# 1 Econ 240A Power Three. 2 Summary: Week One Descriptive Statistics –measures of central tendency –measures of dispersion Distributions of observation.

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1 Econ 240A Power Three

2 Summary: Week One Descriptive Statistics –measures of central tendency –measures of dispersion Distributions of observation values –Histograms: frequency(number) Vs. value Exploratory data Analysis –stem and leaf diagram –box and whiskers diagram

3 Probability The Gambler Kenny Rogers 20 Great Years

4 Outline Why study probability? Random Experiments and Elementary Outcomes Notion of a fair game Properties of probabilities Combining elementary outcomes into events probability statements probability trees

5 Outline continued conditional probability independence of two events

6 Perspectives About Probability Logical Discipline (like economics) –Axiomatic: conclusions follow from assumptions Easier to Understand with Examples –I will use words, symbols and pictures Test Your Understanding By Working Problems

7 Why study probability? Understand the concept behind a random sample and why sampling is important –independence of two or more events understand a Bernoulli event –example; flipping a coin understand an experiment or a sequence of independent Bernoulli trials

8 Cont. Understand the derivation of the binomial distribution, i.e. the distribution of the number of successes, k, in n Bernoulli trials understand the normal distribution as a continuous approximation to the discrete binomial understand the likelihood function, i.e. the probability of a random sample of observations

9 Uncertainty in Life Demography –Death rates –Marriage –divorce

10 Uncertainty in Life: US (CDC)

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12 Probability of First Marriage by Age, Women: US (CDC)

13 Cohabitation: The Path to Marriage?: US(CDC)

14 Race/ethnicity Affects Duration of First Marriage

15 Concepts Random experiments Elementary outcomes example: flipping a coin is a random experiment –the elementary outcomes are heads, tails example: throwing a die is a random experiment –the elementary outcomes are one, two, three, four, five, six

16 Axiomatic Basis or Concepts Elementary outcomes have non-negative probabilities: P(H)>=0, P(T)>=0 The sum of the probabilities over all elementary outcomes equals one: P(H) + P(T) = 1 H H T Flip a coin

17 Axiomatic Basis or Concepts II The probability of two mutually exculsive events is zero: P(H and T) = P(H^T) = 0 The probability of one outcome or the other is the sum of the probabilities of each minus any double counting: P(H or T) = P(H U T) = P(H) + P(T) – P(H^T) The probability of the event not happening is one minus the probability of the event happening:

18 Axiomatic Basis or Concepts III Conditional probability of heads given tails equals the joint probability divided by the probability of tails: P(H/T) = P(H^T)/P(T)

19 Concept A fair game example: the probability of heads, p(h), equals the probability of tails, p(t): p(h) = p(t) =1/2 example: the probability of any face of the die is the same, p(one) = p(two) = p(three) = p(four) =p(five) = p(six) = 1/6

Properties of probabilities Nonnegative –example: p(h) probabilities of elementary events sum to one –example p(h) + p(t) = 1

21 Another Example: Toss Two Coins H1H1 T1T1 H2H2 T2T2 H, H H, T H2H2 T2T2 T, H T, T

22 Flipping a coin twice: 4 elementary outcomes heads tails heads tails heads tails h, h h, t t, h t, t

23 Axiomatic Basis or Concepts Elementary outcomes have non-negative probabilities: P(H, H)>=0, P(H, T)>=0, P(T, H)>=0, P(T, T) >=0 The sum of the probabilities over all elementary outcomes equals one: P(H, H) + P(H, T) + P(T, H) + P(T, T) = 1 The probability of two mutually exculsive events is zero: P[(H, H)^(H, T)] = 0 H

24 Axiomatic Basis or Concepts II The probability of one outcome or the other is the sum of the probabilities of each minus any double counting: P[(H, H) U (H,T)] = P(H, H) + P(H, T) – P[(H, H)^(H, T)] = P(H, H) + P(H, T) The probability of the event not happening is one minus the probability of the event happening:

25 Axiomatic Basis or Concepts III Conditional probability of heads, heads given heads, tails equals the joint probability divided by the probability of heads, tails: P[(H, H)/(H, T)] = P[(H, H)^(H, T)]/P(H, T)

26 Throwing Two Dice, 36 elementary outcomes

27 Larry Gonick and Woollcott Smith, The Cartoon Guide to Statistics

28 Combining Elementary Outcomes Into Events Example: throw two dice: event is white die equals one example: throw two dice and red die equals one example: throw two dice and the sum is three

