Transcript Intermediate Epidemiology

Intermediate methods in
observational epidemiology
2008
Instructor: Moyses Szklo
Measures of Disease Frequency
MEASURES OF RISK
• Absolute measures of event (including
disease) frequency:
– Incidence and Incidence Odds
– Prevalence and Prevalence Odds
What is "incidence"?
Two major ways to define incidence
• Cumulative incidence (probability)
SURVIVAL ANALYSIS (Unit of analysis:
individual)
• Rate or Density
ANALYSIS BASED ON PERSON-TIME (Unit
of analysis: time)
• OBJECTIVE OF SURVIVAL ANALYSIS:
To compare the “cumulative incidence” of an
event (or the proportion surviving eventfree) in exposed and unexposed
(characteristic present or absent) while
adjusting for time to event (follow-up time)
Survival
1.0
Time
• BASIS FOR THE ANALYSIS
• NUMBER of EVENTS
• TIME of occurrence
Need to precisely define:
• “EVENT” (failure):
–
–
–
–
Death
Disease (diagnosis, start of symptoms, relapse)
Quit smoking
Menopause
• “TIME”:
–
–
–
–
–
–
Time from recruitment into the study
Time from employment
Time from diagnosis (prognostic studies)
Time from infection
Calendar time
Age
– Example:
• Follow up of 6 patients (2 yrs)
– 3 Deaths
– 2 censored (lost) before 2 years
– 1 survived 2 years
Question: What is the Cumulative Incidence
(or the Cumulative Survival) up to 2 years?
Person ID
(24)
1
(6)
2
(18)
3
(15)
4
(13)
5
6
(3)
Jan
1999
Jan
2000
Death
Censored observation (lost to follow-up, withdrawal)
( )
Number of months to follow-up
Jan
2001
Crude Survival:
3/6= 50%
Change time scale to “follow-up” time:
Person ID
(24)
1
(6)
2
(18)
3
(15)
4
(13)
5
6
(3)
0
1
Follow-up time (years)
2
One solution:
• Actuarial life table
Assume that censored observations over the period
contribute one-half the persons at risk in the
denominator (censored observations occur uniformly
throughout follow-up interval).
3
3
ID
q2 yrs 
  0.60
1
(24)
1
5
6


