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Econ 240A

Power Four

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Last Time

• Probability

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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.

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Middle Aged men

Bald

P (Bald and MA) = 0.28

Not Bald

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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

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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

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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

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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

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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

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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

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Flipping a Coin Once

Heads, k=1

Tails, k=0

Prob. = p

Prob. = 1-p

The random variable k is the number of headsit is variable because k can equal one or zeroit is random because the value of k depends on probabilities of occurrence, p and 1-p

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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

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Flipping a coin twice: 4 elementary outcomes

heads

tails

heads

tails

heads

tails

h, h

h, t

t, h

t, t

h, h; k=2

h, t; k=1

t, h; k=1

t, t; k=0

Prob =p

Prob =p

Prob =1-p

Prob =1-p

Prob=p

Prob=1-p

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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

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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

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

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Frequency Distribution for the Number of Heads

• A fair coin

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O heads 1 head

1/2

probability

# of heads

One Flip of the Coin

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0 1 2 # of heads

probability

1/4

1/2

Two Flips of a Fair Coin

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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

H

T

H

T

p

H

T

1-p

p

1 - p

p

H

T

H

T

H

T

H

T

p

1-p

p

1-p

Three flips of a coin; 8 elementary outcomes

3 heads

2 heads

2 heads1 head

2 heads1 head

1 head

0 heads

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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

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0 1 2

1/8

2/8

3/8

Probability

3 # of heads

Three Flips of a Coin

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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

n

i

ixpix0

)]([*)(

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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 kj

II) = p(ki

I)*p(kjII)

• Expected number of heads after two flips: E(ki

I + kjII) = (ki

I + kjII)

p(kiI)*p(kj

II)

• E(kiI + kj

II) = kiI p(ki

I)* p(kjII) +

1

0

1

0 ji

1

0i

1

0j

Cont.

• E(kiI + kj

II) = kiI p(ki

I)* p(kjII)

+ kjII *p(kj

II) p(kiI)

• E(kiI + kj

II) = E(kiI) + E(kj

II) = p*1 + p*1 =2p

• So the mean after n flips is n*p

1

0i

1

0j

1

0j

1

0i

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

)]([})]([)({[ 2

0

ixpixEixn

i

Cont.

• VAR(xi) =

• VAR(xi) =

• VAR(xi) =

• So the variance equals the second moment minus the first moment squared

)]([*})]([)({ 2

0

ixpixEixn

i

})([*})]([)(*)]([2)]({[0

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n

i

ixpiExixixEix

2

0

2 )]([)]}([)]([{ ixEixpixn

i

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

)]()([ jyixp

n

j

m

i 00

m

i 0

n

j 0

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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)

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The Field Poll

• In a sample of 731 people, 33% indicate they will vote for Bill Jones as Senator. The others are for Boxer (50%) or otherwise inclined (17% undecided or for others)

• If the poll is an accurate reflection or subset of the population of voters on Nov. 2, what is the expected proportion that will vote for Jones?

• How much uncertainty is in that expectation?

Field Poll

• The estimated proportion, from the sample, that will vote for Jones is:

• where is 0.33 or 33%• k is the number of “successes”, the

number of people sampled who are for Jones, approximately 241

• n is the size of the sample, 731

nkp /ˆ

p̂

Field Poll

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

• = 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, 33% should vote for Jones next November

)ˆ( pE

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 0.33 as an estimate of p,• = 0.33*0.67/731 =0.00030

p̂

)ˆ( pVAR

)ˆ( pVAR

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Field Poll

• So the sampling error should be 0.017 or 1.7%, i.e. the square root of 0.00030

• The Field Poll reports a 95% confidence interval or about two standard errors , I.e 2*1.7%

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Field Poll

• Is it possible that Bill Jones could win? This estimate of 0.33 plus or minus twice the sampling error of 0.017, creates an interval of 0.31 to 035.

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

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Boxer has been slipping, But Jones hasn’t gained much