Probability & Probability Distributions

Transcription

1 Probability & Probability Distributions Carolyn J. Anderson EdPsych 580 Fall 2005 Probability & Probability Distributions p. 1/61

2 Probability & Probability Distributions Elementary Probability Theory Definitions Rules Bayes Theorem Probability Distributions Discrete & continuous variables. Characteristics of distributions. Expectations Probability & Probability Distributions p. 2/61

3 Elementary Probability Theory or How Likely are the results? Probabilities arise when sampling individuals from a population and in experimental situations, because different trials or replications of the same experiment usually result in different outcomes. Probability & Probability Distributions p. 3/61

4 Statistical Experiment A (simple) statistical experiment is some well defined act or process (including sampling) that leads to one well defined outcome. It s repeatable (in principle). There is uncertainty about the results. Uncertainty is modeled by assigning probabilities to the outcomes. Examples... Probability & Probability Distributions p. 4/61

5 Examples of Statistical Experiments Well defined, repeatable, uncertainty, model by probabilities? Flip a coin 5 times & record number of heads. Count the number of blue M& M s in a 9 oz. package. Roll two dice & record the total number of spots. Ask people who they intend to vote for in the next presidential election. Recorded number of correct responses on a test. Probability & Probability Distributions p. 5/61

6 Statistical Experiments A statistical experiment maybe Real (it can actually be done). Conceptualize (completely idealized). Probability & Probability Distributions p. 6/61

7 Definition: Probability The probability of an event is the proportion of times that the event occurs in a large number of trials of the experiment. It is the long-run relative frequency of the event. Probability & Probability Distributions p. 7/61

8 Example Experiment: Draw a card from a standard deck of 52. Sample space: The set of all possible distinct outcomes, S (e.g., 52 cards). Elemenatary event or sample point: a member of the sample space. (e.g., the ace of hearts). Event (or event class): any set of elementary events. e.g., Suit (Hearts), Color (Red), or Number (Ace). Probability & Probability Distributions p. 8/61

9 Example (continued) Probability of an Ace = number of aces number of cards = 4 52 =.0769 Notes: Elementary events are equally likely Denote events by roman letters (e.g., A, B, etc) Denote probability of an event as P(A). Probability & Probability Distributions p. 9/61

10 More Definitions Joint Event is when you consider two (or more events) at a time. e.g., A =heads on penny, B = heads on quarter, and joint event is heads on both coins. Intersection: (A B) = A and B occur at the same time. Union: (A B) = A or B occur Only A occurs. Only B occurs. A and B occur. Probability & Probability Distributions p. 10/61

11 More Definitions Complement of an event is that the event did not occur. Ā not A. e.g., if A =red card, then Ā is a black card (not a red card). Mutually exclusive events are events that cannot occur at the same time. Events have no elementary events in common. e.g., A = heart and B = club. Mutually exclusive and exhaustive events are a complete partition of the sample space. e.g., Suits (hearts, diamonds, clubs, spades) Numbers (A, 2, 3, 4, 5, 6, 7, 8, 9, J, Q, K) Probability & Probability Distributions p. 11/61

12 Formal Defintion of Probability Probability is a number assigned to each and every member in the sample space. Denote by P( ). A probability function is a rule of correspondence that associates with each event A in the sample space S a number P(A) such that 0 P(A) 1, for any event A. The sum of probabilities for all distinct events is 1. If A and B are mutually exclusive events, then P(A or B) = P(A B) = P(A) + P(B) Probability & Probability Distributions p. 12/61

13 Example Let A = number card (i.e., 2 10), B = face card (i.e., J, Q, K), and C = Ace. Probabilities of events: P(A) = 9(4)/52 = 36/52 =.6923 P(B) = 3(4)/52 = 12/52 =.2308 P(C) = 1(4)/52 = 4/52 =.0769 P(A) + P(B) + P(C) = 1 P(A B) = P(A) + P(B) = =.9231 = 48/52. Probability & Probability Distributions p. 13/61

14 Another Example Experiment: Randomly select a third grade student from a Unit 4 public school in Champaign county. Sample Space: All 3rd grade students at Unit 4 public schools. Elementary Event: A characteristic of the child. e.g., brown hair, age (in months), weight, gender, the response very much to question How much do you like school? Probability & Probability Distributions p. 14/61