2
(6)
2
2
(18)
3
(15)
S (2 yrs )  1  q2 yrs  0.40
4
5
6
(13)
(3)
0
1
2
Follow-up time (years)
Cumulative Survival
100
82
It can be also calculated for years 1 and 2
separately:
Year 1: S(Y1)= [1 - {1 ÷ [6 – ½(1)]}= 0.82
Year 2: S(Y2)= [1 – {2 ÷ [4 – ½(1)]}= 0.43
S(2yrs)= 0.82 × 0.43= 0.35
43
Year 1
Year 2
Follow-up time
KAPLAN-MEIER METHOD
E.L. Kaplan and P. Meier, 1958*
Calculate the cumulative probability of event (and survival)
based on conditional probabilities at each event time
Step 1: Sort the survival times from shortest to longest
Person ID
1
(24)
2
3
(6)
(18)
4
5
6
(15)
(13)
(3)
0
1
Follow-up time (years)
*Kaplan EL, Meier P.Nonparametric estimation from incomplete observations.
J Am Stat Assoc 1958;53:457-81.
2
KAPLAN-MEIER METHOD
E.L. Kaplan and P. Meier, 1958*
Calculate the cumulative probability of event (and survival)
based on conditional probabilities at each event time
Step 1: Sort the survival times from shortest to longest
Person ID
6
(3)
(6)
2
5
4
3
1
(13)
(15)
(18)
(24)
0
1
Follow-up time (years)
*Kaplan EL, Meier P.Nonparametric estimation from incomplete observations.
J Am Stat Assoc 1958;53:457-81.
2
Step 2: For each time of occurrence of an event, compute the
conditional survival
Person ID
6
(3)
(6)
2
5
(13)
(15)
4
3
1
(18)
(24)
0
1
Follow-up time (years)
2
When the first event occurs (3 months after beginning of
follow-up), there are 6 persons at risk. One dies at that
point; 5 of the 6 survive beyond that point. Thus:
• Incidence of event at exact time 3 months: 1/6
• Probability of survival beyond 3 months: 5/6
Person ID
6
(3)
(6)
2
5
(13)
(15)
4
3
1
(18)
(24)
0
1
Follow-up time (years)
2
When the second event occurs (13 months),
there are 4 persons at risk. One of them dies at that point; 3
of the 4 survive beyond that point. Thus:
• Incidence of event at exact time 13 months: 1/4
• Probability of survival beyond 13 months: ¾
Person ID
6
(3)
(6)
2
5
(13)
(15)
4
3
1
(18)
(24)
0
1
Follow-up time (years)
2
When the third event occurs (18
months), there are 2 persons at risk. One of them dies at
that point; 1 of the 2 survive beyond that point. Thus:
• Incidence of event at exact time 18 months: 1/2
• Probability of survival beyond 18 months: 1/2
CONDITIONAL PROBABILITY OF AN EVENT (or of
survival)
The probability of an event (or of survival) at time
t (for the individuals at risk at time t), that is,
conditioned on being at risk at exact time t.
Step 3:
For each time of occurrence of an event,
compute the cumulative survival (survival
function), multiplying conditional probabilities of
survival.
3 months:
12 months:
18 months:
S(3)=5/6=0.833
S(13)=5/63/4=0.625
S(18)=5/6 3/41/2 =0.3125
Plotting the survival function:
Survival
1.00
Time (mo)
Si
3
13
18
0.833
0.625
0.3125
0.833
0.80
0.625
0.60
0.3125
0.40
0.3125
0.20
0
5
10
15
20
25
Month of follow-up
The cumulative incidence (up to 24 months): 1-0.3125 = 0.6875 (or 69%)
Time (mo)
Plotting the survival function:
3
13
18
Cumulative Survival
1.00
0.8
0.80
0.6
0.60
0.3
0.40
0.20
0
5
10
15
20
25
Month of follow-up
0.833
0.625
0.3125
EXPERIMENTAL STUDY
Cumulative Hazards for Coronary Heart Disease and Stroke
in the Women’s Health Initiative Randomized Controlled Trial
(The WHI Steering Committee. JAMA 2004;291:1701-1712)
Time (mo)
Plotting the survival function:
3
13
18
Cumulative Survival
1.00
Cumulative Hazard
1.00
0.8
0.7
0.80
0.80
0.6
0.60
0.60
0.4
0.3
0.40
0.40
0.2
0.20
0.20
0
5
10
15
20
25
Month of follow-up
The cumulative incidence (hazard) at the end of 24 months:
1-0.3 = 0.7 (or 70%)
0.833
0.625
0.3125
ACTUARIAL LIFE TABLE VS KAPLAN-MEIER
If N is large and/or if life-table intervals are small, results are similar
•Survival after diagnosis of Ewing’s sarcoma
ASSUMPTIONS IN KAPLAN-MEIER
SURVIVAL ESTIMATES
• (If individuals are recruited over a long period of time)
No secular trends
Calendar time
Follow-up time
ASSUMPTIONS IN SURVIVAL ESTIMATES
(Cont’d)
•
Censoring is independent of survival
(uninformative censoring): Those censored
at time t have the same prognosis as those
remaining.
Types of censoring:
