SC968: Panel Data Methods for Sociologists

Introduction to survival/event history models

Types of outcome

Continuous OLS Linear regression

Binary Binary regressionLogistic or probit regression

Time to event data Survival or event history analysis

Examples of time to event data

Time to death Time to incidence of disease Unemployed - time till find job Time to birth of first child Smokers – time till quit smoking

Time to event data

Set of a finite, discrete states Units (individuals, firms, households

etc.) –in one state Transitions between states

Time until a transition takes place

4 key concepts for survival analysis

States Events Risk period Duration/ time

States

States are categories of the outcome variable of interest Each person occupies exactly one state at any moment in

time Examples

alive, dead single, married, divorced, widowed never smoker, smoker, ex-smoker

Set of possible states called the state space

Events

A transition from one state to another From an origin state to a destination state Possible events depend on the state space Examples

From smoker to ex-smoker From married to widowed

Not all transitions can be events E.g. from smoker to never smoker

Risk period

2 states: A & B Event: transition from A B To be able to undergo this transition, one must be in

state A (if in state B already cannot transition) Not all individuals will be in state A at any given time Example

can only experience divorce if married

The period of time that someone is at risk of a particular event is called the risk period

All subjects at risk of an event at a point in time called the risk set

Time

Various meanings...

Calendar time ...but onset of risk usually not simultaneous

for all units Ex: by age 40, some individuals will have

smoked for 20+ years, other for 1 year Duration=time since onset of risk ...intensity may not be the same

EX: one smoker may smoke 5 cigarettes a day, another 20

1 unit of time -same for all individuals

Duration

Event history analysis is to do with the analysis of the duration of a nonoccurrence of an event or the length of time during the risk period

Examples Duration of marriage Length of life

In practice we model the probability of a transition conditional on being in the risk set

Example data

ID Entry date Died End date

1 01/01/1991 01/01/2008

2 01/01/1991 01/01/2000 01/01/2000

3 01/01/1995 01/01/2005

4 01/01/1994 01/07/2004 01/07/2004

Calendar time

1991 1994 1997 2000 2003 2006 2009

Study follow-up ended

Censoring

Ideally: observe individual since the onset of risk until event has occurred

...very demanding in terms of data collection (ex: risk of death starts when one is born)

Usually– incomplete data censoring An observation is censored if it has incomplete

information Types of censoring

Right censoring Left censoring

Censoring

Right censoring: the person did not experience the event during the time that they were studied

Common reasons for right censoring the study ends the person drops-out of the study

We do not know when the person experiences the event but we do know that it is later than a given time T

Left censoring: the person became at risk before we started observing her We do not know when the person entered the risk set EHA

cannot deal with We know when the person entered the risk set condition on

the person having survived long enough to enter the study Censoring independent of survival processes!!

Study time in years

0 3 6 9 12 15 18

censored

event

censored

event

Why a special set of methods?

duration =continuous variable why not OLS? Censoring

If excluding higher probability to throw out longer durations If treating as complete mis-measurement of duration

Non normality of residuals Time varying co-variates Interested in the probability of a transition at any given

time rather than in the length of complete spells Need to simultaneously take into account:

Whether the event has taken place or not The length of the period at risk before the event ocurred

Survival function

Length of time (duration) before an event occurs (length of ‘spell’-T) probability density function (pdf)- f(t)

f(t)= lim Pr(t<=T<=t+Δt) = δF(t) δt Δt0 Δt

cumulative density function (cdf)- F(t)F(t)= Pr( T<=t) =∫f(t) dt

Survival function: S(t)=1-F(t)

Hazard rate

h(t)= f(t)/ S(t) The exact definition & interpretation of h(t) differs:

duration is continuous duration is discrete

Conditional on having survived up to t, what is the probability of leaving between t and t+Δt

It is a measure of risk intensity h(t) >=0 In principle h(t)= rate; not a probability There is a 1-1 relationship between h(t), f(t), F(t), S(t) EHA analysis:

h(t)= g (t, Xs) g=parametric & semi-parametric specifications

Data

Survival or event history data characterised by 2 variables Time or duration of risk period Failure (event)

• 1 if not survived or event observed• 0 if censored or event not yet occurred

Data structure different: Duration is discrete Duration is continuous

Assume: 2 states; 1 transition; no repeated events

Data structure-Discrete time

ID Entry End date Event X at t0 X at t1 ....

