Econ 240A

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# Econ 240A - PowerPoint PPT Presentation

1. 1. Econ 240A. Power Four. Last Time. Probability. The Big Picture. The Classical Statistical Trail. Rates &amp; Proportions. Inferential Statistics. Application. Descriptive Statistics. Discrete Random Variables. Binomial. Probability. Discrete Probability Distributions; Moments.

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### Econ 240A

Power Four

Last Time
• Probability
The Classical Statistical Trail

Rates &

Proportions

Inferential

Statistics

Application

Descriptive Statistics

Discrete Random

Variables

Binomial

Probability

Discrete Probability Distributions; Moments

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.

P (Bald and MA) = 0.28

Bald

Not Bald

Middle Aged men

P (Bald and MA) = 0.28

P(HA/Bald and MA) = 0.18

P(HA/Not Bald and MA)

= 0.11

Bald

Not Bald

Middle Aged men

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
Random Variables
• There is a natural transition or easy segue from our discussion of probability and Bernoulli trials last time to random variables
• Define k to be the random variable # of heads in 1 flip, 2 flips or n flips of a coin
• We can find the probability that k=0, or k=n by brute force using probability trees. We can find the histogram for k, its central tendency and its dispersion
Outline
• Random Variables & Bernoulli Trials
• example: one flip of a coin
• expected value of the number of heads
• variance in the number of heads
• example: two flips of a coin
• a fair coin: frequency distribution of the number of heads
• one flip
• two flips
Outline (Cont.)
• Three flips of a fair coin, the number of combinations of the number of heads
• The binomial distribution
• frequency distributions for the binomial
• The expected value of a discrete random variable
• the variance of a discrete random variable
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
Flipping a Coin Once

The random variable k is the number of heads

it is variable because k can equal one or zero

it is random because the value of k depends on

probabilities of occurrence, p and 1-p

Prob. = p

Prob. = 1-p

Tails, k=0

Flipping a coin once
• Expected value of the number of heads is the value of k weighted by the probability that value of k occurs
• E(k) = 1*p + 0*(1-p) = p
• variance of k is the value of k minus its expected value, squared, weighted by the probability that value of k occurs
• VAR(k) = (1-p)2 *p +(0-p)2 *(1-p) = VAR(k) = (1-p)*p[(1-p)+p] =(1-p)*p
Flipping a coin twice: 4 elementary outcomes

h, h; k=2

h, h

Prob =p

Prob=1-p

Prob =p

tails

h, t; k=1

h, t

Prob=p

t, h; k=1

t, h

Prob =1-p

tails

Prob =1-p

t, t

tails

t, t; k=0

Flipping a Coin Twice
• E(k)=2*p2 +1*p*(1-p) +1*(1-p)*p + 0*(1-p)2 E(k) = 2*p2 + p - p2 + p - p2 =2p
• so we might expect the expected value of k in n independent flips is n*p
• Variance in k
• VAR(k) = (2-2p)2 *p2 + 2*(1-2p)2 *p(1-p) + (0-2p)2 (1-p)2
Continuing with the variance in k
• VAR(k) = (2-2p)2 *p2 + 2*(1-2p)2 *p(1-p) + (0-2p)2 (1-p)2
• VAR(k) = 4(1-p)2 *p2 +2*(1 - 4p +4p2)*p*(1-p) + 4p2 *(1-p)2
• adding the first and last terms, 8p2 *(1-p)2 + 2*(1 - 4p +4p2)*p*(1-p)
• and expanding this last term, 2p(1-p) -8p2 *(1-p) + 8p3 *(1-p)
• VAR(k) = 8p2 *(1-p)2 + 2p(1-p) -8p2 *(1-p)(1-p)
• so VAR(k) = 2p(1-p) , or twice VAR(k) for 1 flip

One Flip of the Coin

probability

1/2

Two Flips of a Fair Coin

probability

1/2

1/4

0

2

1

Three Flips of a Fair Coin
• It is not so hard to see what the value of the number of heads, k, might be for three flips of a coin: zero, one ,two, three
• But one head can occur two ways, as can two heads
• Hence we need to consider the number of ways k can occur, I.e. the combinations of branching probabilities where order does not count