15 Venn Diagram S A C B A C A B Addition rules... Probability & Probability Distributions p. 15/61

16 Addition Rules Rule 1: If 2 events, B & C, are mutually exclusive (i.e., no overlap) then the probability that one or both occur is P(B or C) = P(B C) = P(B) + P(C) Rule 2: For any 2 events, A & B, the probability that one or both occur is P(A or B) = P(A B) = P(A)+P(B) P(A B) Probability & Probability Distributions p. 16/61

17 Example: Teachers by Region The population consists of all elementary and secondary teachers in US in Level Region Elementary Secondary Northeast 273, , ,700 North Central 314, , ,462 South 240, , ,208 West 279, , ,466 1,107, ,062 2,013,836 Probability & Probability Distributions p. 17/61

18 Example: Teachers by Region Elementary event (or sample point ) is a teacher. Event is any set of teachers. (e.g., region, level, or combination). Simple Experiment: Select 1 teacher at random, 1, 107, 774 P(elementary) = 2, 0138, 836 =.55 P(not elementary) =P(secondary) = 1.55 =.45 Probability & Probability Distributions p. 18/61

19 Example: Addition Rules Rule 1: Events are an elementary teacher from the South & an elementary teacher from the West, P(elementary in S or W) = = P(elementary, South) + P(elementary, West) 240, , 445 = + 2, 013, 836 2, 013, 836 =.26 Probability & Probability Distributions p. 19/61

20 Example: Addition Rules (continued) Rule 2: Events are an elementary teacher and a teacher from the South P(elementary or from South) = = P(elementary) + P(South) P(elementary and South) 1, 107, , , 028 = + 2, 013, 836 2, 013, 836 2, 013, 836 =.64 Probability & Probability Distributions p. 20/61

21 Conditional Probability Conditional Probability equals the probability of an event A given that we know that event B has occurred. P(A B) = P(A B) P(B) = P(A,B) P(B) Example: What is the probability that a teacher is from the South given that he/she is an elementary school teacher? Probability & Probability Distributions p. 21/61

22 Example: Answer P(elementary and South) P(South elementary) = P(elementary) 240, 028/2, 013, 836 = 1, 107, 774/2, 013, , 028 = 1, 107, 774 =.217 Probability & Probability Distributions p. 22/61

23 Example (continued) Note that P(South) = 423, 208 2, 013, 396 =.210 Knowing that a teacher is an elementary school teacher changes the chance that the teacher is also from the south, P(South elementary) P(South) Probability & Probability Distributions p. 23/61

24 Bayes Theorem P(A B) = P(A,B) = P(A B)P(B) P(A B) = P(A,B) = P(B A)P(A) Bayes Theorem: P(A B) = P(B A)P(A) P(B) Example: Monty hall problem. Door A Door B Door C Probability & Probability Distributions p. 24/61

25 Monty Hall Problem Start of Game: Probability of getting the big prise (e.g, car) P(A) = 1 3 P(B) = 1 3 P(C) = 1 3 You pick door A. Monty opens door B and gives you the chance to switch from door A to door C. What should do you do? Monty s trying to not let you win... Probability & Probability Distributions p. 25/61

26 Monty Hall Problem (continued) Choose the door for which has the larger conditional probability, i.e., P(A Monty opened B) or P(C Monty opened B). Use Bayes Theorem... so we need Conditional probabilities that Monty opens door B given the car is behind A, behind B and behind C. Joint probabilities that Monty chooses door B and the car is behind door A, door B and door C. Unconditional probability that Monty chooses door B. Probability & Probability Distributions p. 26/61

27 Monty Hall Problem (continued) Conditional prob. that Monty opens door B: P(Monty opensb car behind A) = P(B Monty A) = 1 2 P(Monty opensb car behind B) = P(B Monty B) = 0 P(Monty opensb car behind C) = P(B Monty C) = 1 Joint probabilities: P(B Monty,A) = P(B Monty A)P(A) = = 1 6 P(B Monty,B) = P(B Monty B)P(B) = = 0 P(B Monty,C) = P(B Monty C)P(C) = = 1 3 Probability & Probability Distributions p. 27/61

28 Monty Hall Problem (continued) (Unconditional) Probability that Monty opens door B: P(B Monty ) = P(B Monty,A) + P(B Monty,B) + P(B Monty,C) = = 1 2 Apply Bayes Theorem... P(A B Monty ) = P(C B Monty ) = P(A) P(B Monty ) P(B Monty A) = 1/3 1/2 1 2 = 1 3 P(C) P(B Monty ) P(B Monty C) = 1/3 1/2 1 = 2 3 Probability & Probability Distributions p. 28/61