• Lost to follow-up
– Migration
– Refusal
• Death (from another cause)
• Administrative withdrawal (study finished)
Calculation of incidence
Strategy #2
ANALYSIS BASED ON PERSON-TIME
CALCULATION OF PERSON-TIME AND INCIDENCE RATES (Unit of analysis:
time)
Example 1
Observe 1st graders, total 500 hours
Observe 12 accidents
Accident rate:
12
R
 0.024 per person - hour
500
IT IS NOT KNOWN WHETHER 500 CHILDREN WERE OBSERVED
FOR 1 HOUR, OR 250 CHILDREN OBSERVED FOR 2 HOURS, OR
100 CHILDREN OBSERVED FOR 5 HOURS… ETC.
CALCULATION OF PERSON-TIME AND INCIDENCE RATES
Example 2
Person ID
6
2
5
4
3
1
(3)
(6)
(13)
(15)
(18)
(24)
0
1
Follow-up time (years)
Step 1: Calculate denominator, i.e. units of time (years) contributed
by each individual, and total:
1st FU year
2nd FU year
Step 2: Calculate rate per
person-year for the total follow-up
period:
R
No. of person-years in
Person ID
2
Total FU
6
2
5
4
3
1
3/12=0.25
6/12=0.50
12/12=1.00
12/12=1.00
12/12=1.00
12/12=1.00
0
0
1/12=0.08
3/12=0.25
6/12=0.50
12/12=1.00
0.25
0.25
1.00
1.25
1.50
2.00
Total
4.75
1.83
6.58
3
 0.46 per person - year
6.58
It is also possible to calculate the
incidence rates per person-year
separately for shorter periods
during the follow-up:
For year 1:
R
1
 0.21 per person - year
4.75
For year 2:
R
2
 1.09 per person - year
1.83
Notes:
• Rates have units (time-1).
• Proportions (e.g., cumulative incidence) are unitless.
• As velocity, rate is an instantaneous concept. The
choice of time unit used to express it is totally
arbitrary.
E.g.:
0.024 per person-hour = 0.576 per person-day
= 210.2 per person-year
0.46 per person-year = 4.6 per person-decade
No. of person-years of follow-up
Person No.
Year 1
Year 2
Total
1
1/12= 0.08 (D)
0
0.08
2
2/12= 0.17 (C)
0
0.17
3
3/12= 0.25 (C)
0
0.25
4
4/12= 0.33 (C)
0
0.33
5
5/12= 0.42 (C)
0
0.42
6
6/12= 0.50 (D)
0
0.50
7
7/12= 0.58 (C)
0
0.58
8
8/12= 0.67 (C)
0
0.67
9
9/12= 0.75 (C)
0
0.75
10
10/12= 0.83 (C)
0
0.83
11
11/12= 0.92 (C)
0
0.92
12
12/12= 1.00 (D)
0
1.00
13
12/12= 1.00 (C)
1/12= 0.08 (C)
1.08
14
12/12 = 1.00 (C)
2/12= 0.17 (C)
1.17
15
12/12 = 1.00 (C)
3/12= 0.25 (D)
1.25
16
12/12 = 1.00
4/12= 0.33 (C)
1.33
17
12/12 = 1.00
5/12= 0.42 (C)
1.42
18
12/12 = 1.00
6/12= 0.50 (C)
1.50
19
12/12 = 1.00
7/12= 0.58 (C)
1.58
20
12/12 = 1.00
8/12= 0.67 (C)
1.67
21
12/12 = 1.00
9/12= 0.75 (D)
1.75
22
12/12 = 1.00
10/12= 0.83 (C)
1.83
23
12/12 = 1.00
11/12= 0.92 (C)
1.92
24
12/12 = 1.00
12/12= 1.00 (C)
2.0
18.5
6.5
25.0
Total
Death rate per person-time
(person-year)
5 deaths/25.0 person-years= 0.20 or
20 deaths per 100 person-years
Death rate per average
population, estimated at midpoint of follow-up
Mid-point (median) population (When
calculating yearly rate in Vital
Statistics) = 12.5
Death rate= 5/12.5 per 2 years= 0.40
Average annual death rate= 0.40/2= 0.20
or 20/100 population
D, deaths
C, censored
No. of person-years of follow-up
Person No.
Year 1
Year 2
Total
1
1/12= 0.08 (D)
0
0.08
2
2/12= 0.17 (C)
0
0.17
3
3/12= 0.25 (C)
0
0.25
4
4/12= 0.33 (C)
0
0.33
5
5/12= 0.42 (C)
0
0.42
6
6/12= 0.50 (D)
0
0.50
7
7/12= 0.58 (C)
0
0.58
8
8/12= 0.67 (C)
0
0.67
9
9/12= 0.75 (C)
0
0.75
10
10/12= 0.83 (C)
0
0.83
11
11/12= 0.92 (C)
0
0.92
12
12/12= 1.00 (D)
0
1.00
13
12/12= 1.00 (C)
1/12= 0.08 (C)
1.08
14
12/12 = 1.00 (C)
2/12= 0.17 (C)
1.17
15
12/12 = 1.00 (C)
3/12= 0.25 (D)
1.25
16
12/12 = 1.00
4/12= 0.33 (C)
1.33
17
12/12 = 1.00
5/12= 0.42 (C)
1.42
18
12/12 = 1.00
6/12= 0.50 (C)
1.50
19
12/12 = 1.00
7/12= 0.58 (C)
1.58
20
12/12 = 1.00
8/12= 0.67 (C)
1.67
21
12/12 = 1.00
9/12= 0.75 (D)
1.75
22
12/12 = 1.00
10/12= 0.83 (C)
1.83
23
12/12 = 1.00
11/12= 0.92 (C)
1.92
24
12/12 = 1.00
12/12= 1.00 (C)
2.0
18.5
6.5
25.0
Total
Death rate per person-time
(person-year)
5 deaths/25.0 person-years= 0.20 or
20 deaths/100 person-years
Death rate per average
population, estimated at midpoint of follow-up
Mid-point (median) population (When
calculating yearly rate in Vital
Statistics) = 12.5
Death rate= 5/12.5 per 2 years= 0.40
Average annual death rate= 0.40/2= 0.20
or 20/100 population
No. of person  time 

