lec1 - Course Work: Homeworks 0 There will be 5 homeworks o...

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Unformatted text preview: Course Work: Homeworks 0 There will be 5 homeworks o Homework 0 will not count towards your grade a You should be able to solve all problems in HWO o If you have trouble in HWO, you will have trouble later CSci 5512: Artificial Intelligence || ° H°mew°rk subml55i°n$ :- Using the submit system. only pdf (encouraged) in Paper copy at the beginning of the class Instructor: Arindam Banerjee o All programming in Matlab, Java, C, Python 0 Follow the instructions for programming assignments o Other languages will not receive any credit January 18, 2012 0 Ok to discuss with others, you have to write on your own, put the name(s) of people you discussed with Instructor: Arindam Banerjee Instructor: Arindam Banerjee — General Information Course Work: Homeworks (Contd.) 0 Course Number: CSci 5512 0 Class: Mon Wed 12:45-02:00 pm 0 Dates: . Location- Keller Ha” 3_111 o HW 0: Posted Jan 18 (Wed), due Jan 23 (Mon) at 12:45 pm ' o HW 1: Jan 30 (Mon), due Feb 13 (Mon) at 12:45 pm 0 HW 2: Feb 20 (Mon), due Mar 05 (Mon) at 12:45 pm _ HW 3: Mar 26 (Mon), due Apr 09 (Mon) at 12:45 pm 0 TA: James Parker, Jparker@cs.umn.edu o HW 4: Apr 16 (Mon), due May 30 (Mon) at 12:45 pm 0 Instructor: Arindam Banerjee, banerjee©cs.umn.edu 0 Late submission policy: Late by 0—24 hrs: 25% deducted from actual score Late by 24-48 hrs: 50% deducted from actual score Late by more than 48 hrs: Will receive a zero All late submissions must be submitted in pdf using Submit 0 Office Hours: a Arindam: EE/CS 6—213 Mon Wed 02:00—03:00 pm a James: EE/CS 2-216 Thu 10:00-11:00 am, Fri 01:00-02:00 pm 0 Web page: http://www—users.itlabs.umn.edu/classes/Spring— 2012/csci5512 Instructor: Arindam Banerjee Instructor: Arindam Banerjee Course Work: Exams, etc. Topics 0 Exams . . . o Mid—Term: Mar 26 (Mon) in class ° Quant'fy'"g uncerta'my :- Final: May ?? (??), ?:00-?:00 o Probabilistic Reasoning ° CI°Sed b°°k' Closed "_°tes exam _ o Probabilistic Reasoning over Time a Allowed 1 sheet for mid-term, 2 sheets for final 0 Making Simple Decisions 0 Participation: 0 Making Complex Decisions 0 Ask questions 0 Learning from Examples ' Participate in discussim‘s o Learning Probabilistic Models 0 Web Links: 0 Reinforcement Learning o Online Submission for homeworks 0 Natural Language Processing 0 Bulletin Board for discussions Instructor: Arindam Banerjee Instructor: Arindam Banerjee — o Uncertainty inherent in decision problems 0 Partial knowledge of environment in Environment may be complex or stochastic o Existence of other agents o Homework: 50 % = 4 x 12.5 % o Mid-Term: 20 % 0 Final: 25 % o First-order logic is inappropriate for such domains 0 Several different events are possible 0 Participation: 5 % 0 Each event 0 Has a different "probability" of happening 0 Has different "utility" or “payoffs” o In order to pass the course: :- Average on exams (midterm and final) must be at least 50%. Rational decisions maximize expected utility ° Overall Sc°re must be at least 50% Decision Theory E Utility Theory + Probability Theory Instructor: Arindam Banerjee Instructor: Arindam Banerjee ° Game of Monopoly 0 Random variables are mappings of events (to real numbers) . i- o MappingX:Qi—>R : i' V? 0 Any event w maps to X(w) 0 Example: o Tossing a coin has two possible outcomes o Denoted by {H, T} or {0,1} ° PUVSUIt With constraints a Fair coin has uniform probabilities :- Chasing in Manhattan - 1 1 o Robotic teams for search/rescue p(X = 0) = _ P(X = 1) = _ 2 2 o The Stock Market 0 Random variables (r.v.s) can be 0 Discrete, e.g., Bernoulli 0 Continuous. e.g., Gaussian Instructor: Arindam Banerjee Instructor: Arindam Banerjee — — o For a continuous r.v. a Distribution function