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

Econ 240A PowerPoint PPT Presentation


  • 66 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


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


Working Problems


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


This time


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

Heads, k=1

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

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

  • 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

  • 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


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


Frequency Distribution for the Number of Heads

  • A fair coin


One Flip of the Coin

probability

1/2

1 head

O heads

# of heads


Two Flips of a Fair Coin

probability

1/2

1/4

0

2

# of heads

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

3 heads

2 heads

2 heads

1 head

2 heads

1 head

1 head

0 heads


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

# of 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

  • 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

  • E(x + y) = E(x) + E(y)


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) +


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 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 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 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 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 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

  • The Binomial Distribution, Numbers & Plots

    • Coin flips: one, two, …ten

    • Die Throws: one, ten ,twenty

  • The Normal Approximation to the Binomial


  • Login