Introduction to Likelihood
Meaningful Modeling of Epidemiologic Data, 2012
AIMS, Muizenberg, South Africa
Steve Bellan, MPH, PhD
Department of Environmental Science, Policy & Management
University of California at Berkeley
In a population of 1,000,000 people with a true prevalence of 30%, the
probability distribution of number of positive individuals if 100 are
sampled:
æ100ö
f (x) = ç ÷(0.3) x (0.7)100-x
è x ø
barplot(dbinom(x = 0:100, size = 100, prob = .3), names.arg = 0:size)
In a population of 1,000,000 people with a true prevalence of 30%, the
probability distribution of number of positive individuals if 100 are
sampled:
æ100ö
f (x) = ç ÷(0.3) x (0.7)100-x
è x ø
We sample 100 people once and 28 are positive:
> rbinom(n = 1, size = 100, prob = .3)
[1] 28
We don’t know the true prevalence!
But we can calculate the probability of
28 or a more extreme value occurring
for a given prevalence.
We sample 100 people once and 28 are positive:
> rbinom(n = 1, size = 100, prob = .3)
[1] 28
0.08
Cumulative Probability & P Values
for 30% prevalence:
0.04
> 2*pbinom(28,100,.3))
0.02
[1] 0.7535564
p-value = 0.74
0.00
probability
0.06
p(28 or a more extreme value occurring) =
0 2 4 6 8
11
14
17 20
23 26
29
32 35 38
41 44
47 50
53
56 59
number HIV+
We sample 100 people once and 28 are positive.
62 65 68
71 74
77
80 83
86 89
92
95 98
0.14
If true prevalence were 15%, then p(28 or more extreme) is
2*pbinom(28−1, 100, 0.15, lower.tail = FALSE)
0.08
0.06
0.04
0.02
x2
0.00
probability
0.10
0.12
p = 0.00123
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
number HIV+
0.14
If true prevalence were 20%, then p(28 or more extreme) is
2*pbinom(28−1, 100, 0.2, lower.tail = FALSE)
0.08
0.06
0.04
0.02
x2
0.00
probability
0.10
0.12
p = 0.0683
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
number HIV+
0.14
If true prevalence were 25%, then p(28 or more extreme) is
2*pbinom(28−1, 100, 0.25, lower.tail = FALSE)
0.08
0.06
0.04
0.02
x2
0.00
probability
0.10
0.12
p = 0.555
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
number HIV+
0.14
If true prevalence were 30%, then p(28 or more extreme) is
2*pbinom(28, 100, 0.3, lower.tail = TRUE)
0.08
0.06
0.04
0.02
x2
0.00
probability
0.10
0.12
p = 0.754
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
number HIV+
0.14
If true prevalence were 35%, then p(28 or more extreme) is
2*pbinom(28, 100, 0.35, lower.tail = TRUE)
0.08
0.06
0.04
0.02
x2
0.00
probability
0.10
0.12
p = 0.17
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
number HIV+
0.14
If true prevalence were 40%, then p(28 or more extreme) is
2*pbinom(28, 100, 0.4, lower.tail = TRUE)
0.08
0.06
0.04
0.02
x2
0.00
probability
0.10
0.12
p = 0.0169
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
number HIV+
p = 0.00123
p = 0.0683
p = 0.555
11
18
25
32
39
53
60
67
74
81
88
95
0.00
0.00
46
0
5
11
18
25
32
39
46
53
60
67
74
81
88
95
0
5
11
18
25
32
39
46
53
60
67
74
81
88
95
hypothetical prevalence: 30 %
hypothetical prevalence: 35 %
hypothetical prevalence: 40 %
p = 0.754
p = 0.17
p = 0.0169
x2
11
18
25
32
39
46
53
60
number HIV+
67
74
81
88
95
0.00
x2
0.00
5
0.08
probability
0.04
0.08
probability
0.04
0.08
0.04
0.00
0.12
number HIV+
0.12
number HIV+
0.12
number HIV+
x2
0
0.08
probability
0.04
0.08
probability
0.04
0.08
probability
0.04
0.00
5
x2
x2
x2
0
probability
0.12
hypothetical prevalence: 25 %
0.12
hypothetical prevalence: 20 %
0.12
hypothetical prevalence: 15 %
0
5
11
18
25
32
39
46
53
60
number HIV+
67
74
81
88
95
0
5
11
18
25
32
39
46
53
60
number HIV+
67
74
81
88
95
Which hypotheses do we reject?
IF GIVEN THE HYPOTHESIS
p value < cutoff
THEN REJECT HYPOTHESIS
Cutoff usually chosen as α = 0.05
Which hypotheses do we reject?