F(x) = P(X S x) 9 Corresponding density function f(x)dx = dF(x) 0 Sample space 9 of events 0 Each "event" to E (2 has an associated "measure" a Probability of the event P(w) a Note that x o Axioms of Probability: F(X) =/ f(t)dt 0 Va), P(w) 6 [0,1] t=—oo o P(Q) = 1 _ , p(w1 Uwz) : pom) + p(w2) _ P(wl mm) o For a discrete r.v. o Probability mass function f(x) = P(X = x) = P(X) I . . a We will call this the probability of a discrete event ° N°te- We are bang '"f0’ma' . Distribution function P(X) = P(X g x) Instructor: Arindam Banerjee Instructor: Arindam Banerjee Joint Distributions, Marginals Independence o For two continuous r.v.s X1,X2 0 Joint distribution F(X1,X2) = P(X1 S X1,X2 S x2) o Joint density function f(x1,X2) can be defined as before ° JOInt Pr0bablllty P(Xl = X17X2 = X2) 9 The marginal probability density o X1,X2 are different dice 0 X1 denotes if grass is wet, X2 denotes if sprinkler was on 00 f(X1) = / f(X1,X2)dX2 X2=—°° 0 Two r.v.s are independent if c For two discrete r.v.s X1,X2 0 Joint probability f(X1,X2) = = X1,X2 2 X2) = P(X1,X2) o The marginal probability P X = X = P X = X ,X = X a Two different dice are independent ( 1 1) x22 ( 1 1 2 2) o If sprinkler was on, then grass will be wet => dependent P(X1 = x1,X2 = x2) = P(Xl = x1)P(X2 2 X2) 0 Can be extended to joint distribution over several r.v.s 0 Many hard problems involve computing marginals Instructor: Arindam Banerjee Instructor: Arindam Banerjee — Grass Wet Grass Dry Sprinkler On 0.4 Sprinkler Off 0.2 o The expected value of a r.v. X o For continuous r.v.s E[X] = fx xp(x)dx o Inference problems: ° For discrete r'V' E[X] = Z:iX’pi o Given 'grass wet' what is P('sprinkler on'|‘grass wet') o Expectation is a linear operator o Given ‘symptom' what is P(‘disease'|‘symptom') o For any r.v.s X, Y, the conditional probability P(xly) = P) 0 Since P(x, y) = P(y|x)P(x), we have P(ylx) = —P (1?: (y) o Expressing ‘posterior' in terms of 'conditional': Bayes Rule Instructor: Arindam Banerjee Instructor: Arindam Banerjee E[aX—l— bY + c] = aE[X] + bE[Y] + c Product Rule & Independence Conditional Independence 0 Product Rule: 0 X and Y are conditionally independent given Z ° For X1,X2.P(X1iX2)= P(X1)P(X2IX1) . For X1,X2,X3, P(X1,X2,X3) = P(X1)P(X2|X1)P(X3|X1,X2) P(Xi VIZ) = P(XIZ)P(Y|Z) o In general, the chain rule n P(X1,--- ,X,,) = HP(X,-|X1,...,X,-_1) ° Example: [:1 P( Toothache, Catch| Cavity) = P( Toothache| Cavity)P(Catch| Cavity) 0 Joint distribution of n Boolean variables 0 Specification requires 2" — 1 parameters 0 Recall Independence: _, . .. . . . . . O For X1,X2' P(thz) = P(X1)P(X2) o Conditional Independence Simplifies Jomt distributions a In general " o Often reduces from exponential to linear in n PX,---,X,, = PX,- “ I I I P(X,Y,Z)=P(Z)P(X|Z)P(Y|Z) 0 Independence reduces specification to n parameters Instructor: Arindam Banerjee Instructor: Arindam Banerjee — — Independence Naive Bayes Model 0 If X1, . . . ,X,, are independent given Y A / I Cavity ’1 CaVIty decomposesinm {cothache Catct> P(Y, X1, _ _ _ a X") = H y) [:1 Toothache Catch ‘ \V/ Weather 0 Exa m p le: P(Cavity, Toothache, Catch) 0 Consider 4 variables: Toothache,Catch,Cavity,Weather = P(Cavity)P( Toothache|Cavity)P(Catch|Cavity) 0 Independence implies o More generally I1 P(TOOthaCheacatCha caViWa weather) P(Cause, Effectl, . . . , Effect”) = P(Cause) H P(EffectilCause) = P(Toothache, Catch, Cavity)P(Weather) i=1 0 In terms of the joint distribution: ® ® in 32 parameters reduced to 12 o For boolean variables 2" — 1 reduces to n ’ I ‘ 0 Absolute independence powerful but rare ' ' ° Instructor: Arindam Banerjee Instructor: Arindam Banerjee ...
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lec1 - Course Work: Homeworks 0 There will be 5 homeworks o...

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