1 01/01/1991 01/01/2008 01/01/2002

2 01/01/1991 01/01/2008

ID Date Duration (t) Event X

1 01/01/1991 1 01 01/01/1992 2 0... ..... .... .....1 01/01/2002 11 12 01/01/1991 1 0... .... .... ....2 01/01/2008 17 0

Data structure-Discrete time

The row is a an individual period An individual has as many rows as the number of

periods he is observed to be at risk No longer at risk when

Experienced event No longer under observation (censored)

For each period (row)- explanatory variable X very easy to incorporate time varying co-variates

Stata: reshape long

Data structure-continuous time

ID Entry Died End date Duration Event X

1 01/01/1991 01/01/2008 17.0 0 02 01/01/1991 01/01/2002 01/01/2002 11.0 1 03 01/01/1995 01/01/2000 5.0 0 03 01/01/2000 01/01/2005 01/01/2005 5.0 1 1

Data structure-continuous time

The row is a person Indicator for observed events/ censored cases Calculate duration= exit date – entry date Exit date=

Failure date Censoring date

If time-varying covariates- Split the period an individual is under observation by the

number of times time-varying Xs change If many Xs-change often- multiple rows

Worked example

Random 20% sample from BHPS Waves 1 – 15 One record per person/wave Outcome: Duration of cohabitation Conditions on cohabiting in first wave Survival time: years from entry to the study in 1991

till year living without a partner

The data

+----------------------------+ | pid wave mastat | |----------------------------| | 10081798 1 married | | 10081798 2 married | | 10081798 3 married | | 10081798 4 married | | 10081798 5 married | | 10081798 6 married | | 10081798 7 widowed | | 10081798 8 widowed | | 10081798 9 widowed | | 10081798 10 widowed | | 10081798 11 widowed | | 10081798 12 widowed | | 10081798 13 widowed | | 10081798 14 widowed | | 10081798 15 widowed | |----------------------------|

Duration = 6 years

Event = 1

Ignore data after event = 1

The data (continued)

+----------------------------+ | pid wave mastat | |----------------------------| | 10162747 1 living a | | 10162747 2 living a | | 10162747 3 living a | | 10162747 4 living a | | 10162747 5 living a | | 10162747 6 living a | | 10162747 10 separate | | 10162747 11 . | | 10162747 12 . | | 10162747 13 . | | 10162747 14 never ma | | 10162747 15 never ma | +----------------------------+

Note missing waves before event

Preparing the data

. sort pid wave . generate skey=1 if wave==1&(mastat==1|mastat==2) . by pid: replace skey=skey[_n-1] if wave~=1 . keep if skey==1 . drop skey . . stset wave,id(pid) failure(mastat==3/6) id: pid failure event: mastat == 3 4 5 6 obs. time interval: (wave[_n-1], wave] exit on or before: failure ------------------------------------------------------------------------------ 15058 total obs. 1628 obs. begin on or after (first) failure ------------------------------------------------------------------------------ 13430 obs. remaining, representing 1357 subjects 270 failures in single failure-per-subject data 13612 total analysis time at risk, at risk from t = 0 earliest observed entry t = 0 last observed exit t = 15

