with that steel thingie, and it catches… How do we update our belief in Cavity? P(Cavity|TA, Catch) = P(TA,Catch| Cavity) * P(Cavity) P(TA,Catch) = a P(TA,Catch|Cavity) * P(Cavity)

26. Written as A||B

27. Conditional Independence Assertions • We write X || Y | Z to say that the set of variables X is conditionally independent of the set of variables Y given evidence on the set of variables Z (where X,Y,Z are subsets of the set of all random variables in the domain model) • We saw that Bayes Rule computations can exploit conditional independence assertions. Specifically, • X || Y| Z implies • P(X & Y|Z) = P(X|Z) * P(Y|Z) • P(X|Y, Z) = P(X|Z) • P(Y|X,Z) = P(Y|Z) • If A||B|C then P(A,B,C)=P(A|B,C)P(B,C) =P(A|B,C)P(B|C)P(C) =P(A|C)P(B|C)P(C) (Can get by with 1+2+2=5 numbers instead of 8) Why not write down all conditional independence assertions that hold in a domain?

28. Cond. Indep. Assertions (Contd) • Idea: Why not write down all conditional independence assertions (CIA) (X || Y | Z) that hold in a domain? • Problem: There can be exponentially many conditional independence assertions that hold in a domain (recall that X, Y and Z are all subsets of the domain variables). • Many of them might well be redundant • If X||Y|Z, then X||Y|Z+U for all U • Brilliant Idea: May be we should implicitly specify the CIA by writing down the “local dependencies” between variables using a graphical model • A Bayes Network is a way of doing just this. • The Bayes Net is a Directed Acyclic Graph whose nodes are random variables, and the immediate dependencies between variables are represented by directed arcs • The topology of a bayes network shows the inter-variable dependencies. Given the topology, there is a way of checking if any Cond. Indep. Assertion. holds in the network (the Bayes Ball algorithm and the D-Sep idea)

30. CIA implicit in Bayes Nets • So, what conditional independence assumptions are implicit in Bayes nets? • Local Markov Assumption: • A node N is independent of its non-descendants (including ancestors) given its immediate parents. (So if P are the immediate paretnts of N, and A is a subset of of Ancestors and other non-descendants, then {N} || A| P ) • (Equivalently) A node N is independent of all other nodes given its markov blanket (parents, children, children’s parents) • Given this assumption, many other conditional independencies follow. For a full answer, we need to appeal to D-Sep condition and/or Bayes Ball reachability

31. Topological Semantics Independence from Every node holds Given markov blanket Independence from Non-descedants holds Given just the parents Markov Blanket Parents; Children; Children’s other parents These two conditions are equivalent Many other conditional indepdendence assertions follow from these