1. November 2004CSA4050: Crash Concepts in Probability1 CSA4050: Advanced Topics in NLP Probability I Experiments/Outcomes/Events Independence/Dependence Bayes’ Rule Conditional Probability/Chain Rule

2. November 2004 CSA4050: Crash Concepts in Probability 2 Acknowledgement Much of this material is based on material by Mary Dalrymple, Kings College, London

3. November 2004 CSA4050: Crash Concepts in Probability 3 Experiment, Basic Outcome, Sample Space Probability theory is founded upon the notion of an experiment. An experiment is a situation which can have one or more different basic outcomes. Example: if we throw a die, there are six possible basic outcomes. A Sample Space Ω is a set of all possible basic outcomes. For example,  If we toss a coin, Ω = {H,T}  If we toss a coin twice, Ω = {HT,TH,TT,HH}  if we throw a die, Ω = {1,2,3,4,5,6}

4. November 2004 CSA4050: Crash Concepts in Probability 4 Event An Event A  Ω is a set of basic outcomes e.g.  tossing two heads {HH}  throwing a 6, {6}  getting either a 2 or a 4, {2,4}. Ω itself is the certain event, whilst { } is the impossible event. Event Space ≠ Sample Space

5. November 2004 CSA4050: Crash Concepts in Probability 5 Probability distribution A probability distribution of an experiment is a function that assigns a number (or probability) between 0 and 1 to each basic outcome such that the sum of all the probabilities = 1. The probability p(E) of an event E is the sum of the probabilities of all the basic outcomes in E. Uniform distribution is when each basic outcome is equally likely.

6. November 2004 CSA4050: Crash Concepts in Probability 6 Probability of an Event: die example Sample space = set of basic outcomes = {1,2,3,4,5,6} If the die is not loaded, distribution is uniform. Thus for each basic outcome, e.g. {6} (throwing a six) is assigned the same probability = 1/6. So p({3,6}) = p({3}) + p({6}) = 2/6 = 1/3

7. November 2004 CSA4050: Crash Concepts in Probability 7 Estimating Probability Repeat experiment T times and count frequency of E. Estimated p(E) = count(E)/count(T) This can be done over m runs, yielding estimates p 1 (E),...p m (E). Best estimate is (possibly weighted) average of individual p i (E)

8. November 2004 CSA4050: Crash Concepts in Probability 8 3 times coin toss Ω = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} Cases with exactly 2 tails = {HTT, THT,TTH} Experiment i = 1000 cases (3000 tosses).  c 1 (E)= 386, p 1 (E) =.386  c 2 (E)= 375, p 2 (E) =.375  p mean (E)= (.386+.375)/2 =.381 Uniform distribution is when each basic outcome is equally likely. Assuming uniform distribution, p(E) = 3/8 =.375

9. November 2004 CSA4050: Crash Concepts in Probability 9 Word Probability General Problem: What is the probability of the next word/character/phoneme in a sequence, given the first N words/characters/phonemes. To approach this problem we study an experiment whose sample space is the set of possible words. N.B. The same approach could be used to study the the probability of the next character or phoneme.

10. November 2004 CSA4050: Crash Concepts in Probability 10 Word Probability Approximation 1: all words are equally probable Then probability of each word = 1/N where N is the number of word types. But all words are not equally probable Approximation 2: probability of each word is the same as its frequency of occurrence in a corpus.

11. November 2004 CSA4050: Crash Concepts in Probability 11 Word Probability Estimate p(w) - the probability of word w: Given corpus C p(w)  count(w)/size(C) Example  Brown corpus: 1,000,000 tokens  the: 69,971 tokens  Probability of the: 69,971/1,000,000 .07  rabbit: 11 tokens  Probability of rabbit: 11/1,000,000 .00001  conclusion: next word is most likely to be the Is this correct?

12. November 2004 CSA4050: Crash Concepts in Probability 12 A counter example Given the context: Look at the cute... is the more likely than rabbit? Context matters in determining what word comes next. What is the probability of the next word in a sequence, given the first N words?

13. November 2004 CSA4050: Crash Concepts in Probability 13 Independent Events A: eggs B: monday sample space

14. November 2004 CSA4050: Crash Concepts in Probability 14 Sample Space (eggs,mon)(cereal,mon) (nothing,mon) (eggs,tue)(cereal,tue) (nothing,tue) (eggs,wed)(cereal,wed) (nothing,wed) (eggs,thu)(cereal,thu) (nothing,thu) (eggs,fri)(cereal,fri) (nothing,fri) (eggs,sat)(cereal,sat) (nothing,sat) (eggs,sun)(cereal,sun) (nothing,sun)

15. November 2004 CSA4050: Crash Concepts in Probability 15 Independent Events Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. When two events, A and B, are independent, the probability of both occurring p(A,B) is the product of the prior probabilities of each, i.e. p(A,B) = p(A) · p(B)

16. November 2004 CSA4050: Crash Concepts in Probability 16 Dependent Events Two events, A and B, are dependent if the occurrence of one affects the probability of the occurrence of the other.

17. November 2004 CSA4050: Crash Concepts in Probability 17 Dependent Events A B sample space A  B

18. November 2004 CSA4050: Crash Concepts in Probability 18 Conditional Probability The conditional probability of an event A given that event B has already occurred is written p(A|B) In general p(A|B)  p(B|A)

19. November 2004 CSA4050: Crash Concepts in Probability 19 Dependent Events: p(A|B)≠ p(B|A) A B sample space A  B

20. November 2004 CSA4050: Crash Concepts in Probability 20 Example Dependencies Consider fair die example with  A = outcome divisible by 2  B = outcome divisible by 3  C = outcome divisible by 4 p(A|B) = p(A  B)/p(B) = (1/6)/(1/3) = ½ p(A|C) = p(A  C)/p(C) = (1/6)/(1/6) = 1

21. November 2004 CSA4050: Crash Concepts in Probability 21 Conditional Probability Intuitively, after B has occurred, event A is replaced by A  B, the sample space Ω is replaced by B, and probabilities are renormalised accordingly The conditional probability of an event A given that B has occurred (p(B)>0) is thus given by p(A|B) = p(A  B)/p(B). If A and B are independent, p(A  B) = p(A) · p(B) so p(A|B) = p(A) · p(B) /p(B) = p(A).

22. November 2004 CSA4050: Crash Concepts in Probability 22 Bayesian Inversion For A and B to occur, either B must occur first, then B, or vice versa. We get the following possibilites: p(A|B) = p(A  B)/p(B) p(B|A) = p(A  B)/p(A) Hence p(A|B) p(B) = p(B|A) p(A) We can thus express p(A|B) in terms of p(B|A) p(A|B) = p(B|A) p(A)/p(B) This equivalence, known as Bayes’ Theorem, is useful when one or other quantity is difficult to determine

23. November 2004 CSA4050: Crash Concepts in Probability 23 Bayes’ Theorem p(B|A) = p(B  A)/p(A) = p(A|B) p(B)/p(A) The denominator p(A) can be ignored if we are only interested in which event out of some set is most likely. Typically we are interested in the value of B that maximises an observation A, i.e. arg max B p(A|B) p(B)/p(A) = arg max B p(A|B) p(B)

24. November 2004 CSA4050: Crash Concepts in Probability 24 The Chain Rule We can use the definition of conditional probability to more than two events p(A1 ...  An) = p(A1) * p(A2|A1) * p(A3|A1  A2)..., p(An|A1 ...  An-1) The chain rule allows us to talk about the probability of sequences of events p(A1,...,An).