Events ( D)

Population ( N )  Time ( N )
Events ( D)
Population ( N )
Time ( N )
D, deaths
C, censored
Notes: Rates have an undesirable
statistical property
• Rates can be more than 1.0 (100%):
– 1 person dies exactly after 6 months:
• No. of person-years: 1 x 0.5 years= 0.5 person-years
1
Rate   2.0 per PY  200 per 100 PYs
0.5
Use of person-time to account for changes in
exposure status (Time-dependent exposures)
Example: Adjusting for age, are women after menopause at a higher risk for
myocardial infarction?
ID 1
1
2
3
4
5
6
2
Year of follow-up
3 4 5 6 7

8
9 10
C

C

Number of PY in each group
No. PY
PRE meno
No. PY
POST meno
3
0
6
0
5
3
17
4
5
0
1
5
3
18
: Myocardial Infarction; C: censored observation.
Note: Event is assigned to exposure status when it occurs
Rates per person-year:
Pre-menopausal = 1/17 = 0.06 (6 per 100 py)
Post-menopausal = 2/18 = 0.11 (11 per 100 py)
Rate ratio = 0.11/0.06 = 1.85
ASSUMPTIONS IN PERSON-TIME ESTIMATES
Risk is constant within each interval for which
person-time units are estimated (no
cumulative effect):
– N individuals followed for t time  t individuals
followed for N time
– However,Rate
are 10
forsmokers
1st Year=followed
0.21/PYfor 1 year
comparable to 1 smoker followed for 10 years
(both: 10Rate
person-years)
for 2nd Year= 1.09/ PY
• No secular trends (if individuals are recruited
Total forlong
2 years
over a relatively
time= 0.46/PY
interval)
• Losses are independent from survival
ASSUMPTIONS IN PERSON-TIME ESTIMATES
Risk is constant within each interval/period for
which person-time units are estimated (no
cumulative effect):
– N individuals followed for t time  t individuals
followed for N time
– However, are 10 smokers followed for 1 year
comparable to 1 smoker followed for 10 years
(both: 10 person-years)
• No secular trends (if individuals are recruited
over a relatively long time interval)
• Losses are independent of survival
SUMMARY OF ESTIMATES
Method
Estimate
Value
Life-table
Life-table
Kaplan-Meier
q (2 years)
q(Y1) × q(Y2)
q (2 years)
0.60
0.65
0.64
Person-year
Midpoint (median) population
Rate (yearly)
0.46/py
0.43 per year
POINT REVALENCE
Point Prevalence
“The number of affected persons present at the
population at a specific time divided by the
number of persons in the population at that time”
Gordis, 2000, p.33
Relation with incidence --- Usual formula:
Point Prevalence = Incidence x Duration*
P=IxD
True formula:
Prevalence
 Incidence  Duration
1  Prevalence
* Average duration (survival) after disease onset.
ODDS
Odds
The ratio of the probabilities of an event to that of
the non-event.
Prob
Odds 
1- Prob
Example: The probability of an event (e.g., death, disease,
recovery, etc.) is 0.20, and thus the odds is:
Odds 
0.20
0.20

 1: 4 (or 0.25)
1- 0.20 0.80
That is, for every person with the event, there
are 4 persons without the event.