Three flips of a coin; 8 elementary outcomes

Three Flips of a Coin
• There is only one way of getting three heads or of getting zero heads
• But there are three ways of getting two heads or getting one head
• One way of calculating the number of combinations is Cn(k) = n!/k!*(n-k)!
• Another way of calculating the number of combinations is Pascal’s triangle

Three Flips of a Coin

Probability

3/8

2/8

1/8

0

1

2

3

The Probability of Getting k Heads
• The probability of getting k heads (along a given branch) in n trials is: pk *(1-p)n-k
• The number of branches with k heads in n trials is given by Cn(k)
• So the probability of k heads in n trials is Prob(k) = Cn(k) pk *(1-p)n-k
• This is the discrete binomial distribution where k can only take on discrete values of 0, 1, …k
Expected Value of a discrete random variable
• E(x) =
• the expected value of a discrete random variable is the weighted average of the observations where the weight is the frequency of that observation
Expected Number of Heads After Two Flips
• Because of independence p(kiI and kjII) = p(kiI)*p(kjII)
• Expected number of heads after two flips: E(kiI + kjII) = (kiI + kjII) p(kiI)*p(kjII)
• E(kiI + kjII) = kiI p(kiI)* p(kjII) +
Cont.
• E(kiI + kjII) = kiI p(kiI)* p(kjII) + kjII *p(kjII) p(kiI)
• E(kiI + kjII) = E(kiI) + E(kjII) = p*1 + p*1 =2p
• So the mean after n flips is n*p
Variance of a discrete random variable
• VAR(xi) =
• the variance of a discrete random variable is the weighted sum of each observation minus its expected value, squared,where the weight is the frequency of that observation
Cont.
• VAR(xi) =
• VAR(xi) =
• VAR(xi) =
• So the variance equals the second moment minus the first moment squared
The variance of the sum of discrete random variables
• VAR[xi + yj] = E[xi + yj - E(xi + yj)]2
• VAR[xi + yj] = E[(xi - Exi) + (yj - Eyj)]2
• VAR[xi + yj] = E[(xi - Exi)2 + 2(xi - Exi) (yj - Eyj) + (yj - Eyj)2]
• VAR[xi + yj] = VAR[xi] + 2 COV[xi*yj] + VAR[yj]
The variance of the sum if x and y are independent
• COV [xi*yj] = E(xi - Exi) (yj - Eyj)
• COV [xi*yj]= (xi - Exi) (yj - Eyj)
• COV [xi*yj]= (xi - Exi) p[x(i)]* (yj - Eyj)* p[y(j)]
• COV [xi*yj] = 0
Variance of the number of heads after two flips
• Since we know the variance of the number of heads on the first flip is p*(1-p)
• and ditto for the variance in the number of heads for the second flip
• then the variance in the number of heads after two flips is the sum, 2p(1-p)
• and the variance after n flips is np(1-p)
Application
• Rates and Proportions
Field Poll
• The estimated proportion, from the sample, that will vote for Guliani is:
• where is 0.35 or 35%
• k is the number of “successes”, the number of likely voters sampled who are for Guliani, approximately 122
• n is the size of the sample, 348
Field Poll
• What is the expected proportion of voters Nov. 7 who will vote for Guliani?
• = E(k)/n = np/n = p, where from the binomial distribution, E(k) = np
• So if the sample is representative of voters and their preferences, 35% should vote for Guliani next February
Field Poll
• How much dispersion is in this estimate, i.e. as reported by the Field Poll, what is the sampling error?
• The sampling error is calculated as twice the standard deviation or square root of the variance in
• = VAR(k)/n2 = np(1-p)/n2 =p(1-p)/n
• and using 0.35 as an estimate of p,
• = 0.35*0.65/348 =0.000654
Field Poll
• So the sampling error should be 2*0.026 or 5.2%.
• The Field Poll reports a 95% confidence interval or about two standard errors , I.e 2*2.6% ~ 5.4%
Field Poll
• Is it possible that Guliani might get 50% of the vote or more? Not likely since the probabilty of Guliani reciving more then 40% of the vote is only 2.5%
• Based on a normal approximation to the binomial, the true proportion voting for Guliani should fall between 29.5% and 40.5% with probability of about 95%, unless sentiments change.
Lab Two
• The Binomial Distribution, Numbers & Plots
• Coin flips: one, two, …ten
• Die Throws: one, ten ,twenty
• The Normal Approximation to the Binomial