1 bernoulli and binomial distributions. 2 bernoulli random variables setting: –finite population...
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Bernoulli and Binomial Distributions
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Bernoulli Random Variables
• Setting:– finite population – each subject has a categorical response
with one of 2 possible values (0/1) – pick a simple random sample of n=1
subject
• Y random variable representing response (a Bernoulli random variable)
E Y p
var 1Y p p
Prob(Y=1)
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Bernoulli Random Variables
• Example: Finite population of 100 subjects, where 40 are normal weight and 60 are overweight.
• Response: • 0 normal weight • 1 overweight
1
1 600.6
100
N
ss
y pN
Population Parameters:
Mean
2 2 2
1
22
22
1 140 0 60 1
100
40 601
100 100
1 1
1 1
1
N
ss
y p pN
p p
p p p p
p p p p
p p
Variance
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Bernoulli Random Variables
• Example: Finite population of 100 subjects, where 40 are normal weight and 60 are overweight.
Values: • 0 normal weight • 1 overweight
0.6p Population Parameters:
Mean 2 1 0.24p p Variance
Pick a single subject at random:
Y
a Bernoulli Random Variable
1n
10
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Bernoulli Random Variables
• Example: Finite population of 100 subjects, where 40 are normal weight and 60 are overweight.
Values: • 0 normal weight • 1 overweight
0.6p 2 1 0.24p p
ProbabilityY a Bernoulli Random
Variable
Event y P(y)
Normal 0 1-p
Overwt 1 p
Total 1
•events are mutually exclusive•exhaustive•probabilities sum to 1
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Bernoulli Random Variables
• Example: Finite population of 100 subjects, where 40 are normal weight and 60 are overweight.
Values: • 0 normal weight • 1 overweight
0.6p 2 1 0.24p p
Y a Bernoulli Random Variable
Event y P(y)
Normal 0 1-p
Overwt 1 p
Total 1
1 0 1events
E Y p y y
p p
p
2
2 2
var
1 0 1
1
events
Y p y y E y
p p p p
p p
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Bernoulli Random Variables
• Example: Finite population of 100 subjects, where 40 are normal weight and 60 are overweight.
Values: • 0 normal weight • 1 overweight
0.6p 2 1 0.24p p
Y a Bernoulli Random Variable
E Y p var 1Y p p
Simple random sample of n=1
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Binomial Random Variable
• Binomial Random Variable: The sum of independent identically distributed Bernoulli random variables.
• Example: Finite population of 100 subjects, where 40 are normal weight and 60 are overweight.
Values: • 0 normal weight • 1 overweight
• Select a simple random sample of size n with replacement– the random variable representing each selection is a Bernoulli
Random variables– the random variables are independent– the random variables are identically distributed
• iid = independent and identically distributed (always occurs for random variables representing selections using simple random sampling with replacement)
1
n
ii
X Y
a Binomial Random Variable
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Independent Variables
Are the two random variables independent?
1Y first selection in a sample2Y second selection in a sample
(with Rep)
Two random variables are independent if for any realized value of the firstrandom variable, the probability is unchanged for any realized value of the second random variable.
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Independent Variables
Are the two random variables independent?
1Y first selection in a sample2Y second selection in a sample
(with Rep)
Suppose
1 0Y
22
2
0 with 0 1
1 with 1
p Y pY
p Y p
2 12
2 1
0 with 0 | 0 1
1 with 1| 0
p Y Y pY
p Y Y p
1 1Y
2 1
22 1
0 with 0 | 1 1
1 with 1| 1
p Y Y pY
p Y Y p
Conclusion: The RV’s are independent
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Independent Variables
Are the two random variables independent?
1Y first selection in a sample2Y second selection in a sample
(without Rep)
1
11
0 with 0 1
1 with 1
p Y pY
p Y p
Y
a Bernoulli Random Variable
10
10
0.6
N
p
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Independent Variables
Are the two random variables independent?
1Y first selection in a sample2Y second selection in a sample
(without Rep)Suppose
1 0Y
2 1
2
2 1
30 with 0 | 0
96
1 with 1| 09
p Y YY
p Y Y
1 1Y
2 1
2
2 1
40 with 0 | 1
95
1 with 1| 19
p Y YY
p Y Y
Conclusion: The RV’s are not independent
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Binomial Random Variable
• Binomial Random Variable: The sum of independent identically distributed (iid) Bernoulli random variables.