Select records for respondents who were cohabiting in 1991

Declare that you want to set the data to survival time

Important to check that you have set data as intended

Checking the data setup. list pid wave mastat _st _d _t _t0 if pid==10081798,sepby(pid) noobs +-------------------------------------------------+ | pid wave mastat _st _d _t _t0 | |-------------------------------------------------| | 10081798 1 married 1 0 1 0 | | 10081798 2 married 1 0 2 1 | | 10081798 3 married 1 0 3 2 | | 10081798 4 married 1 0 4 3 | | 10081798 5 married 1 0 5 4 | | 10081798 6 married 1 0 6 5 | | 10081798 7 widowed 1 1 7 6 | | 10081798 8 widowed 0 . . . | | 10081798 9 widowed 0 . . . | | 10081798 10 widowed 0 . . . | | 10081798 11 widowed 0 . . . | | 10081798 12 widowed 0 . . . | | 10081798 13 widowed 0 . . . | | 10081798 14 widowed 0 . . . | | 10081798 15 widowed 0 . . . | +-------------------------------------------------+ 1 if observation is to be used

and 0 otherwise

1 if event, 0 if censoring orevent not yet occurred

time of exit

time of entry

Checking the data setup

. list pid wave mastat _st _d _t _t0 if pid==10162747,sepby(pid) noobs +--------------------------------------------------+ | pid wave mastat _st _d _t _t0 | |--------------------------------------------------| | 10162747 1 living a 1 0 1 0 | | 10162747 2 living a 1 0 2 1 | | 10162747 3 living a 1 0 3 2 | | 10162747 4 living a 1 0 4 3 | | 10162747 5 living a 1 0 5 4 | | 10162747 6 living a 1 0 6 5 | | 10162747 10 separate 1 1 10 6 | | 10162747 11 . 0 . . . | | 10162747 12 . 0 . . . | | 10162747 13 . 0 . . . | | 10162747 14 never ma 0 . . . | | 10162747 15 never ma 0 . . . | +--------------------------------------------------+ How do we know when

this person separated?

Trying again!

. fillin pid wave . stset wave,id(pid) failure(mastat==3/6) exit(mastat==3/6 .) id: pid failure event: mastat == 3 4 5 6 obs. time interval: (wave[_n-1], wave] exit on or before: mastat==3 4 5 6 . ------------------------------------------------------------------------------ 20355 total obs. 7524 obs. begin on or after exit ------------------------------------------------------------------------------ 12831 obs. remaining, representing 1357 subjects 234 failures in single failure-per-subject data 12831 total analysis time at risk, at risk from t = 0 earliest observed entry t = 0 last observed exit t = 15

. list pid wave mastat _st _d _t _t0 if pid==10162747,sepby(pid) noobs +--------------------------------------------------+ | pid wave mastat _st _d _t _t0 | |--------------------------------------------------| | 10162747 1 living a 1 0 1 0 | | 10162747 2 living a 1 0 2 1 | | 10162747 3 living a 1 0 3 2 | | 10162747 4 living a 1 0 4 3 | | 10162747 5 living a 1 0 5 4 | | 10162747 6 living a 1 0 6 5 | | 10162747 7 . 1 0 7 6 | | 10162747 8 . 0 . . . | | 10162747 9 . 0 . . . | | 10162747 10 separate 0 . . . | | 10162747 11 . 0 . . . | | 10162747 12 . 0 . . . | | 10162747 13 . 0 . . . | | 10162747 14 never ma 0 . . . | | 10162747 15 never ma 0 . . . | +--------------------------------------------------+

Checking the new data setup

Now censored instead of an event

Summarising time to event data

Individuals followed up for different lengths of time So can’t use prevalence rates (% people who have

an event) Use rates instead that take account of person years

at risk Incidence rate per year Death rate per 1000 person years

Summarising time to event data

Number of observationsPerson-years Rate per year

<25% of sample had event by 15 elapsed years

. stsum failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid | incidence no. of |------ Survival time -----| | time at risk rate subjects 25% 50% 75% ---------+--------------------------------------------------------------------- total | 12831 .0182371 1357 . . .