29 Event: white die equals one is the bottom row Event: red die equals one is the right hand column

30 Combining Elementary Outcomes Into Events Example: throw two dice: event is white die equals one P(W1) =P(W1^R1) + P(W1^R2) + P(W1^R3) + P(W1^R4) + P(W1^R5) + P(W1^R6) = 6/36 example: throw two dice and red die equals one example: throw two dice and the sum is three

31 Event: 2 dice sum to three is lower diagonal

Operations on events The event A and the event B both occur: Either the event A or the event B occurs or both do: The event A does not occur, i.e.not A:

Probability statements Probability of either event A or event B –if the events are mutually exclusive, then probability of event B

34 Probability of a white one or a red one: p(W1) + p(R1) double counts

Two dice are thrown: probability of the white die showing one and the red die showing one

36 Probability 2 dice add to 6 or add to 3 are mutually exclusive events Probability of not rolling snake eyes is easier to calculate as one minus the probability of rolling snake eyes

Problem What is the probability of rolling at least one six in two rolls of a single die? –At least one six is one or two sixes –easier to calculate the probability of rolling zero sixes: (5/36 + 5/36 + 5/36 + 5/36 + 5/36) = 25/36 –and then calculate the probability of rolling at least one six: 1- 25/36 = 11/36

38 1 2 3 4 5 6 1 2 3 4 5 6 Probability tree 2 rolls of a die: 36 elementary outcomes, of which 11 involve one or more sixes

39 Conditional Probability Example: in rolling two dice, what is the probability of getting a red one given that you rolled a white one? –P(R1/W1) ?

40 In rolling two dice, what is the probability of getting a red one given that you rolled a white one?

Conditional Probability Example: in rolling two dice, what is the probability of getting a red one given that you rolled a white one? –P(R1/W1) ?

Independence of two events p(A/B) = p(A) –i.e. if event A is not conditional on event B –then

43 Concept Bernoulli Trial –two outcomes, e.g. success or failure –successive independent trials –probability of success is the same in each trial Example: flipping a coin multiple times

Problem 6.28 Distribution of a retail store purchases classified by amount and method of payment

45 Problem (Cont.) A. What proportion of purchases was paid by debit card? B. Find the probability a credit card purchase was over \$100 C. Determine the proportion of purchases made by credit card or debit card

Problem 6.28

47 Problem (Cont.) A. What proportion of purchases was paid by debit card? 0.36 B. Find the probability a credit card purchase was over \$100 C. Determine the proportion of purchases made by credit card or debit card

48 Problem (Cont.) A. What proportion of purchases was paid by debit card? B. Find the probability a credit card purchase was over \$100 p(>\$100/credit card) = 0.23/0.47 = 0.489 C. Determine the proportion of purchases made by credit card or debit card

49 Problem (Cont.) A. What proportion of purchases was paid by debit card? B. Find the probability a credit card purchase was over \$100 C. Determine the proportion of purchases made by credit card or debit card –note: credit card and debit card purchases are mutually exclusive –p(credit or debit) = p(credit) + p (debit) = 0.47 + 0.36

50 Problem 6.61 A survey of middle aged men reveals that 28% of them are balding at the crown of their head. Moreover, it is known that such men have an 18% probability of suffering a heart attack in the next ten years. Men who are not balding in this way have an 11% probability of a heart attack. Find the probability that a middle aged man will suffer a heart attack in the next ten years.

51 Middle Aged men Bald P (Bald and MA) = 0.28 Not Bald

52 Middle Aged men Bald P (Bald and MA) = 0.28 Not Bald P(HA/Bald and MA) = 0.18 P(HA/Not Bald and MA) = 0.11

53 Probability of a heart attack in the next ten years P(HA) = P(HA and Bald and MA) + P(HA and Not Bald and MA) P(HA) = P(HA/Bald and MA)*P(BALD and MA) + P(HA/Not BALD and MA)* P(Not Bald and MA) P(HA) = 0.18*0.28 + 0.11*0.72 = 0.054 +.0792 = 0.1296

54 Summary: Probability Rules Addition: P(A or B) = P(A) + P(B) – P(A and B) –If A and B are mutually exclusive, P(A and B) = 0 Subtraction: P(E) = 1 – P( not E) Multiplication: P(A and B) = P(A/B) P(B) –If A and B are independent, then P(A/B) = P(B)

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