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PROBABILITY AND BAYES THEOREM

PROBABILITY AND BAYES THEOREM. PROBABILITY. POPULATION. SAMPLE. STATISTICAL INFERENCE. PROBABILITY: A numerical value expressing the degree of uncertainty regarding the occurrence of an event. A measure of uncertainty.

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PROBABILITY AND BAYES THEOREM

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  1. PROBABILITY AND BAYES THEOREM

  2. PROBABILITY POPULATION SAMPLE STATISTICAL INFERENCE

  3. PROBABILITY: A numerical value expressing the degree of uncertainty regarding the occurrence of an event. A measure of uncertainty. • STATISTICAL INFERENCE: The science of drawing inferences about the population based only on a part of the population, sample.

  4. PROBABILITY • CLASSICAL INTERPRETATION If a random experiment is repeated an infinite number of times, the relative frequency for any given outcome is the probability of this outcome. Probability of an event: Relative frequency of the occurrence of the event in the long run. • Example: Probability of observing a head in a fair coin toss is 0.5 (if coin is tossed long enough). • SUBJECTIVE INTERPRETATION The assignment of probabilities to event of interest is subjective • Example: I am guessing there is 50% chance of raining today.

  5. PROBABILITY • Random experiment • a random experiment is a process or course of action, whose outcome is uncertain. • Examples Experiment Outcomes • Flip a coin Heads and Tails • Record a statistics test marks Numbers between 0 and 100 • Measure the time to assemble Numbers from zero and abovea computer

  6. PROBABILITY • Performing the same random experiment repeatedly, may result in different outcomes, therefore, the best we can do is consider the probability of occurrence of a certain outcome. • To determine the probabilities, first we need to define and list the possible outcomes

  7. Sample Space • Determining the outcomes. • Build an exhaustive list of all possible outcomes. • Make sure the listed outcomes are mutually exclusive. • The set of all possible outcomes of an experiment is called a sample space and denoted byS.

  8. Sample Space Uncountable (Continuous ) Countable Finite number of elements Infinite number of elements

  9. EXAMPLES • Countable sample space examples: • Tossing a coin experiment S : {Head, Tail} • Rolling a dice experiment S : {1, 2, 3, 4, 5, 6} • Determination of the sex of a newborn child S : {girl, boy} • Uncountable sample space examples: • Life time of a light bulb S : [0, ∞) • Closing daily prices of a stock S : [0, ∞)

  10. EXAMPLES • Examine 3 fuses in sequence and note the results of each experiment, then an outcome for the entire experiment is any sequence of N’s (non-defectives) and D’s (defectives) of length 3. Hence, the sample space is S : { NNN, NND, NDN, DNN, NDD, DND, DDN, DDD}

  11. Assigning Probabilities • Given a sample space S ={O1,O2,…,Ok}, the following characteristics for the probability P(Oi) of the simple event Oi must hold: • Probability of an event: The probability P(A), of event A is the sum of the probabilities assigned to the simple events contained in A.

  12. Assigning Probabilities • P(A) is the proportion of times the event A is observed.

  13. Intersection • The intersection of event A and B is the event that occurs when both A and B occur. • The intersection of events A and B is denoted by (A and B) or AB. • The joint probability of A and B is the probability of the intersection of A and B, which is denoted by P(A and B) or P(AB).

  14. Union • The union event of A and B is the event that occurs when either A or B or both occur. • At least one of the events occur. • It is denoted “A or B” OR AB

  15. Complement Rule • The complement of event A (denoted by AC) is the event that occurs when event A does not occur. • The probability of the complement event is calculated by A and AC consist of all the simple events in the sample space. Therefore,P(A) + P(AC) = 1 P(AC) = 1 - P(A)

  16. MUTUALLY EXCLUSIVE EVENTS • Two events A and B are said to be mutually exclusive or disjoint, if A and B have no common outcomes. That is, A and B =  (empty set) • The events A1,A2,… are pairwise mutually exclusive (disjoint), if • Ai  Aj =  for all i  j.