stvary-check whether a variable varies within individuals and over time

Descriptive analysis

To recap…. pdf= probability that a spell has a length of

exactly Tf(t)= lim Pr(t<=T<=t+Δt) = δF(t) δt

Δt0 Δt cdf=probability that a spell has a length<=T F(t)= Pr( T<=t) =∫f(t) dt Survival function S(t)=1-F(t)

Kaplan-Meier estimates of survival time

The Kaplan-Meier cumulative probability of an individual surviving to any time, t

Analysis can be made by subgroup Nonparametric method First period: S1=1-d1/n1 exit rate After t periods: St=(1-d1/n1)*(1-d2/n2)*……*(1-dt/nt) Survival function estimated only at times where

you observe exits!!! Last t that can be estimated highest non-censored

time observed

Survival/ failure function

Describing the survival/ failure function. sts list, failure failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Beg. Net Failure Std. Time Total Fail Lost Function Error [95% Conf. Int.] ------------------------------------------------------------------------------- 2 1357 29 162 0.0214 0.0039 0.0149 0.0306 3 1166 33 89 0.0491 0.0061 0.0384 0.0625 4 1044 16 64 0.0636 0.0070 0.0513 0.0789 5 964 35 58 0.0976 0.0088 0.0818 0.1164 6 871 12 34 0.1101 0.0094 0.0931 0.1300 7 825 20 24 0.1316 0.0103 0.1128 0.1534 8 781 14 17 0.1472 0.0109 0.1271 0.1701 9 750 12 30 0.1609 0.0115 0.1398 0.1848 10 708 15 23 0.1786 0.0121 0.1563 0.2038 11 670 9 32 0.1897 0.0125 0.1666 0.2155 12 629 8 16 0.2000 0.0128 0.1762 0.2266 13 605 13 24 0.2172 0.0134 0.1922 0.2449 14 568 8 24 0.2282 0.0138 0.2025 0.2566 15 536 10 526 0.2426 0.0143 0.2160 0.2719 -------------------------------------------------------------------------------

Kaplan-Meier graphs

Can read off the estimated probability of surviving a relationship at any time point on the graph E.g. at 5 years 88% are still cohabiting

The survival probability only changes when an event occurs graph not smooth but (irregular) stepwise

sts graph, survival

0.00

0.25

0.50

0.75

1.00

0 5 10 15analysis time

Kaplan-Meier survival estimate

0.00

0.25

0.50

0.75

1.00

0 5 10 15time in years

Kaplan-Meier survival estimate

0.00

0.25

0.50

0.75

1.00

0 5 10 15analysis time

sex = male sex = female

Comparing survival by group using Kaplan-Meier graphs

Testing equality of survival curves among groups

The log-rank test

A non –parametric test that assesses the null hypothesis that there are no differences in survival times between groups

. sts test sex, logrank failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Log-rank test for equality of survivor functions | Events Events sex | observed expected -------+------------------------- male | 98 113.59 female | 136 120.41 -------+------------------------- Total | 234 234.00 chi2(1) = 4.25 Pr>chi2 = 0.0392

Log-rank test example

Significant difference between men and women

More elaborate models…

Modeling the hazard rate not survival time directly h(t)=transitioning at time t, having survived up to t Time:

Continuous- parametric• Exponential• Weibull• Log-logistic

Continuous-semi-parametric• Cox

Discrete• Logistic• Complementary log-log

Some hazard shapes

Increasing Onset of Alzheimer's

Decreasing Survival after surgery

U-shaped Age specific mortality

Constant Time till next email arrives

Proportional-hazards (PH) models

h(t) is separable into h0(t) and the effects of Xs

h0(t)=‘baseline’ hazard that depends on t but not on individual characteristics

h(t)=h0(t)exp(βX) Absolute differences in X proportional

differences in h(t) ~scaling of h0(t)

