Econ 240a
This presentation is the property of its rightful owner.
Sponsored Links
1 / 51

Econ 240A PowerPoint PPT Presentation


  • 64 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

Econ 240A

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Econ 240a

1

1

Econ 240A

Power Four


Last time

Last Time

  • Probability


The big picture

The Big Picture


The classical statistical trail

The Classical Statistical Trail

Rates &

Proportions

Inferential

Statistics

Application

Descriptive Statistics

Discrete Random

Variables

Binomial

Probability

Discrete Probability Distributions; Moments


Working problems

Working Problems


Problem 6 61

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.


Econ 240a

P (Bald and MA) = 0.28

Bald

Not Bald

Middle Aged men


Econ 240a

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

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


This time

This time


Random variables

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

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

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

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

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

Heads, k=1

Prob. = p

Prob. = 1-p

Tails, k=0


Flipping a coin once1

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

Flipping a coin twice: 4 elementary outcomes

h, h; k=2

heads

h, h

Prob =p

heads

Prob=1-p

Prob =p

tails

h, t; k=1

h, t

Prob=p

heads

t, h; k=1

t, h

Prob =1-p

tails

Prob =1-p

t, t

tails

t, t; k=0


Flipping a coin twice

Flipping a Coin Twice

  • Expected number of heads

    • 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

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


Econ 240a

  • So we might expect the variance in n flips to be np(1-p)


Frequency distribution for the number of heads

Frequency Distribution for the Number of Heads

  • A fair coin


Econ 240a

One Flip of the Coin

probability

1/2

1 head

O heads

# of heads


Econ 240a

Two Flips of a Fair Coin

probability

1/2

1/4

0

2

# of heads

1


Three flips of a fair coin

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


Econ 240a

Three flips of a coin; 8 elementary outcomes

3 heads

2 heads

2 heads

1 head

2 heads

1 head

1 head

0 heads


Three flips of a coin

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


Econ 240a

Three Flips of a Coin

Probability

3/8

2/8

1/8

0

1

2

3

# of heads


The probability of getting k heads

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

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 value of the sum of random variables

Expected Value of the sum of random variables

  • E(x + y) = E(x) + E(y)


Expected number of heads after two flips

Expected Number of Heads After Two Flips

  • Flip One: kiI heads

  • Flip Two: kjII heads

  • 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) +


Econ 240a

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

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


Econ 240a

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

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

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

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

Application

  • Rates and Proportions


Field poll

Field Poll

  • The estimated proportion, from the sample, that will vote for Prop. 74 is:

  • where is 0.46 or 46%

  • k is the number of “successes”, the number of likely voters sampled who are for Prop 74, approximately 144

  • n is the size of the sample, 314


Field poll1

Field Poll

  • What is the expected proportion of voters Nov. 2 that will vote for Prop 74?

  • = 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, 46% should vote for Prop. 74 next November


Field poll2

Field Poll

  • How much dispersion is in this estimate, i.e. as reported in newspapers, what is the margin of sampling error?

  • The margin of sampling error is calculated as the standard deviation or square root of the variance in

  • = VAR(k)/n2 = np(1-p)/n2 =p(1-p)/n

  • and using 0.46 as an estimate of p,

  • = 0.46*0.54/314 =0.000791


Field poll3

Field Poll

  • So the sampling error should be 0.028 or 2.8%, i.e. the square root of 0.000791

  • The Field Poll reports a 95% confidence interval or about two standard errors , I.e 2*2.8% = 5.6%


Field poll4

Field Poll

  • Is it possible that Prop 74 could win? This estimate of 0.46 plus or minus twice the sampling error of 0.028, creates an interval of 0.40 to 0. 52

  • Based on a normal approximation to the binomial, the true proportion voting for Prop.74 should fall in this interval with probability of about 95%, unless sentiments change.


Lab two

Lab Two

  • The Binomial Distribution, Numbers & Plots

    • Coin flips: one, two, …ten

    • Die Throws: one, ten ,twenty

  • The Normal Approximation to the Binomial


  • Login