  17. EXAMPLE • The number of spots turning up when a six-sided dice is tossed is observed. Consider the following events. A: The number observed is at most 2. B: The number observed is an even number. C: The number 4 turns up.

  18. S 2 1 1 1 A A A 3 5 2 2 B B B C 4 4 6 6 4 6 AB 2 VENN DIAGRAM • A graphical representation of the sample space. AB AC = A and C are mutually exclusive

  19. AXIOMS OF PROBABILTY(KOLMOGOROV AXIOMS) Given a sample space S, the probability function is a function P that satisfies 1) For any event A, 0  P(A)  1. 2) P(S) = 1. 3) If A1, A2,… are pairwise disjoint, then

  20. THE CALCULUS OF PROBABILITIES • If P is a probability function and A is any set, then a. P()=0 b. P(A)  1 c. P(AC)=1  P(A)

  21. THE CALCULUS OF PROBABILITIES • If P is a probability function and A and B any sets, then • P(B  AC) = P(B)P(A  B) • If A  B, then P(A)  P(B) c. P(A  B)  P(A)+P(B)  1 (Bonferroni Inequality) d. (Boole’s Inequality)

  22. EQUALLY LIKELY OUTCOMES • The same probability is assigned to each simple event in the sample space, S. • Suppose that S={s1,…,sN} is a finite sample space. If all the outcomes are equally likely, then P({si})=1/N for every outcome si.

  23. Addition Rule For any two events A and B P(A  B) = P(A) + P(B) - P(A  B)

  24. ODDS • The odds of an event A is defined by • It tells us how much more likely to see the • occurrence of event A. • P(A)=3/4P(AC)=1/4 P(A)/P(AC) = 3. • That is, the odds is 3. It is 3 times more likely that A occurs as it is that it does not.

  25. CONDITIONAL PROBABILITY • (Marginal) Probability: P(A): How likely is it that an event A will occur when an experiment is performed? • Conditional Probability: P(A|B): How will the probability of event A be affected by the knowledge of the occurrence or nonoccurrence of event B? • If two events are independent, then P(A|B)=P(A)

  26. CONDITIONAL PROBABILITY

  27. Example • Roll two dice • S=all possible pairs ={(1,1),(1,2),…,(6,6)} • Let A=first roll is 1; B=sum is 7; C=sum is 8 • P(A|B)=?; P(A|C)=? • Solution: • P(A|B)=P(A and B)/P(B) P(B)=P({1,6} or {2,5} or {3,4} or {4,3} or {5,2} or {6,1}) = 6/36=1/6 P(A|B)= P({1,6})/(1/6)=1/6 =P(A) A and B are independent

  28. Example • P(A|C)=P(A and C)/P(C)=P(Ø)/P(C)=0 A and C are disjoint Out of curiosity: P(C)=P({2,6} or {3,5} or {4,4} or {5,3} or {6,2}) = 5/36

  29. BAYES THEOREM • Suppose you have P(B|A), but need P(A|B). • Can be generalized to more than two events.

  30. Example • Let: • D: Event that person has the disease; • T: Event that medical test results positive • Given: • Previous research shows that 0.3 % of all Turkish population carries this disease; i.e., P(D)= 0.3 % = 0.003 • Probability of observing a positive test result for someone with the disease is 95%; i.e., P(T|D)=0.95 • Probability of observing a positive test result for someone without the disease is 4%; i.e. P(T| )= 0.04 • Find: probability of a randomly chosen person having the disease given that the test result is positive.

  31. Example • Solution: Need P(D|T). Use Bayes Thm. P(D|T)=P(T|D)*P(D)/P(T) P(T)=P(D and T)+P( and T) = 0.95*0.003+0.04*0.997 = 0.04273 P(D|T) =0.95*0.003 / 0.04273 = 6.67 % Test is not very reliable!

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