29 Monty Hall Problem (continued) I got this example from: Gill, J. (2002). Bayesian Methods for the Social and Behavioral Sciences. Chapman & Hall. Other sources on The Monty Hall Problem. History. Use today. Probability & Probability Distributions p. 29/61

30 Independence If the conditional and unconditional probabilities are identical, then the two events are Independent. For Independent events, P(A B) = P(A) P(B A) = P(B) P(A and B) = P(A B) = P(A)P(B) = the multiplicative rule. Probability & Probability Distributions p. 30/61

31 Conditional Independence (continued) Conditional probabilities and Conditional Independence: two very important concepts. Conditional probability and regression. Conditional Independence: explaining dependency (e.g., classic example: Cal graduate admissions) Demonstration: Toss penny and quarter and give information about what happens. Probability & Probability Distributions p. 31/61

32 Are Events Conditionally Independence? Physical considerations physically unrelated events.. Deduced from observations. Probability & Probability Distributions p. 32/61

33 Independent: Physical Considerations Examples: Toss a penny & a quarter: P(penny = head & quarter = head) = P(penny = head)p(quarter = head) = (.5)(.5) =.25 Role two dice: P(die1 = 5 & die2 = 6) = P(die1 = 5)P(die2 = 6) = (1/6)(1/6) = 1/36 =.0278 Probability & Probability Distributions p. 33/61

34 Independent: Physical Considerations Examples: Administer an test that measures attitude toward gun control to 2 randomly drawn adults in the US population. P(Score 1 = 50 and Score 2 = 55) = P(Score 1 = 50)P(Score 2 = 55) Probability & Probability Distributions p. 34/61

35 Independence: Deduction Whether events are independent can sometimes be deduced from observations, e.g., Mendal s experiments. Mendal postulated that existence of genes that are recessive and dominant. Experiment: Bred pure strains of yellow peas & green peas. 1st generation: Cross the yellow and green peas. 2nd generation: Cross plants from 1st generation with each other and found... Probability & Probability Distributions p. 35/61

36 Mendal s Experiments (continued) Results: About 75% yellow and About 25% green. Results were very regular and replicable (with other traits and plants). Part of explanation involves assumption of independence. Probability & Probability Distributions p. 36/61

37 Mendal s Experiments: Explanation There exist genes which when paired up control seed color according to rules: y/g yellow g/y yellow y/y yellow g/g green 1st generation: Pure yellow strain (y/y) could only give a y gene and pure green strain (g/g) could only give a g gene. 2nd generation: About 1/2 of parent plants contribute a y, about 1/2 contribute a g, and pairing in random (independent). Probability & Probability Distributions p. 37/61

38 Mendal s Experiments: Explanation Maternal y g y y/y y/g Paternal (.25) (.25) (.50) g g/y g/g (.25) (.25) (.50) (.50) (.50) (1.00) Probability of each cell = (.50)(.50) = this is the independence part of the theory. Probability of phenotype: P(yellow pea) = P(y/y) + P(y/g) + P(g/y) =.75 P(green pea) = P(g/g) =.25 Probability & Probability Distributions p. 38/61

39 Mendal s Experiments: Explanation Mendal s theory is an example where an abstract probability theory is applied to observed data. The postulated probability distribution of seed color for 2nd generation Probability & Probability Distributions p. 39/61

40 Basic Logic Assumed some things to be true (e.g., Mendal s theory). Make deductions about what should be true in the long-run (e.g., 2nd generation: 75% yellow and 25% green). It s physically impossible to do all possible experiments, so we do some ( sample ). By chance the results will differ from what should be true; however, in the long-run, it would be exactly equal/true. Probability & Probability Distributions p. 40/61

41 Probability Distributions From Hayes: Any statement of a function associating each of a set of mutually exclusive and exhaustive events with its probability is a probability distribution Let X represent a function that associates a Real number with each and every elementary event in some sample space S. Then X is called a random variable on the sample space S. Probability & Probability Distributions p. 41/61

42 Random Variables If random variable can only equal a finite number of values, it is a discrete random variable. Probability distribution is known as a probability mass function. If a random variable can equal an infinite (or really really large) number of values, then it is a continuous random variable. Probability distribution is know as a probability density function. Probability & Probability Distributions p. 42/61

43 Discrete Random Variables From Mendal s theory, assign event to real number (arbitary): { 1 if yellow Y = 0 if green Probability Mass Function: Probability & Probability Distributions p. 43/61