The Cox regression model

Cox regression model

Regression model for survival analysis Can model time invariant and time varying

explanatory variables Produces estimated hazard ratios (sometimes

called rate ratios or risk ratios) Regression coefficients are on a log scale

Exponentiate to get hazard ratio Similar to odds ratios from logistic models

Cox regression equation (i)

).......exp()()( 22110 inniii xxxthth

)(0 th

)(thi

is the baseline hazard function and can take any formIt is estimated from the data (non parametric)

is the hazard function for individual i

inii xxx ,....,, 21

n ,....,, 21

are the covariates

are the regression coefficients estimated from the data

PH assumption neededEstimate βs without estimating h0(t) semi parametric model

Cox regression equation (ii)

If we divide both sides of the equation on the previous slide by h0(t) and take logarithms, we obtain:

We call h(t) / h0(t) the hazard ratio The coefficients bi...bn are estimated by Cox regression, and can

be interpreted in a similar manner to that of multiple logistic regression

exp(bi) is the instantaneous relative risk of an event

inniii xxxthth

.......

)()(

ln 22110

Cox regression in Stata

Will first model a time invariant covariate (sex) on risk of partnership ending

Then will add a time dependent covariate (age) to the model

Cox regression in Stata

. stcox female failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Cox regression -- Breslow method for ties No. of subjects = 1357 Number of obs = 12337 No. of failures = 234 Time at risk = 12337 LR chi2(1) = 4.18 Log likelihood = -1574.5782 Prob > chi2 = 0.0409 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | 1.30913 .1734699 2.03 0.042 1.009699 1.697358 ------------------------------------------------------------------------------

Interpreting output from Cox regression

Cox model has no intercept It is included in the baseline hazard

In our example, the baseline hazard is when sex=1 (male) The hazard ratio is the ratio of the hazard for a unit

change in the covariate HR = 1.3 for women vs. men The risk of partnership breakdown is increased by 30% for women

compared with men Hazard ratio assumed constant over time

At any time point, the hazard of partnership breakdown for a woman is 1.3 times the hazard for a man

Interpreting output from Cox regression (ii)

The hazard ratio is equivalent to the odds that a female has a partnership breakdown before a man

The probability of having a partnership breakdown first is = (hazard ratio) / (1 + hazard ratio)

So in our example, a HR of 1.30 corresponds to aprobability of 0.57 that a woman will experience a partnership breakdown first

The probability or risk of partnership breakdown can be different each year but the relative risk is constant

So if we know that the probability of a man having a partnership breakdown in the following year is 1.5% then the probability of a woman having a partnership breakdown in the following year is

0.015*1.30 = 1.95%

0.0

5.1

.15

.2.2

5

0 5 10 15_t

sex = women sex = men

Estimated cumulative hazard: men vs. women

.012

.014

.016

.018

.02

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ooth

ed h

azar

d fu

nctio

n

4 6 8 10 12analysis time

hazard function varying over time

Cox proportional hazards regression:

Time dependent covariates

Examples Current age group rather than age at baseline GHQ score may change over time and predict break-ups

Will use age to predict duration of cohabitation Nonlinear relationship hypothesised Recode age into 8 equally spaced age groups

Cox regression with time dependent covariates

. xi: stcox female i.agecat i.agecat _Iagecat_0-7 (naturally coded; _Iagecat_0 omitted) failure _d: mastat == 3 4 5 6 analysis time _t: wave exit on or before: mastat==3 4 5 6 . id: pid Cox regression -- Breslow method for ties No. of subjects = 1357 Number of obs = 12337 No. of failures = 234 Time at risk = 12337 LR chi2(8) = 78.44 Log likelihood = -1537.4472 Prob > chi2 = 0.0000 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | 1.3705 .1842481 2.34 0.019 1.05304 1.783666 _Iagecat_1 | .5838602 .1883578 -1.67 0.095 .3102449 1.098786 _Iagecat_2 | .311325 .1039311 -3.50 0.000 .1618279 .5989281 _Iagecat_3 | .2136714 .0737986 -4.47 0.000 .1085813 .4204725 _Iagecat_4 | .2225187 .0811395 -4.12 0.000 .1088888 .4547261 _Iagecat_5 | .4770023 .1691695 -2.09 0.037 .238035 .9558732 _Iagecat_6 | 1.203702 .4306775 0.52 0.604 .5969856 2.427023 _Iagecat_7 | 1.644141 .9677715 0.84 0.398 .518688 5.21161 ------------------------------------------------------------------------------

