Title: Econ 240A
1Econ 240A
1
1
2Last Time
3The Big Picture
4The Classical Statistical Trail
Rates Proportions
Inferential Statistics
Application
Descriptive Statistics
Discrete Random Variables
Binomial
Probability
Discrete Probability Distributions Moments
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9Working Problems
10Problem 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.
11P (Bald and MA) 0.28
Bald
Not Bald
Middle Aged men
12P (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
13Probability 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) 0.180.28 0.110.72 0.054 .0792
0.1296
14This time
15Random 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 k0, or kn by
brute force using probability trees. We can find
the histogram for k, its central tendency and its
dispersion
16Outline
- 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
17Outline (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
18Concept
- 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
19Flipping 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, k1
Prob. p
Prob. 1-p
Tails, k0
20Flipping 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) 1p 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
21Flipping a coin twice 4 elementary outcomes
h, h k2
heads
h, h
Prob p
heads
Prob1-p
Prob p
tails
h, t
h, t k1
Probp
heads
t, h
t, h k1
Prob 1-p
tails
t, t
Prob 1-p
tails
t, t k0
22Flipping a Coin Twice
- Expected number of heads
- E(k)2p2 1p(1-p) 1(1-p)p 0(1-p)2 E(k)
2p2 p - p2 p - p2 2p - so we might expect the expected value of k in n
independent flips is np - Variance in k
- VAR(k) (2-2p)2 p2 2(1-2p)2 p(1-p)
(0-2p)2 (1-p)2
23Continuing 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
24- So we might expect the variance in n flips to be
np(1-p)
25Frequency Distribution for the Number of Heads
26One Flip of the Coin
probability
1/2
1 head
O heads
of heads
27Two Flips of a Fair Coin
probability
1/2
1/4
0
2
of heads
1
28Three 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 three 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
29Three flips of a coin 8 elementary outcomes
3 heads
2 heads
2 heads
1 head
2 heads
1 head
1 head
0 heads
30Three 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 Pascals triangle
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32Three Flips of a Coin
Probability
3/8
2/8
1/8
0
1
2
3
of heads
33The 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
34Expected 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
35Expected Value of the sum of random variables
36Expected 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)
37Cont.
- E(kiI kjII) kiI p(kiI) p(kjII)
kjII p(kjII) p(kiI) - E(kiI kjII) E(kiI) E(kjII) p1 p1
2p - So the mean after n flips is np
38Variance 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
39Cont.
- VAR(xi)
- VAR(xi)
- VAR(xi)
- So the variance equals the second moment minus
the first moment squared
40The variance of the sum of discrete random
variables
- VARxi yj Exi yj - E(xi yj)2
- VARxi yj E(xi - Exi) (yj - Eyj)2
- VARxi yj E(xi - Exi)2
2(xi - Exi) (yj - Eyj) (yj -
Eyj)2 - VARxi yj VARxi 2 COVxiyj VARyj
41The variance of the sum if x and y are independent
- COV xiyj E(xi - Exi) (yj - Eyj)
- COV xiyj (xi - Exi) (yj - Eyj)
- COV xiyj (xi - Exi) px(i) (yj -
Eyj) py(j) - COV xiyj 0
42Variance 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)
43Application
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46Field 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
47Field 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
48Field 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,
- 0.350.65/348 0.000654
49Field Poll
- So the sampling error should be 20.026 or 5.2.
- The Field Poll reports a 95 confidence interval
or about two standard errors , I.e 22.6 5.4
50Field 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.
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53Lab Two
- The Binomial Distribution, Numbers Plots
- Coin flips one, two, ten
- Die Throws one, ten ,twenty
- The Normal Approximation to the Binomial