1
n
ii
X Y
a Binomial Random Variable
1
21 1 1 n
n
Y
YX
Y
1 Y
1
2
n
Y
Y
Y
a vector of Random Variables
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Expected Value and Variance of a Vector of Random Variables
1 2 1 1 2 2 1 1 2 2 possible
values
cov , ,all
Y Y p Y y Y y y E Y y E Y
1
2
n
Y
Y
Y
a vector of Random Variables
1 1
2 2
n n
Y E Y
Y E YE
Y E Y
1 1 1 2 1
2 2 1 2 2
1 2
var cov , cov ,
cov , var cov ,var
cov , cov , var
n
n
n n n n
Y Y Y Y Y Y
Y Y Y Y Y Y
Y Y Y Y Y Y
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Expected Value and Variance of a Vector of Random Variables
1
2
n
Y
Y
Y
a vector of independent Random Variables
1 1
2 2
n n
Y
YE
Y
21 1
22 2
2
0 0
0 0var
0 0n n
Y
Y
Y
a vector of independent and identically distributed (iid)Random Variables
1
2
1
1
1
n
n
Y
YE
Y
1
21
22 2
2
0 0
0 0var
0 0
n
n
Y
Y
Y
I
zero covariances
identity matrix
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Expected Value and Variance of a Linear Combination of Random Variables
nE X E1 Y
a Binomial Random Variable
1
2
1
n
i n ni
n
Y
YX Y
Y
1 1 Y
a vector of independent and identically distributed Bernoulli Random Variables
1
2n
n
Y
YE E p
Y
Y 1
1
2
1 0 0
0 1 0var 1 1
0 0 1
n
n
Y
Yp p p p
Y
I
var varn nX 1 Y 1
1
n
i ii
X cY
c Y
In general
E X Ec Y var varX c Y c
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Variance of a Binomal Random Variables
2
2
2
2
2
2
var var
0 0
0 0
0 0
1 0 0
0 1 01 1 1
0 0 1
1 1 1
n n
n n
n
n
X
n
1 Y 1
1 1
1
1
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Expected Value and Variance of a Binomal Random Variable
E X np
a Binomial Random Variable
1
n
i ni
X Y
1 Y
a vector of independent and identically distributed Bernoulli Random Variables
nE pY 1
var 1 np p Y I
var 1X np p
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Binomial Distribution
see table A.1 in Appendix of Textn k=X 0.4=p
X=x
4 0 0.1785
1 0.3456
2 0.3456
3 0.1536
4 0.0256
2 | 0.4, 4 0.3456E X p n
3 | 0.4, 4 0.1792E X p n
3 | 0.4, 4 1 3 | 0.4, 4
1 0.1792
0.8208
E X p n E X p n
1| 0.6, 4 ?E X p n
P X x
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Binomial Distribution
see table A.1 in Appendix of Textn k 0.6
4 4 0.1785
3 0.3456
2 0.3456
1 0.1536
0 0.0256
1| 0.6, 4 ?
0.1536
E X p n
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SRS with rep: Seasons Study
With Seasons Study, define High Total Cholesterol: TC>240
Select SRS with replacement:
Run SAS program: ejs09b540p46.sas
Example: Change Program to get 5 samples of size n=10
For each, calculate total TC>240
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Binomial DistributionFigure 2. Histogram of Totals for Sample (Prop with TC>240) based on Samples of n=20
Source: ejs09b540p47.sas 12/2/2009 by ejs
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0
5
10
15
20
25
30
35
40
Pe
rce
nt
x1_sum
What if 10,000 Sampleswere selected?
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Binomial Distribution
( ) !!( )!nn
x x n x=
-
P(X=x=# with TC>240)=
=(# ways of ways of picking samples with x)Pr(x ‘success’)P(n-x ‘failures’)
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Binomial DistributionLikelihood
We select a srs with replacement of n=10 and observe x=4. What is p?
64
64
64
4 | , 10 1
101
4
10 9 8 71
4 3 2 1
210 1
n xxnP X p n p p
x
p p
p p
p p
This is a function of p
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Binomial DistributionLikelihood
We select a srs with replacement of n=10 and observe x=4. What is p?
644 | , 10 210 1P X p n p p
Likelihood: 64210 1L p p p
Use table to find values for p:p L(p) p L(p)
0.05 0.001 0.40 0.2508
0.10 0.0112 0.45 0.2384
0.15 0.0401 0.50 0.2051
0.20 0.0881 0.55 0.1596
0.25 0.1460 0.60 0.1115
0.30 0.2001 0.65 0.0689
0.35 0.2377 etc
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Binomial DistributionMaximum LikelihoodLikelihood: 64210 1L p p p
p L(p) p L(p)
0.05 0.001 0.40 0.2508
0.10 0.0112 0.45 0.2384
0.15 0.0401 0.50 0.2051
0.20 0.0881 0.55 0.1596
0.25 0.1460 0.60 0.1115
0.30 0.2001 0.65 0.0689
0.35 0.2377 etc
L p
0.05
0.1
0.2
0.2 0.3 0.4 0.5
MaximumLikelihood
ˆ 0.4x
pn
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Binomial Distribution- Differences in Use
Mean
Usually report “total” instead of “mean”.
Total
EstimateVariance
Estimated Variance
Use Normal CLT
P̂ Y ˆnP nY
1P P
n
1nP P
2ˆ ˆ1
ˆ pP P
n
ˆ ˆ1nP P
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Binomial Distribution- Differences in Use
Mean
Total
Use Normal Dist for Interval Estimates
0.975ˆ ˆ pP z 0.975
ˆ ˆ pnP z n
Approximation good when
5np 1 5n p and
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Binomial Distribution- Differences in Use
0
0 0
ˆ
1cal
p pz
p p
n
Use hypothesized p for variance when
5np 1 5n p and
0 0:H p p
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Binomial Distribution- CI for Difference in Prop.
Diff in Means
(Proportions see 14.6)
1 1 0.975 1 2ˆ ˆ ˆ ˆvarP P z P P
1 1 2 2
1 21 2
ˆ ˆ ˆ ˆ1 1ˆ ˆvar
P P P PP P
n n
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Binomial Distribution- Hyp. Test for Difference in Prop.
1 2 1 2ˆ ˆ
ˆcald
P P p pz
1 2
ˆ ˆ ˆ ˆ1 1ˆd
P P P P
n n
1 1 2 2
1 2
ˆ ˆˆ n P n PP
n n
Pooled prob
0 1 2:H p p
1 2:aH p p0 1 2: 0H p p
1 2: 0aH p p
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Chi-Square Distribution Hyp. Test for Difference in Prop.
22cal calz
Under the null hypothesis,this statistic follows a chi-square distribution with 1 degree of freedom.
0 1 2:H p p
1 2:aH p p0 1 2: 0H p p
1 2: 0aH p p