Cox regression assumptions

Assumption of proportional hazards No censoring patterns True starting time Plus assumptions for all modelling

Sufficient sample size, proper model specification, independent observations, exogenous covariates, no high multicollinearity, random sampling, and so on

Proportional hazards assumption

Cox regression with time-invariant covariates assumes that the ratio of hazards for any two observations is the same across time periods

This can be a false assumption, for example using age at baseline as a covariate

If a covariate fails this assumption for hazard ratios that increase over time for that covariate,

relative risk is overestimated for ratios that decrease over time, relative risk is

underestimated standard errors are incorrect and significance tests are

decreased in power

Testing the proportional hazards assumption

Graphical methods Comparison of Kaplan-Meier observed & predicted curves

by group. Observed lines should be close to predicted Survival probability plots (cumulative survival against time

for each group). Lines should not cross Log minus log plots (minus log cumulative hazard against

log survival time). Lines should be parallel

Testing the proportional hazards assumption

Formal tests of proportional hazard assumption

Include an interaction between the covariate and a function of time. Log time often used but could be any function. If significant then assumption violated

Test the proportional hazards assumption on the basis of partial residuals. Type of residual known as Schoenfeld residuals.

When assumptions are not met

If categorical covariate, include the variable as a strata variable

Allows underlying hazard function to differ between categories and be non proportional

Estimates separate underlying baseline hazard for each stratum

When assumptions are not met

If a continuous covariate

Consider splitting the follow-up time. For example, hazard may be proportional within first 5 years, next 5-10 years and so on

Could covariate be included as time dependent covariate? There are different survival regression methods (e.g.

parametric models) that do not assume PH

Censoring assumptions

Censored cases must be independent of the survival distribution. There should be no pattern to these cases, which instead should be missing at random.

If censoring is not independent, then censoring is said to be informative

You have to judge this for yourself Usually don’t have any data that can be used to test the

assumption Think carefully about start and end dates Always check a sample of records

True starting time

The ideal model for survival analysis would be where there is a true zero time

If the zero point is arbitrary or ambiguous, the data series will be different depending on starting point. The computed hazard rate coefficients could differ, sometimes markedly

Conduct a sensitivity analysis to see how coefficients may change according to different starting points

Other extensions to survival analysis

Discrete (interval-censored) survival times Repeated events Multi-state models (more than 1 event type)-

competing risks Transition from employment to unemployment or leaving

labour market Modelling type of exit from cohabiting relationship-

separation/divorce/widowhood Frailty (unobserved heterogeneity)

Could you use logistic regression instead?

May produce similar results for short or fixed follow-up periods Examples

• everyone followed-up for 7 years• maximum follow-up 5 years

Results may differ if there are varying follow-up times

If dates of entry and dates of events are available then better to use Cox regression

Finally….

This is just an introduction to survival/ event history analysis

Only reviewed the Cox regression model Also parametric survival methods But Cox regression likely to suit type of analyses of

interest to sociologists

Consider an intensive course if you want to use survival analysis in your own work

Some Resources

Stephen Jenkins’s course on survival analysis: https://

www.iser.essex.ac.uk/files/teaching/stephenj/ec968/pdfs/ec968lnotesv6.pdf

Allison, Paul D. (1984) Event History Analysis: Regression for Longitudinal Event Data, Sage

Cleves, M., W. Gould, and R. Gutierrez. 2004. An Introduction to Survival Analysis Using Stata. Rev. ed. Stata Press: College Station, Texas