p = 0.00123
p = 0.0683
p = 0.555
11
18
25
32
39
46
53
60
67
74
81
88
95
0
5
11
18
25
32
39
46
53
60
67
74
81
88
95
0
5
11
18
25
32
39
46
53
60
67
74
81
88
95
number HIV+
hypothetical prevalence: 30 %
hypothetical prevalence: 35 %
hypothetical prevalence: 40 %
p = 0.754
p = 0.17
p = 0.0169
5
11
18
25
32
39
46
53
60
number HIV+
67
74
81
88
95
0.08
probability
0.00
0.04
0.08
probability
0.00
0.04
0.08
0.04
0.12
number HIV+
0.12
number HIV+
0.00
0
0.08
probability
0.00
0.04
0.08
probability
0.00
0.04
0.08
probability
0.04
0.00
5
0.12
0
probability
0.12
hypothetical prevalence: 25 %
0.12
hypothetical prevalence: 20 %
0.12
hypothetical prevalence: 15 %
0
5
11
18
25
32
39
46
53
60
number HIV+
67
74
81
88
95
0
5
11
18
25
32
39
46
53
60
number HIV+
67
74
81
88
95
0.6
95% CI includes HIV prevalences of 19.5% to 37.8%
0.378
0.2
0.4
0.195
0.0
p−value
0.8
1.0
Which hypotheses do we NOT reject:
CONFIDENCE INTERVAL
0.0
0.2
0.4
0.6
hypothetical prevalence
0.8
1.0
Let’s take another approach
We don’t know the true prevalence, but
the probability that we had exactly 28/100
with 30% prevalence is:
> dbinom(x = 28, size = 100, prob = .3)
[1] 0.08041202
We sample 100 people once and 28 are positive:
> rbinom(n = 1, size = 100, prob = .3)
[1] 28
Which prevalence gives the greatest probability
of observing exactly 28/100?
Which of these prevalence values is most likely
given our data?
Maximum Likelihood Estimate
parameter value giving greatest probability
of the data having occurred.
What do you think is the MLE here?
MLE = 28/100 = 0.28
true unknown value = 0.30
different null hypotheses
Defining Likelihood
• L(parameter | data) = p(data | parameter)
• Not a probability
distribution.
• Probabilities
taken from many
different
distributions.
function of x
PDF:
æ nö x
f (x | p) = ç ÷ p (1- p) n-x
è xø
LIKELIHOOD:
æ nö x
L( p | x) = ç ÷ p (1- p) n-x
è xø
function of p
Deriving the Maximum Likelihood Estimate
maximize
æ nö x
L( p) = ç ÷ p (1- p) n-x
è xø
maximize
éæ n ö x
ù
n-x
log(L( p) = logêç ÷ p (1- p) ú
ëè x ø
û
minimize
éæ n ö x
ù
n-x
l( p) = -logêç ÷ p (1- p) ú
ëè x ø
û
Deriving the Maximum Likelihood Estimate
éæ nö x
ù
n-x
l( p) = -log(L( p)) = -logêç ÷ p (1- p) ú
ëè x ø
û
æ nö
l( p) = -logç ÷ - log( px ) - log((1- p) n-x )
è xø
æ nö
l( p) = -logç ÷ - x log( p) - (n - x)log(1- p)
è xø
Deriving the Maximum Likelihood Estimate
æ nö
l( p) = -logç ÷ - x log( p) - (n - x)log(1- p)
è xø
dl( p)
x -(n - x)
=0- dp
p
1- p
0 = -x + pˆ n
x n-x
0=- +
pˆ 1- pˆ
-x(1- pˆ ) + pˆ (n - x)
0=
pˆ (1- pˆ )
ˆx
0 = -x + pˆXx + pˆ n - pX
x
pˆ =
n
The proportion of positives!
Maximum Likelihood Estimate
x 28
pˆ = =
= 0.28
n 100
Maximum Likelihood Estimate
x 28
pˆ = =
= 0.28
n 100
Building Confidence Intervals
Likelihood Ratio Test
If the null hypothesis were true then
L(null hypothesis)
-2log(
~ c df2 =1
L(alternative hypothesis)
2lalternative - 2lnull ~ c df2 =1
So if our α = .05, then we reject any null hypothesis for which
2lMLE - 2lnull > c df2 =1, a = 0.05 = 3.84
> qchisq(p = .95, df = 1)
[1] 3.841459
2lMLE - 2lnull > 3.84
lMLE - lnull >1.92
When lMLE - lnull > 1.92,
we reject that null hypothesis.
Building Confidence Intervals
Likelihood Ratio Test
Maximum Likelihood Estimate
x 28
pˆ = =
= 0.28
n 100
Let’s zoom in…
Building Confidence Intervals
Likelihood Ratio Test
Maximum Likelihood Estimate
x 28
pˆ = =
= 0.28
n 100
Building Confidence Intervals
Likelihood Ratio Test
2lMLE - 2lnull > c df2 =1, a = 0.05 = 3.84
lMLE - lnull >1.92
Building Confidence Intervals
Likelihood Ratio Test
0.8
0.4
0.195
0.378
95% CI includes HIV prevalences of 19.5% to 37.8%
0.0
p−value
Comparing Confidence Intervals
0.0
0.2
0.4
0.6
0.8
1.0
7
5 6
0.195
0.378
4
95% CI includes HIV prevalences of 19.5% to 37.8%
2 3
−log(likelihood)
hypothetical prevalence (null hypothesis)
0.0
0.2
0.4
0.6
potential prevalences (our models)
0.8
1.0
Advantages of Likelihood
• Practical method for
estimating parameters
estimating variance of our estimates
• Easily adaptable to different probability
distributions & dynamic models
This presentation is made available through a Creative Commons Attribution-Noncommercial
license. Details of the license and permitted uses are available at
http://creativecommons.org/licenses/by-nc/3.0/
© 2010 Steve Bellan and the Meaningful Modeling of Epidemiological Data Clinic
Title: Introduction to Likelihood
Attribution: Steve Bellan, Clinic on the Meaningful Modeling of Epidemiological Data
Source URL: http://lalashan.mcmaster.ca/theobio/mmed/index.php/
For further information please contact Steve Bellan (sbellan@berkeley.edu).