44 Lottery Spinner Color Y P(Y ) Yellow Blue 5.20 Red 0.50 Green Tan Probability & Probability Distributions p. 44/61

45 Lottery Spinner Probability & Probability Distributions p. 45/61

46 Lottery Spinner Probability & Probability Distributions p. 46/61

47 Continuous Random Variables When a numerical variable is continuous, it s probability distribution is represented by a curve known as a probability density function or just p.d.f. Denote a p.d.f by f(y). P(x 1 Y x 2 ) = area under curve. Probability = area under curve. P(Y = y) = 0 Probability & Probability Distributions p. 47/61

48 Continuous Random Variable The event is how many miles a randomly selected graduate student attending UIUC is from home. What would c.d.f look like? Probability & Probability Distributions p. 48/61

49 Continuous Random Variable Probability that a graduate student attending UIUC is 2,000 or more miles from home corresponds to the shaded area. Probability & Probability Distributions p. 49/61

50 Continuous Random Variables The event is temperature outside the education building on January 27th. P(22.0 o ) = 0 P(19.5 o temperature 21 o ) =black area. Probability & Probability Distributions p. 50/61

51 Examples of p.d.f. s Where s the mean, median and mode? Probability & Probability Distributions p. 51/61

52 Examples of p.d.f. s Where s the mean, median and mode? Probability & Probability Distributions p. 52/61

53 Characteristics of Distributions Discrete or continuous Shape Central tendency Dispersion (variability) Probability & Probability Distributions p. 53/61

54 Expected Value If you played this game what would you expect to win or lose? Color Y P(Y ) Yellow Blue 5.20 Red 0.50 Green Tan µ Y = E(Y ) =.1( 100) +.2( 5) +.5(0) +.1(10) +.1(100) = 0 Probability & Probability Distributions p. 54/61

55 Expectations are Means For discrete random variable, E[Y ] = µ y n y i P(y i ) i=1 For continuous variables, E[Y ] = µ y yf(y)d(y) Variance is the mean squared deviation, σy 2 = E[(y µ y ) 2 ] = E[y 2 2yµ y + µ 2 y] = E[y 2 ] 2µ y E[y] + µ 2 y] = E[y 2 ] µ 2 y Probability & Probability Distributions p. 55/61

56 Expectations are Means (continued) Example: The variance of lottery spinner: σ 2 = E[(y µ y ) 2 ] 5 = (y i µ) 2 P(Y i ) i=1 =.1( 100 0) 2 +.2( 5 0) 2 +.5(0 0) 2.1(10 0) 2 +.1(100 0) 2 = 2, So how much would you pay to take a spin? Probability & Probability Distributions p. 56/61

57 The Algebra of Expectations Why? We don t have to deal with calculus & it s used alot in statistics. From Hayes Appendix B, Rule 1: If a is a constant, then E(a) = a Rule 2: If a is a constant real number and Y is a random variable with expectation E(Y ), then E(aY ) = ae(y ) Probability & Probability Distributions p. 57/61

58 The Algebra of Expectations Rule 3: If a is a constant real number and Y is a random variable with expectation E(Y ), then E(Y + a) = E(Y ) + a Rule 4: If X and Y are random variables with expectations E(X) and E(Y ), respectively, then E(X + Y ) = E(X) + E(Y ) Probability & Probability Distributions p. 58/61

59 The Algebra of Expectations Rule 5: Given a finite number of random variables, the expectation of the sum of those variables is the sum of their individual expectations, e.g. Variances: E(X + Y + Z) = E(X) + E(Y ) + E(Z) Rule 6: If a is a constant and if Y is a random variable with variance σ 2 y, then the random variable (Y + a) has variance σ 2 y. Probability & Probability Distributions p. 59/61

60 The Algebra of Expectations Rule 7: If a is a constant and if Y is a random variable with variance σ 2 y, then the random variable (ay ) has variance a 2 σ 2 y. Rule 8: If X and Y are independent random variables with variances σx 2 and σ2 y, then the variance of X + Y is σ 2 (x+y) = σ2 x + σ 2 y What about variance of (X y)? Probability & Probability Distributions p. 60/61

61 The Algebra of Expectations Independence Rule 9a: Given random variables X and Y with expectations E(X) and E(Y ), respectively, then X and Y are independent if E(XY ) = E(X)E(Y ) Rule 9b: If E(XY ) E(X)E(Y ), the variables X and Y are not independent.... that s enough for now. o Probability & Probability Distributions p. 61/61